$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 4.5: Logarithmic Properties

Logarithmic Properties
In this section, you will:
• Use the product rule for logarithms.
• Use the quotient rule for logarithms.
• Use the power rule for logarithms.
• Expand logarithmic expressions.
• Condense logarithmic expressions.
• Use the change-of-base formula for logarithms.
<figure class="small" id="CNX_Precalc_Figure_04_05_001" style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: auto; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 1; word-spacing: 0px; -webkit-text-stroke-width: 0px;"> <figcaption>The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan)</figcaption> </figure>

In chemistry, pH is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered acidic, and substances with a pH greater than 7 are said to be alkaline. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances:

• Battery acid: 0.8
• Stomach acid: 2.7
• Orange juice: 3.3
• Pure water: 7 (at 25° C)
• Human blood: 7.35
• Fresh coconut: 7.8
• Sodium hydroxide (lye): 14

To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is the concentration of hydrogen ion in the solution

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>pH</mtext><mo>=</mo><mo>−</mo><mi>log</mi><mo stretchy="false">(</mo><mo stretchy="false">[</mo><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] H + ])      =log( 1 [ H + ] )

The equivalence of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo>−</mo><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] [ H + ] ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 [ H + ] ) is one of the logarithm properties we will examine in this section.

# Using the Product Rule for Logarithms

Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><msub><mi>log</mi></msub></mtd></mtr></mtable></annotation-xml></semantics>[/itex] b 1=0 log b b=1

For example,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 5 1=0 since<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] 5 0 =1. And<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 5 5=1 since<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] 5 1 =5.

Next, we have the inverse property.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] b ( b x )=x     b log b x =x,x>0

For example, to evaluate<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 100 ),we can rewrite the logarithm as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 10 ( 10 2 ), and then apply the inverse property<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log b ( b x )=x to get<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 10 ( 10 2 )=2.

To evaluate<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] e ln( 7 ) ,we can rewrite the logarithm as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] e log e 7 , and then apply the inverse property<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b log b x =x to get<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] e log e 7 =7.

Finally, we have the one-to-one property.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b M= log b N  if and only if  M=N

We can use the one-to-one property to solve the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 3 ( 3x )= log 3 ( 2x+5 ) for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>:</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>3</mn><mi>x</mi><mo>=</mo><mn>2</mn><mi>x</mi><mo>+</mo><mn>5</mn><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] Set the arguments equal.     x=5 Subtract 2x.

But what about the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 3 ( 3x )+ log 3 ( 2x+5 )=2? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation.

Recall that we use the product rule of exponents to combine the product of exponents by adding:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] x a x b = x a+b . We have a similar property for logarithms, called the product rule for logarithms, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.

Given any real number<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and positive real numbers<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>M</mi><mo>,</mo><mi>N</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>≠</mo><mn>1</mn><mo>,</mo></mrow></annotation-xml></semantics>[/itex] we will show

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log b ( MN )= log b ( M )+ log b ( N ).

Let<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>m</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics>[/itex] log b M and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics>[/itex] log b N. In exponential form, these equations are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b m =M and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b n =N. It follows that

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] b ( MN ) = log b ( b m b n )       Substitute for M and N. = log b ( b m+n ) Apply the product rule for exponents. =m+nApply the inverse property of logs. = log b ( M )+ log b ( N ) Substitute for m and n.

Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log b (wxyz). Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b (wxyz)= log b w+ log b x+ log b y+ log b z
The Product Rule for Logarithms

The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b (MN)= log b ( M )+ log b ( N ) for b>0

Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.

1. Factor the argument completely, expressing each whole number factor as a product of primes.
2. Write the equivalent expression by summing the logarithms of each factor.
Using the Product Rule for Logarithms

Expand <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 3 ( 30x( 3x+4 ) ).

We begin by factoring the argument completely, expressing<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>30</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]as a product of primes.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 3 ( 30x( 3x+4 ) )= log 3 ( 2⋅3⋅5⋅x⋅( 3x+4 ) )

Next we write the equivalent equation by summing the logarithms of each factor.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 3 ( 30x( 3x+4 ) )= log 3 ( 2 )+ log 3 ( 3 )+ log 3 ( 5 )+ log 3 ( x )+ log 3 ( 3x+4 )

Expand<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log b (8k).

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b 2+ log b 2+ log b 2+ log b k=3 log b 2+ log b k

# Using the Quotient Rule for Logarithms

For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to combine the quotient of exponents by subtracting:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] x a b = x a−b . The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.

Given any real number<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and positive real numbers<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>M</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex]<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>N</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] and <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>≠</mo><mn>1</mn><mo>,</mo></mrow></annotation-xml></semantics>[/itex] we will show

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b ( M N )= log b ( M )− log b ( N ).

Let<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>m</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics>[/itex] log b M and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics>[/itex] log b N. In exponential form, these equations are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b m =M and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b n =N. It follows that

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] b ( M N ) = log b ( b m b n )      Substitute for M and N. = log b ( b m−n ) Apply the quotient rule for exponents. =m−nApply the inverse property of logs. = log b ( M )− log b ( N ) Substitute for m and n.

For example, to expand<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2 x 2 +6x 3x+9 ),we must first express the quotient in lowest terms. Factoring and canceling we get,

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>log</mi><mrow><mo>(</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 2 x 2 +6x 3x+9 )=log( 2x(x+3) 3(x+3) ) Factor the numerator and denominator.                        =log( 2x 3 )Cancel the common factors.

Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>log</mi><mrow><mo>(</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 2x 3 )=log(2x)−log(3)              =log(2)+log(x)−log(3)
The Quotient Rule for Logarithms

The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b ( M N )= log b M− log b N

Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.

1. Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
2. Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
3. Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.
Using the Quotient Rule for Logarithms

Expand<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 2 ( 15x(x−1) (3x+4)(2−x) ).

First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 2 ( 15x(x−1) (3x+4)(2−x) )= log 2 ( 15x(x−1) )− log 2 ( (3x+4)(2−x) )

Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 2 (15x(x−1))− log 2 ((3x+4)(2−x))=[ log 2 (3)+ log 2 (5)+ log 2 (x)+ log 2 (x−1)]−[ log 2 (3x+4)+ log 2 (2−x)]                                                                 = log 2 (3)+ log 2 (5)+ log 2 (x)+ log 2 (x−1)− log 2 (3x+4)− log 2 (2−x)
Analysis

There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 4 3  and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>=</mo><mn>2.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>></mo><mn>0</mn><mo>,</mo></mrow></annotation-xml></semantics>[/itex]<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi><mo>></mo><mn>1</mn><mo>,</mo></mrow></annotation-xml></semantics>[/itex]<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi><mo>></mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 4 3 , and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo><</mo><mn>2.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.

Expand<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 3 ( 7 x 2 +21x 7x( x−1 )( x−2 ) ).

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 3 ( x+3 )− log 3 ( x−1 )− log 3 ( x−2 )

# Using the Power Rule for Logarithms

We’ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] x 2 ? One method is as follows:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] b ( x 2 ) = log b ( x⋅x ) = log b x+ log b x =2 log b x

Notice that we used the product rule for logarithms to find a solution for the example above. By doing so, we have derived the power rule for logarithms, which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>100</mn><mo>=</mo><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 10 2       3 = 3 1 2       1 e = e −1
The Power Rule for Logarithms

The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b ( M n )=n log b M

Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm.

1. Express the argument as a power, if needed.
2. Write the equivalent expression by multiplying the exponent times the logarithm of the base.
Expanding a Logarithm with Powers

Expand<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 2 x 5 .

The argument is already written as a power, so we identify the exponent, 5, and the base,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 2 ( x 5 )=5 log 2 x

Expand<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>ln</mi><msup/></mrow></annotation-xml></semantics>[/itex] x 2 .

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>2</mn><mi>ln</mi><mi>x</mi></mrow></annotation-xml></semantics>[/itex]

Rewriting an Expression as a Power before Using the Power Rule

Expand<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 3 ( 25 ) using the power rule for logs.

Expressing the argument as a power, we get<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 3 ( 25 )= log 3 ( 5 2 ).

Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 3 ( 5 2 )=2 log 3 ( 5 )

Expand<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 x 2 ).

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mn>2</mn><mi>ln</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics>[/itex]

Using the Power Rule in Reverse

Rewrite<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>4</mn><mi>ln</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]using the power rule for logs to a single logarithm with a leading coefficient of 1.

Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>4</mn><mi>ln</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>,</mo></mrow></annotation-xml></semantics>[/itex]we identify the factor, 4, as the exponent and the argument,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] as the base, and rewrite the product as a logarithm of a power:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>4</mn><mi>ln</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mi>ln</mi><mo stretchy="false">(</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 4 ).

Rewrite<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><msub/></mrow></annotation-xml></semantics>[/itex] log 3 4 using the power rule for logs to a single logarithm with a leading coefficient of 1.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 3 16

# Expanding Logarithmic Expressions

Taken together, the product rule, quotient rule, and power rule are often called “laws of logs.” Sometimes we apply more than one rule in order to simplify an expression. For example:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] b ( 6x y ) = log b ( 6x )− log b y = log b 6+ log b x− log b y

We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] b ( A C ) = log b ( A C −1 ) = log b ( A )+ log b ( C −1 ) = log b A+(−1) log b C = log b A− log b C

We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.

With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots—never with addition or subtraction inside the argument of the logarithm.

Expanding Logarithms Using Product, Quotient, and Power Rules

Rewrite<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x 4 y 7 ) as a sum or difference of logs.

First, because we have a quotient of two expressions, we can use the quotient rule:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x 4 y 7 )=ln( x 4 y )−ln(7)

Then seeing the product in the first term, we use the product rule:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x 4 y )−ln(7)=ln( x 4 )+ln(y)−ln(7)

Finally, we use the power rule on the first term:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x 4 )+ln(y)−ln(7)=4ln(x)+ln(y)−ln(7)

Expand<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x 2 y 3 z 4 ).

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>2</mn><mi>log</mi><mi>x</mi><mo>+</mo><mn>3</mn><mi>log</mi><mi>y</mi><mo>−</mo><mn>4</mn><mi>log</mi><mi>z</mi></mrow></annotation-xml></semantics>[/itex]

Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression

Expand<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ).

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>log</mi><mrow><mo>(</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] x ) =log x ( 1 2 ) = 1 2 logx

Expand<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x 2 3 ).

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>2</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 3 lnx

Can we expand<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x 2 + y 2 )?

No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.

Expanding Complex Logarithmic Expressions

Expand<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 6 ( 64 x 3 ( 4x+1 ) ( 2x−1 ) ).

We can expand by applying the Product and Quotient Rules.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 6 ( 64 x 3 (4x+1) (2x−1) ) = log 6 64+ log 6 x 3 + log 6 (4x+1)− log 6 (2x−1) Apply the Quotient Rule. = log 6 2 6 + log6 x 3 + log 6 (4x+1)− log 6 (2x−1) Simplify by writing  64 as 2 6 . =6 log 6 2+3 log 6 x+ log 6 (4x+1)− log 6 (2x−1)Apply the Power Rule.

Expand<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] (x−1) (2x+1) 2 ( x 2 −9) ).

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 2 ln( x−1 )+ln( 2x+1 )−ln( x+3 )−ln( x−3 )

# Condensing Logarithmic Expressions

We can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing.

Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm.

1. Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.
2. Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.
3. Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.
Using the Product and Quotient Rules to Combine Logarithms

Write<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 3 ( 5 )+ log 3 ( 8 )− log 3 ( 2 ) as a single logarithm.

Using the product and quotient rules

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 3 ( 5 )+ log 3 ( 8 )= log 3 ( 5⋅8 )= log 3 ( 40 )

This reduces our original expression to

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 3 (40)− log 3 (2)

Then, using the quotient rule

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 3 ( 40 )− log 3 ( 2 )= log 3 ( 40 2 )= log 3 ( 20 )

Condense<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>log</mi><mn>3</mn><mo>−</mo><mi>log</mi><mn>4</mn><mo>+</mo><mi>log</mi><mn>5</mn><mo>−</mo><mi>log</mi><mn>6.</mn></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3⋅5 4⋅6 ); can also be written<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 5 8 ) by reducing the fraction to lowest terms.

Condensing Complex Logarithmic Expressions

Condense<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 2 ( x 2 )+ 1 2 log 2 ( x−1 )−3 log 2 ( ( x+3 ) 2 ).

We apply the power rule first:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 2 ( x 2 )+ 1 2 log 2 ( x−1 )−3 log 2 ( ( x+3 ) 2 )= log 2 ( x 2 )+ log 2 ( x−1 )− log 2 ( ( x+3 ) 6 )

Next we apply the product rule to the sum:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 2 ( x 2 )+ log 2 ( x−1 )− log 2 ( ( x+3 ) 6 )= log 2 ( x 2 x−1 )− log 2 ( ( x+3 ) 6 )

Finally, we apply the quotient rule to the difference:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 2 ( x 2 x−1 )− log 2 ( ( x+3 ) 6 )= log 2 x 2 x−1 ( x+3 ) 6
Rewriting as a Single Logarithm

Rewrite<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>log</mi><mi>x</mi><mo>−</mo><mn>4</mn><mi>log</mi><mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mn>5</mn><mo stretchy="false">)</mo><mo>+</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 x log( 3x+5 ) as a single logarithm.

We apply the power rule first:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>2</mn><mi>log</mi><mi>x</mi><mo>−</mo><mn>4</mn><mi>log</mi><mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mn>5</mn><mo stretchy="false">)</mo><mo>+</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 x log( 3x+5 )=log( x 2 )−log( ( x+5 ) 4 )+log( (3x+5) x −1 )

Next we apply the product rule to the sum:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x 2 )−log( ( x+5 ) 4 )+log( (3x+5) x −1 )=log( x 2 )−log( ( x+5 ) 4 ( 3x+5 ) x −1 )

Finally, we apply the quotient rule to the difference:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x 2 )−log( ( x+5 ) 4 ( 3x+5 ) x −1 )=log( x 2 ( x+5 ) 4 ( ( 3x+5 ) x −1 ) )

Rewrite<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 5 )+0.5log( x )−log( 7x−1 )+3log( x−1 ) as a single logarithm.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 5 ( x−1 ) 3 x ( 7x−1 ) )

Condense<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>4</mn><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3log( x )+log( x+5 )−log( 2x+3 ) ).

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>log</mi><mfrac/></mrow></annotation-xml></semantics>[/itex] x 12 ( x+5 ) 4 ( 2x+3 ) 4 ; this answer could also be written<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>log</mi><msup/></mrow></annotation-xml></semantics>[/itex] ( x 3 ( x+5 ) ( 2x+3 ) ) 4 .

Applying of the Laws of Logs

Recall that, in chemistry,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>pH</mtext><mo>=</mo><mo>−</mo><mi>log</mi><mo stretchy="false">[</mo><msup/></mrow></annotation-xml></semantics>[/itex] H + ]. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?

Suppose<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is the original concentration of hydrogen ions, and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>P</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is the original pH of the liquid. Then<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>P</mi><mo>=</mo><mo>–</mo><mi>log</mi><mo stretchy="false">(</mo><mi>C</mi><mo stretchy="false">)</mo><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]If the concentration is doubled, the new concentration is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>C</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex] Then the pH of the new liquid is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>pH</mtext><mo>=</mo><mo>−</mo><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2C )

Using the product rule of logs

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>pH</mtext><mo>=</mo><mo>−</mo><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2C )=−( log(2)+log(C) )=−log(2)−log(C)

Since<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>P</mi><mo>=</mo><mo>–</mo><mi>log</mi><mo stretchy="false">(</mo><mi>C</mi><mo stretchy="false">)</mo><mo>,</mo></mrow></annotation-xml></semantics>[/itex]the new pH is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>pH</mtext><mo>=</mo><mi>P</mi><mo>−</mo><mi>log</mi><mo stretchy="false">(</mo><mn>2</mn><mo stretchy="false">)</mo><mo>≈</mo><mi>P</mi><mo>−</mo><mn>0.301</mn></mrow></annotation-xml></semantics>[/itex]

When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.

How does the pH change when the concentration of positive hydrogen ions is decreased by half?

The pH increases by about 0.301.

# Using the Change-of-Base Formula for Logarithms

Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than 10 or<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>e</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex]we use the change-of-base formula to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.

To derive the change-of-base formula, we use the one-to-one property and power rule for logarithms.

Given any positive real numbers<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>M</mi><mo>,</mo><mi>b</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] and <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mo>≠</mo><mn>1</mn><mo> </mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>≠</mo><mn>1</mn><mo>,</mo></mrow></annotation-xml></semantics>[/itex]we show

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log b M= log n M log n b

Let<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics>[/itex] log b M. By taking the log base<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]of both sides of the equation, we arrive at an exponential form, namely<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b y =M. It follows that

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] n ( b y ) = log n M         Apply the one-to-one property.   y log n b = log n M  Apply the power rule for logarithms.           y= log n M log n b Isolate y.     log b M = log n M log n b Substitute for y.

For example, to evaluate<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 5 36 using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 5 36 = log( 36 ) log( 5 )    Apply the change of base formula using base 10. ≈2.2266 Use a calculator to evaluate to 4 decimal places.
The Change-of-Base Formula

The change-of-base formula can be used to evaluate a logarithm with any base.

For any positive real numbers<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>M</mi><mo>,</mo><mi>b</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mo>≠</mo><mn>1</mn><mo> </mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>≠</mo><mn>1</mn><mo>,</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b M= log n M log n b .

It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b M= lnM lnb

and

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b M= logM logb

Given a logarithm with the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log b M, use the change-of-base formula to rewrite it as a quotient of logs with any positive base<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mo>≠</mo><mn>1.</mn></mrow></annotation-xml></semantics>[/itex]

1. Determine the new base<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] remembering that the common log,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ), has base 10, and the natural log,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ),has base<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>e</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]
2. Rewrite the log as a quotient using the change-of-base formula
• The numerator of the quotient will be a logarithm with base<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and argument<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>M</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]
• The denominator of the quotient will be a logarithm with base<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and argument<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]
Changing Logarithmic Expressions to Expressions Involving Only Natural Logs

Change<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 5 3 to a quotient of natural logarithms.

Because we will be expressing<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 5 3 as a quotient of natural logarithms, the new base,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mo>=</mo><mi>e</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] b M = lnM lnb     log 5 3 = ln3 ln5

Change<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 0.5 8 to a quotient of natural logarithms.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>ln</mi><mn>8</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] ln0.5

Can we change common logarithms to natural logarithms?

Yes. Remember that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>log</mi><mn>9</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]means<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 10 9. So,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>log</mi><mn>9</mn><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] ln9 ln10 .

Using the Change-of-Base Formula with a Calculator

Evaluate<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 2 (10) using the change-of-base formula with a calculator.

According to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>e</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 2 10= ln10 ln2 Apply the change of base formula using base e.                  ≈3.3219Use a calculator to evaluate to 4 decimal places.

Evaluate<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 5 (100) using the change-of-base formula.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>ln</mi><mn>100</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] ln5 ≈ 4.6051 1.6094 =2.861

Access these online resources for additional instruction and practice with laws of logarithms.

# Key Equations

 The Product Rule for Logarithms log[/itex] b (MN)= log b ( M )+ log b ( N ) The Quotient Rule for Logarithms log[/itex] b ( M N )= log b M− log b N The Power Rule for Logarithms log[/itex] b ( M n )=n log b M The Change-of-Base Formula log[/itex] b M= log n M log n b          n>0,n≠1,b≠1

# Key Concepts

• We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms. See [link].
• We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms. See [link].
• We can use the power rule for logarithms to rewrite the log of a power as the product of the exponent and the log of its base. See [link], [link], and [link].
• We can use the product rule, the quotient rule, and the power rule together to combine or expand a logarithm with a complex input. See [link], [link], and [link].
• The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm. See [link], [link], [link], and [link].
• We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula. See [link].
• The change-of-base formula is often used to rewrite a logarithm with a base other than 10 and <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>e</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] as the quotient of natural or common logs. That way a calculator can be used to evaluate. See [link].

# Section Exercises

## Verbal

How does the power rule for logarithms help when solving logarithms with the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log b ( x n )?

Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log b ( x 1 n )= 1 n log b (x).

What does the change-of-base formula do? Why is it useful when using a calculator?

## Algebraic

For the following exercises, expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b ( 7x⋅2y )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b ( 2 )+ log b ( 7 )+ log b ( x )+ log b ( y )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3ab⋅5c )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b ( 13 17 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b ( 13 )− log b ( 17 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 4 (   x z   w )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 4 k )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mi>k</mi><mi>ln</mi><mo stretchy="false">(</mo><mn>4</mn><mo stretchy="false">)</mo></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 2 ( y x )

For the following exercises, condense to a single logarithm if possible.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 7 )+ln( x )+ln( y )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 7xy )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 3 (2)+ log 3 (a)+ log 3 (11)+ log 3 (b)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b (28)− log b (7)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] b (4)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] a )−ln( d )−ln( c )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><msub/></mrow></annotation-xml></semantics>[/itex] log b ( 1 7 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mtext>log</mtext></mrow></msub></mrow></annotation-xml></semantics>[/itex] b ( 7 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 3 ln( 8 )

For the following exercises, use the properties of logarithms to expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x 15 y 13 z 19 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>15</mn><mi>log</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>+</mo><mn>13</mn><mi>log</mi><mo stretchy="false">(</mo><mi>y</mi><mo stretchy="false">)</mo><mo>−</mo><mn>19</mn><mi>log</mi><mo stretchy="false">(</mo><mi>z</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] a −2 b −4 c 5 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x 3 y −4 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>3</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 2 log(x)−2log(y)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] y y 1−y )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x 2 y 3 x 2 y 5 3 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>8</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 3 log(x)+ 14 3 log(y)

For the following exercises, condense each expression to a single logarithm using the properties of logarithms.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2 x 4 )+log( 3 x 5 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>ln</mi><mo stretchy="false">(</mo><mn>6</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 9 )−ln(3 x 2 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>ln</mi><mo stretchy="false">(</mo><mn>2</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 7 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>2</mn><mi>log</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>+</mo><mn>3</mn><mi>log</mi><mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mn>1</mn><mo stretchy="false">)</mo></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>log</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 2 log(y)+3log(z)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x z 3 y )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>4</mn><msub/></mrow></annotation-xml></semantics>[/itex] log 7 ( c )+ log 7 ( a ) 3 + log 7 ( b ) 3

For the following exercises, rewrite each expression as an equivalent ratio of logs using the indicated base.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 7 ( 15 ) to base<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>e</mi></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 7 ( 15 )= ln( 15 ) ln( 7 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 14 ( 55.875 ) to base<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>10</mn></mrow></annotation-xml></semantics>[/itex]

For the following exercises, suppose<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 5 ( 6 )=a and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 5 ( 11 )=b. Use the change-of-base formula along with properties of logarithms to rewrite each expression in terms of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Show the steps for solving.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 11 ( 5 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 11 ( 5 )= log 5 ( 5 ) log 5 ( 11 ) = 1 b

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 6 ( 55 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 11 ( 6 11 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 11 ( 6 11 )= log 5 ( 6 11 ) log 5 ( 11 ) = log 5 ( 6 )− log 5 ( 11 ) log 5 ( 11 ) = a−b b = a b −1

## Numeric

For the following exercises, use properties of logarithms to evaluate without using a calculator.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 3 ( 1 9 )−3 log 3 ( 3 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>6</mn><msub/></mrow></annotation-xml></semantics>[/itex] log 8 ( 2 )+ log 8 ( 64 ) 3 log 8 ( 4 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mn>3</mn></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>2</mn><msub/></mrow></annotation-xml></semantics>[/itex] log 9 ( 3 )−4 log 9 ( 3 )+ log 9 ( 1 729 )

For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 3 ( 22 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>2.81359</mn></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 8 ( 65 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 6 ( 5.38 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>0.93913</mn></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 4 ( 15 2 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 1 2 ( 4.7 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mn>2.23266</mn></mrow></annotation-xml></semantics>[/itex]

## Extensions

Use the product rule for logarithms to find all<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]values such that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 12 ( 2x+6 )+ log 12 ( x+2 )=2. Show the steps for solving.

Use the quotient rule for logarithms to find all<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]values such that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 6 ( x+2 )− log 6 ( x−3 )=1. Show the steps for solving.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi><mo>=</mo><mn>4</mn><mo>;</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]By the quotient rule:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mrow><mi>log</mi></mrow></msub></mrow></annotation-xml></semantics>[/itex] 6 ( x+2 )− log 6 ( x−3 )= log 6 ( x+2 x−3 )=1.

Rewriting as an exponential equation and solving for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>:</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msup><mn>6</mn></msup></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 1 = x+2 x−3  0 = x+2 x−3 −6  0 = x+2 x−3 − 6( x−3 ) ( x−3 )  0 = x+2−6x+18 x−3  0 = x−4 x−3 ​ x =4

Checking, we find that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 6 ( 4+2 )− log 6 ( 4−3 )= log 6 ( 6 )− log 6 ( 1 ) is defined, so<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>=</mo><mn>4.</mn></mrow></annotation-xml></semantics>[/itex]

Can the power property of logarithms be derived from the power property of exponents using the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b x =m? If not, explain why. If so, show the derivation.

Prove that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log b ( n )= 1 log n ( b )  for any positive integers<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>></mo><mn>1</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mo>></mo><mn>1.</mn></mrow></annotation-xml></semantics>[/itex]

Let<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]be positive integers greater than<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>1.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Then, by the change-of-base formula,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log b ( n )= log n ( n ) log n ( b ) = 1 log n ( b ) .

Does<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] log 81 ( 2401 )= log 3 ( 7 )? Verify the claim algebraically.

## Glossary

change-of-base formula
a formula for converting a logarithm with any base to a quotient of logarithms with any other base.
power rule for logarithms
a rule of logarithms that states that the log of a power is equal to the product of the exponent and the log of its base
product rule for logarithms
a rule of logarithms that states that the log of a product is equal to a sum of logarithms
quotient rule for logarithms
a rule of logarithms that states that the log of a quotient is equal to a difference of logarithms