4.5: Logarithmic Properties
 Page ID
 31105
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 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 changeofbase formula for logarithms.
In chemistry, the pH scale is used as a measure of the acidity or alkalinity of a substance. 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 \(25^\circ C\)
 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 \([H^+]\) is the concentration of hydrogen ion in the solution
\[\begin{align} {pH}&=−{\log}([H^+]) \label{eq1} \\[5pt] &={\log}\left(\dfrac{1}{[H^+]}\right) \label{eq2} \end{align}\]
The equivalence of Equations \ref{eq1} and \ref{eq2} is one of the logarithm properties we will examine in this section.
Figure \(\PageIndex{1}\): The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan).
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.
\[{\log}_b1=0\]
\[{\log}_bb=1\]
For example, \({\log}_51=0\) since \(5^0=1\). And \({\log}_55=1\) since \(5^1=5\).
Next, we have the inverse property.
\[{\log}_b(b^x)=x\]
\[b^{{\log}_bx}=x, \,x>0\]
For example, to evaluate \({\log(100)}\),we can rewrite the logarithm as \({\log}_{10}({10}^2)\), and then apply the inverse property \({\log}_b(b^x)=x\) to get \({\log}_{10}({10}^2)=2\).
To evaluate \(e^{\ln(7)}\),we can rewrite the logarithm as \(e^{{\log}_e7}\), to see that we can apply the inverse property \(b^{{\log}_bx}=x\) to get \(e^{{\log}_e 7}=7\).
Finally, we have the onetoone property.
\[\begin{align} \mbox{if }\log_bM={\log}_bN,\text{ then} M=N \end{align}\]
We can use the onetoone property to solve the equation \({\log}_3(3x)={\log}_3(2x+5)\) for \(x\). Since the bases are the same, we can apply the onetoone property by setting the arguments equal and solving for \(x\):
\(3x=2x+5\) Set the arguments equal.
\(x=5\) Subtract \(2x\).
But what about the equation \({\log}_3(3x)+{\log}_3(2x+5)=2\)? The onetoone 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: \(x^ax^b=x^{a+b}\). 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. The product rule for logarithms can be used to rewrite a logarithm of a product as the sum of individual 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.
Theorem \(\PageIndex{1}\) THE PRODUCT RULE FOR LOGARITHMS
Given any real number \(x\) and positive real numbers \(M\), \(N\), and \(b\), where \(b≠1\), \[\log_b(MN)=\log_b(M)+\log_b(N).\]
 Proof

Let \(m={\log}_bM\) and \(n={\log}_bN\). In exponential form, these equations are \(b^m=M\) and \(b^n=N\). It follows that
\[\begin{align*} {\log}_b(MN)&= {\log}_b(b^mb^n) \qquad \text{Substitute for M and N}\\ &= {\log}_b(b^{m+n}) \qquad \text{Apply the product rule for exponents}\\ &= m+n \qquad \text{Apply the inverse property of logs}\\ &= {\log}_b(M)+{\log}_b(N) \qquad \text{Substitute for m and n} \end{align*}\]
Extra credit: Does this rule work if \(M\) and \(N\) are both negative? Why, or why not? Bring your explanation to your professor, and you will get extra credit.
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 \({\log}_b(wxyz)\). Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:
\({\log}_b(wxyz)={\log}_bw+{\log}_bx+{\log}_by+{\log}_bz\)
Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms
 Factor the argument completely, expressing each whole number factor as a product of primes.
 Write the equivalent expression by summing the logarithms of each factor.
Example \(\PageIndex{1}\): Using the Product Rule for Logarithms
Expand \({\log}_3(30x(3x+4))\).
Solution
We begin by factoring the argument completely, expressing \(30\) as a product of primes.
\({\log}_3(30x(3x+4))={\log}_3(2⋅3⋅5⋅x⋅(3x+4))\)
Next we write the equivalent equation by summing the logarithms of each factor.
\({\log}_3(30x(3x+4))={\log}_3(2)+{\log}_3(3)+{\log}_3(5)+{\log}_3(x)+{\log}_3(3x+4)\)
\(\PageIndex{1}\)
Expand \({\log}_b(8k)\).
 Answer

\({\log}_b2+{\log}_b2+{\log}_b2+{\log}_bk=3{\log}_b2+{\log}_bk\)
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: \(x^{\frac{a}{b}}=x^{a−b}\). The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms. The quotient rule for logarithms can be used to rewrite a logarithm of a quotient as the difference of individual logarithms.
Just as with the product rule, we can use the inverse property to derive the quotient rule.
Theorem \(\PageIndex{2}\) THE QUOTIENT RULE FOR LOGARITHMS
Given any real number \(x\) and positive real numbers \(M\), \(N\), and b, b, where \(b≠1\), \[\log_b\left(\dfrac{M}{N}\right)={\log}_b(M)−{\log}_b(N).\]
 Proof

Let \(m={\log}_bM\) and \(n={\log}_bN\). In exponential form, these equations are \(b^m=M\) and \(b^n=N\). It follows that
\[\begin{align*} {\log}_b\left (\dfrac{M}{N} \right )&= {\log}_b\left(\dfrac{b^m}{b^n}\right) \qquad \text{Substitute for M and N}\\ &= {\log}_b(b^{mn}) \qquad \text{Apply the quotient rule for exponents}\\ &= mn \qquad \text{Apply the inverse property of logs}\\ &= {\log}_b(M){\log}_b(N) \qquad \text{Substitute for m and n} \end{align*}\]
For example, to expand \({\log}\left (\dfrac{2x^2+6x}{3x+9} \right )\), we must first express the quotient in lowest terms. Factoring and canceling we get,
\[\begin{align*} {\log}\left (\dfrac{2x^2+6x}{3x+9} \right )&= {\log}\left (\dfrac{2x(x+3)}{3(x+3)} \right ) \qquad \text{Factor the numerator and denominator}\\ &= {\log}\left (\dfrac{2x}{3} \right ) \qquad \text{Cancel the common factors} \end{align*}\]
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.
\[ \begin{align*} {\log}\left(\dfrac{2x}{3}\right) &={\log}(2x)−{\log}(3) \\[5pt] &={\log}(2)+{\log}(x)−{\log}(3) \end{align*}\]
Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms
 Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
 Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
 Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.
Example \(\PageIndex{2}\): Using the Quotient Rule for Logarithms
Expand \(\log_2\left(\dfrac{15x(x−1)}{(3x+4)(2−x)}\right)\).
Solution
First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.
\(\log_2\left(\dfrac{15x(x−1)}{(3x+4)(2−x)}\right)={\log}_2(15x(x−1))−{\log}_2((3x+4)(2−x))\)
Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of 15 are 3 and 5.
\[\begin{align*} {\log}_2(15x(x1)){\log}_2((3x+4)(2x))&= [{\log}_2(3)+{\log}_2(5)+{\log}_2(x)+{\log}_2(x1)][{\log}_2(3x+4)+{\log}_2(2x)]\\ &= {\log}_2(3)+{\log}_2(5)+{\log}_2(x)+{\log}_2(x1){\log}_2(3x+4){\log}_2(2x) \end{align*}\]
Analysis
There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for \(x=−\dfrac{4}{3}\) and \(x=2\). Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that \(x>0\), \(x>1\), \(x>−\dfrac{4}{3}\), and \(x<2\). Combining these conditions yields the condition \(1<x<2.\)
\(\PageIndex{2}\)
Expand \({\log}_3\left(\dfrac{7x^2+21x}{7x(x−1)(x−2)}\right)\).
 Answer

\({\log}_3(x+3)−{\log}_3(x−1)−{\log}_3(x−2)\)
: Can we use the product or quotient rules for logarithms to rewrite \((\log_bM)(\log_bN)\) or \(\dfrac{\log_bM}{\log_bN}\)?
No. These rules apply if we are applying a logarithm to a product or to a quotient, not when we are multiplying or dividing logarithms.
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 \(x^2\)? One method is as follows:
\[\begin{align*} {\log}_b(x^2)&= {\log}_b(x\cdot x)\\ &= {\log}_bx+{\log}_bx\\ &= 2{\log}_bx \end{align*}\]
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. 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,
\(100={10}^2 \quad \sqrt{3}=3^{\frac{1}{2}} \quad \dfrac{1}{e}=e^{−1}\)
THE POWER RULE FOR LOGARITHMS
The power rule for logarithms can be used to rewrite the logarithm of a positive number \(M\) raised to an exponent as the product of the exponent times the logarithm of \(M.\)
\[{\log}_b(M^n)=n{\log}_bM\]
Given the logarithm of a power, use the power rule of logarithms to write an equivalent product
 Express the argument as a power, if needed.
 Write the equivalent expression by multiplying the exponent times the logarithm of the base.
Example \(\PageIndex{3}\): Rewriting a logarithmic Expression Using the Power Rule
Rewrite \({\log}_3(25)\) using the power rule for logs.
Solution
Expressing the argument as a power, we get \({\log}_3(25)={\log}_3(5^2)\).
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.
\({\log}_3(5^2)=2{\log}_3(5)\)
\(\PageIndex{3}\)
Rewrite \(\ln\left (\dfrac{1}{x^2} \right )\) using the power rule for logs.
 Answer

\(−2\ln(x)\)
Example \(\PageIndex{4}\): Using the Power Rule in Reverse
Rewrite \(4\ln(x)\) using the power rule for logs to a single logarithm with a leading coefficient of \(1\).
Solution
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 \(4\ln(x)\), we identify the factor, \(4\), as the exponent and the argument, \(x\), as the base, and rewrite the product as a logarithm of a power: \(4\ln(x)=\ln(x^4)\).
\(\PageIndex{4}\)
Rewrite \(2{\log}_34\) using the power rule for logs to a single logarithm with a coefficient of \(1\).
 Answer

\({\log}_316\)
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:
\[\begin{align*} {\log}_b \left (\dfrac{6x}{y} \right )&= {\log}_b(6x){\log}_by\\ &= {\log}_b6+{\log}_bx{\log}_by \end{align*}\]
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:
\[\begin{align*} {\log}_b\left (\dfrac{A}{C} \right )&= {\log}_b(AC^{1})\\ &= {\log}_b(A)+{\log}_b(C^{1})\\ &= {\log}_bA+(1){\log}_bC\\ &= {\log}_bA−{\log}_bC \end{align*}\]
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.
Example \(\PageIndex{5}\): Expanding Logarithms Using Product, Quotient, and Power Rules
Rewrite \(ln \left (\dfrac{x^4y}{7} \right )\) as a sum or difference of logs.
Solution
First, because we have a quotient of two expressions, we can use the quotient rule:
\(\ln \left (\dfrac{x^4y}{7} \right )=\ln(x^4y)−\ln(7)\)
Then seeing the product in the first term, we use the product rule:
\(\ln(x^4y)−\ln(7)=\ln(x^4)+\ln(y)−\ln(7)\)
Finally, we use the power rule on the first term:
\(\ln(x^4)+\ln(y)−\ln(7)=4\ln(x)+\ln(y)−\ln(7)\)
\(\PageIndex{5}\)
Expand \(\log \left (\dfrac{x^2y^3}{z^4} \right )\).
 Answer

\(2\log x+3\log y−4\log z\)
Example \(\PageIndex{6}\): Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression
Expand \(\log(\sqrt{x})\).
Solution
\[\begin{align*} \log(\sqrt{x})&= \log \left(x^{1/2}\right)\\ &= \dfrac{1}{2}\log x \end{align*}\]
\(\PageIndex{6}\)
Expand \(\ln(\sqrt[3]{x^2})\).
 Answer

\(\dfrac{2}{3}\ln x\)
: Can we expand \(\ln(x^2+y^2)\)?
No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.
Example \(\PageIndex{7}\): Expanding Complicated Logarithmic Expressions
Expand \({\log}_6 \left (\dfrac{64x^3(4x+1)}{(2x−1)} \right )\).
Solution
We can expand by applying the Product and Quotient Rules.
\[\begin{align*} {\log}_6\left (\dfrac{64x^3(4x+1)}{(2x1)} \right )&= {\log}_664+{\log}_6x^3+{\log}_6(4x+1){\log}_6(2x1)\qquad \text{Apply the Quotient Rule}\\ &= {\log}_62^6+{\log}_6x^3+{\log}_6(4x+1){\log}_6(2x1) \qquad \text{Simplify by writing 64 as } 2^6\\ &= 6{\log}_62+3{\log}_6x+{\log}_6(4x+1){\log}_6(2x1) \qquad \text{Apply the Power Rule} \end{align*}\]
\(\PageIndex{7}\)
Expand \(\ln \left (\dfrac{\sqrt{(x−1){(2x+1)}^2}}{(x^2−9)}\right )\).
 Answer

\(\dfrac{1}{2}\ln(x−1)+\ln(2x+1)−\ln(x+3)−\ln(x−3)\)
Good job! You completed an exercise that has relatively little meaning as a useful exercise. Tell your professor to give you extra credit!
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
 Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.
 Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.
 Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.
Example \(\PageIndex{8}\): Using the Product and Quotient Rules to Combine Logarithms
Write \({\log}_3(5)+{\log}_3(8)−{\log}_3(2)\) as a single logarithm.
Solution
Using the product and quotient rules
\({\log}_3(5)+{\log}_3(8)={\log}_3(5⋅8)={\log}_3(40)\)
This reduces our original expression to
\({\log}_3(40)−{\log}_3(2)\)
Then, using the quotient rule
\({\log}_3(40)−{\log}_3(2)={\log}_3 \left (\dfrac{40}{2} \right )={\log}_3(20)\)
\(\PageIndex{8}\)
Condense \(\log 3−\log 4+\log 5−\log 6\).
 Answer

\(\log \left (\dfrac{3⋅5}{4⋅6} \right)\); can also be written \(\log \left (\dfrac{5}{8} \right )\) by reducing the fraction to lowest terms.
Example \(\PageIndex{9}\): Condensing Complicated Logarithmic Expressions
Condense \({\log}_2(x^2)+\dfrac{1}{2}{\log}_2(x−1)−3{\log}_2({(x+3)}^2)\).
Solution
We apply the power rule first:
\({\log}_2(x^2)+\dfrac{1}{2}{\log}_2(x−1)−3{\log}_2({(x+3)}^2)={\log}_2(x^2)+{\log}_2(\sqrt{x−1})−{\log}_2({(x+3)}^6)\)
Next we apply the product rule to the sum:
\({\log}_2(x^2)+{\log}_2(\sqrt{x−1})−{\log}_2({(x+3)}^6)={\log}_2(x^2\sqrt{x−1})−{\log}_2({(x+3)}^6)\)
Finally, we apply the quotient rule to the difference:
\({\log}_2(x^2\sqrt{x−1})−{\log}_2({(x+3)}^6)={\log}_2\left(\dfrac{x^2\sqrt{x−1}}{{(x+3)}^6}\right)\)
\(\PageIndex{9}\)
Condense \(4(3\log(x)+\log(x+5)−\log(2x+3))\).
 Answer

\(\log\left(\dfrac{x^{12}{(x+5)}^4}{{(2x+3)}^4}\right)\); this answer could also be written \(\log{ \left (\dfrac{x^3(x+5)}{(2x+3)} \right )}^4\)
Example \(\PageIndex{10}\): Applying of the Laws of Logs
Recall that, in chemistry, \({pH}=−\log[H+]\). If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?
Solution
Suppose \(C\) is the original concentration of hydrogen ions, and \(P\) is the original pH of the liquid. Then \(P=–\log(C)\). If the concentration is doubled, the new concentration is \(2C\). Then the pH of the new liquid is
\(pH=−\log(2C)\)
Using the product rule of logs
\(pH=−\log(2C)=−(\log(2)+\log(C))=−\log(2)−\log(C)\)
Since \(P=–\log(C)\),the new pH is
\(pH=P−\log(2)≈P−0.301\)
\(\PageIndex{10}\)
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?
 Answer

The pH increases by about \(0.301\).
Using the ChangeofBase Formula for Logarithms
Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than \(10\) or e, we use the changeofbase formula, which allows us to rewrite a logarithm base \(b\) as the quotient of logarithms of any other base. When using a calculator, we can change any logarithm to common or natural logs.
To derive the changeofbase formula, we use the onetoone property and power rule for logarithms.
Theorem \(\PageIndex{3}\) THE CHANGEOFBASE FORMULA
Given any positive real numbers \(M\), \(b\), and \(n\), where \(n≠1\) and \(b≠1\), \[\log_bM=\dfrac{{\log}_nM}{{\log}_nb.\]
 Proof

Let \(y={\log}_bM\). By taking the log base \(n\) of both sides of the equation, we arrive at an exponential form, namely \(b^y=M\). It follows that
\[\begin{align*} {\log}_n(b^y)&= {\log}_nM \qquad \text{Apply the onetoone property}\\ y{\log}_nb&= {\log}_nM \qquad \text{Apply the power rule for logarithms}\\ y&= \dfrac{{\log}_nM}{{\log}_nb} \qquad \text{Isolate y}\\ {\log}_bM&= \dfrac{{\log}_nM}{{\log}_nb} \qquad \text{Substitute for y} \end{align*}\]
The changeofbase formula can be used to calculate a logarithm with any base. For example, to evaluate \({\log}_536\) using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.
\[\begin{align*} {\log}_536&= \dfrac{\log(36)}{\log(5)} \qquad \text{Apply the change of base formula using base 10}\\ &\approx 2.2266 \qquad \text{Use a calculator to evaluate to 4 decimal places} \end{align*}\]
Given a logarithm with the form \({\log}_bM\), use the changeofbase formula to rewrite it as a quotient of logs with any positive base \(n\), where \(n≠1\)
 Choose the new base \(n\), remembering that the common log, \(\log(x)\), has base 10, and the natural log, \(\ln(x)\),has base \(e\).
 Rewrite the log as a quotient using the changeofbase formula:
 The numerator of the quotient will be a logarithm with base \(n\) and argument \(M\).
 The denominator of the quotient will be a logarithm with base \(n\) and argument \(b\).
Example \(\PageIndex{11}\): Changing Logarithmic Expressions to Expressions Involving Only Natural Logs
Change \({\log}_53\) to a quotient of natural logarithms and evaluate to 4 decimal places.
Solution
Because we will be expressing \({\log}_53\) as a quotient of natural logarithms, the new base, \(n=e\).
We rewrite the log as a quotient using the changeofbase 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.
\({\log}_bM=\dfrac{\ln M}{\ln b}\)
\({\log}_53=\dfrac{\ln3}{\ln5} \approx 0.6826\)
\(\PageIndex{11}\)
Change \(\log_{0.5}8\) to a quotient of natural logarithms and evaluate.
 Answer

\(\dfrac{\ln8}{\ln 0.5} = 3.0000\)
: Can we change common logarithms to natural logarithms?
Yes. Remember that \(\log9\) means \({\log}_{10}9\). So, \(\log9=\dfrac{\ln9}{\ln10}\).
\(\PageIndex{12}\)
Evaluate \({\log}_5(100)\) using the changeofbase formula.
 Answer

\(\dfrac{\ln100}{\ln5} \approx 2.861\)
Media
Access these online resources for additional instruction and practice with laws of logarithms.
Key Equations
The Product Rule for Logarithms  \({\log}_b(MN)={\log}_b(M)+{\log}_b(N)\) 
The Quotient Rule for Logarithms  \({\log}_b(\dfrac{M}{N})={\log}_bM−{\log}_bN\) 
The Power Rule for Logarithms  \({\log}_b(M^n)=n{\log}_bM\) 
The ChangeofBase Formula  \({\log}_bM=\dfrac{{\log}_nM}{{\log}_nb}\) \(n>0\), \(n≠1\), \(b>0, b≠1\) 
Key Concepts
 We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms. See Example \(\PageIndex{1}\).
 We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms. See Example \(\PageIndex{2}\).
 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 Example \(\PageIndex{3}\).
 We can use the product rule, the quotient rule, and the power rule together to expand a logarithm with a complex input. See Example \(\PageIndex{5}\), Example \(\PageIndex{6}\), and Example \(\PageIndex{7}\).
 The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm. See Example \(\PageIndex{8}\) and Example \(\PageIndex{9}\).
 We can convert a logarithm with any base to a quotient of logarithms with any other base using the changeofbase formula. See Example \(\PageIndex{11}\).
Contributors
Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.