5.6: Laws of Logarithms
- Page ID
- 89909
- 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.
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 \([\ce{H^{+}}]\) is the concentration of hydrogen ions in the solution
\[\begin{align*} {pH}&=−{\log}([\ce{H^{+}}]) \label{eq1} \\[4pt] &={\log}\left(\dfrac{1}{[\ce{H^{+}}]}\right) \label{eq2} \end{align*}\nonumber\]
These two equations for\(pH\) look different but, in fact, are equivalent. This is one of the logarithm properties we will examine in this section.
Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. First, let's recall the properties that we have learned thus far:
- \(log_b(b^x) =x\)
- \(b^{\log_b x}=x\) where \(x>0\)
- \(\log_b1 =0\)
- \({\log}_bb=1\)
The first two properties listed above are called inverse properties.
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. 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 \(x\) and positive real numbers \(M\), \(N\), and \(b\), where \(b≠1\), we will show
\({\log}_b(MN)={\log}_b(M)+{\log}_b(N)\).
Let \(m={\log}_bM\) and \(n={\log}_bN\). In exponential form, these equations are \(b^m=M\) and \(b^n=N\). It follows that
\[\begin{align*} {\log}_b(MN)&= {\log}_b(b^mb^n) \qquad \text{Substitute for M and N}\\[4pt] &= {\log}_b(b^{m+n}) \qquad \text{Apply the product rule for exponents}\\[4pt] &= m+n \qquad \text{Apply the inverse property of logs}\\[4pt] &= {\log}_b(M)+{\log}_b(N) \qquad \text{Substitute for m and n} \end{align*}\]
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:
\({\log}_b(wxyz)={\log}_bw+{\log}_bx+{\log}_by+{\log}_bz\)
The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.
\[\begin{align*} {\log}_b(MN)={\log}_b(M)+{\log}_b(N)\text{ for } b> 0 \end{align*}\nonumber\]
We can use this rule as well as others we will discuss shortly, to expand a single logarithm into sums and differences of simpler logarithms or we can use this rule to combine logarithms into a single logarithm. This is a useful things to be able to do as it will allow us a way to solve certain kings of logarithmic equations as well as determine behavior of logarithmic functions. As you will see in calculus, these rules are also very useful for making the process of taking derivatives and integrating much easier. Let's now look at some examples of applying this rule.
Expand \(\log_5(6a)\).
Solution
Notice that the argument, \(6a\), is a product. Using the product rule, we can expand to write this as a sum of logarithms, \[\log_5(6a) =\log_5(6)+\log_5(a)\nonumber\]
Remember that is is a common practice to write single term arguments without parentheses. For the last example, it would be acceptable to also write the answer as \(log_56+\log_5a\). However, it would not be correct to write \(\log_5(6a) \text{ as } \log_56a\) since the argument is a product. \[\begin{align*}\log_56a &=\log_5(6)(a) \qquad \text{a product of log base 5 of 6 which is multiplied by a} \\[6pt] &=a\log_56\ \end{align*}\]When we have a product of a logarithm with a single term, it is common practice to write the logarithm last.
A common mistake is to not put parentheses around arguments of products. \[\log_b(mn) \neq \log_bmn\nonumber\]
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.
\({\log}_3(30x(3x+4))={\log}_3(2)+{\log}_3(3)+{\log}_3(5)+{\log}_3(x)+{\log}_3(3x+4)\)
We discussed earlier that it is acceptable to write a logarithm of a single term argument without parentheses. This is not the case with an argument that is a product or an argument that is a sum or difference.
Expand \({\log}_b(8k)\).
- Answer
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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 simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.
\[{\log}_b\left(\dfrac{M}{N}\right)={\log}_bM−{\log}_bN\nonumber\]
Just as with the product rule, we can use the inverse property to derive the quotient rule.
Given any real number \(x\) and positive real numbers \(M\), \(N\), and b, b, where \(b≠1\), we will show
\({\log}_b\left(\dfrac{M}{N}\right)={\log}_b(M)−{\log}_b(N)\).
Let \(m={\log}_bM\) and \(n={\log}_bN\). In exponential form, these equations are \(b^m=M\) and \(b^n=N\). It follows that
\[\begin{align*} {\log}_b\left (\dfrac{M}{N} \right )&= {\log}_b\left(\dfrac{b^m}{b^n}\right) \qquad \text{Substitute for M and N}\\[4pt] &= {\log}_b(b^{m-n}) \qquad \text{Apply the quotient rule for exponents}\\[4pt] &= m-n \qquad \text{Apply the inverse property of logs}\\[4pt] &= {\log}_b(M)-{\log}_b(N) \qquad \text{Substitute for m and n} \end{align*}\]
Expand \(\log_7\left(\dfrac{3}{x}\right)\).
Solution
Notice that the argument is a quotient. Using the quotient rule are are able to write this is as a difference of logarithms, \[\log_7\left(\dfrac{3}{x}\right)=\log_7(3)-\log_7(x)\nonumber\]
A note about notation here. In the previous example, the argument is a quotient. Notice that the logarithm is aligned with the fraction bar in the argument. Unlike with arguments that are products, sometimes you will see arguments that are quotients written without parentheses, \(\log_7\left(\dfrac{3}{x}\right)=\log_7\dfrac{3}{x}\). Logarithms are functions and functions have inputs. For the statement, \(\log_7\dfrac{3}{x}\), to make sense the argument must be the single term \(\dfrac{3}{x}\).
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 the factor 15 are 3 and 5.
\[\begin{align*} {\log}_2(15x(x-1))-{\log}_2((3x+4)(2-x))&= [{\log}_2(3)+{\log}_2(5)+{\log}_2(x)+{\log}_2(x-1)]-[{\log}_2(3x+4)+{\log}_2(2-x)]\\[4pt] &= {\log}_2(3)+{\log}_2(5)+{\log}_2(x)+{\log}_2(x-1)-{\log}_2(3x+4)-{\log}_2(2-x) \end{align*}\]
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 is beyond the scope of this section, and we will not consider them here or in subsequent exercises.
Expand \({\log}_3\left(\dfrac{7x^2+21x}{7x(x−1)(x−2)}\right)\).
- Answer
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\({\log}_3(x+3)−{\log}_3(x−1)−{\log}_3(x−2)\)
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 \(x^2\)? One method is as follows:
\[\begin{align*} {\log}_b(x^2)&= {\log}_b(x\cdot x)\\[4pt] &= {\log}_bx+{\log}_bx\\[4pt] &= 2{\log}_bx \end{align*}\]
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, 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,
\(100={10}^2\) \(\sqrt{3}=3^{\dfrac{1}{2}}\) \(\dfrac{1}{e}=e^{−1}\)
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.
\[{\log}_b(M^n)=n{\log}_bM\nonumber\]
Note that since \(M^n\) is a single term that \({\log}_b(M^n)={\log}_bM^n\nonumber\).
Expand \({\log}_2x^5\).
Solution
The argument is already written as a power, so we identify the exponent, 5, and the base, \(x\), and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.
\({\log}_2(x^5)=5{\log}_2x\)
Notice in the previous example there were no parentheses around the argument. The argument is a single term so it is ok to do this.
Expand \(\ln x^2\).
- Answer
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\(2\ln x\)
Expand \({\log}_3(25)\) 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.
\({\log}_3(25)=2{\log}_3(5)\)
Expand \(\ln\left (\dfrac{1}{x^2} \right )\).
- Answer
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\(−2\ln(x)\)
(remember to simplify the answer, \(\ln1=0\))
Now that we are getting good with expanding logarithms, let's look at using this rule in the reverse order.
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: \(4\ln(x)=\ln(x^4)\).
Rewrite \(2{\log}_34\) using the power rule for logs to a single logarithm with a leading coefficient of \(1\).
- Answer
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\({\log}_316\)
Expanding Logarithmic Expressions Using Multiple Rules
Taken together, the product rule, quotient rule, and power rule are often called Laws of Logarithms. Sometimes we apply more than one rule in order to simplify an expression. For example:
\[\begin{align*} {\log}_b \left (\dfrac{6x}{y} \right )&= {\log}_b(6x)-{\log}_by\\[4pt] &= {\log}_b6+{\log}_bx-{\log}_by \end{align*}\]
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:
\[\begin{align*} {\log}_b\left (\dfrac{A}{C} \right )&= {\log}_b(AC^{-1})\\[4pt] &= {\log}_b(A)+{\log}_b(C^{-1})\\[4pt] &= {\log}_bA+(-1){\log}_bC\\[4pt] &= {\log}_bA−{\log}_bC \end{align*}\]
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.
Expand \(\ln \left (\dfrac{x^4y}{7} \right )\).
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:
\(\ln(x^4)+\ln(y)−\ln(7)=4\ln(x)+\ln(y)−\ln(7)\)
So, \(ln \left (\dfrac{x^4y}{7} \right)=4\ln(x)+\ln(y)−\ln(7)\)
Expand \(\log \left (\dfrac{x^2y^3}{z^4} \right )\).
- Answer
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\(2\log x+3\log y−4\log z\)
Expand \(\log\sqrt{x}\).
Solution
\[\begin{align*} \log\sqrt{x} &= \log\left(x^\tfrac{1}{2} \right) \\ &= \dfrac{1}{2}\log x \end{align*}\]
Expand \(\ln(\sqrt[3]{x^2})\).
- Answer
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\(\dfrac{2}{3}\ln x\)
Expand \({\log}_6 \left (\dfrac{64x^3(4x+1)}{(2x−1)} \right )\).
Solution
We can expand by applying the Product and Quotient Rules.
\[\begin{align*} {\log}_6\left (\dfrac{64x^3(4x+1)}{(2x-1)} \right )&= {\log}_664+{\log}_6x^3+{\log}_6(4x+1)-{\log}_6(2x-1)\qquad \text{Apply the Quotient Rule}\\[4pt] &= {\log}_62^6+{\log}_6x^3+{\log}_6(4x+1)-{\log}_6(2x-1) \qquad \text{Simplify by writing 64 as } 2^6\\[4pt] &= 6{\log}_62+3{\log}_6x+{\log}_6(4x+1)-{\log}_6(2x-1) \qquad \text{Apply the Power Rule} \end{align*}\]
Expand \(\ln \left (\dfrac{\sqrt{(x−1){(2x+1)}^2}}{(x^2−9)}\right )\).
- Answer
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\(\dfrac{1}{2}\ln(x−1)+\ln(2x+1)−\ln(x+3)−\ln(x−3)\)
Condensing Logarithmic Expressions Using Multiple Rules
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.
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
\({\log}_3(40)−{\log}_3(2)={\log}_3 \left (\dfrac{40}{2} \right )={\log}_3(20)\)
Write \({\log}3−{\log}4+{\log}5−{\log}6\) as a single logarithm.
- Answer
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\(\log \left (\dfrac{5}{8} \right )\) after reducing the fraction to lowest terms.
Condense \({\log}_2(x^2)+\dfrac{1}{2}{\log}_2(x−1)−3{\log}_2{(x+3)}^2\) into a single logarithm.
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:
\({\log}_2(x^2\sqrt{x−1})−{\log}_2{(x+3)}^6={\log}_2\dfrac{x^2\sqrt{x−1}}{{(x+3)}^6}\)
Rewrite \(2\log x−4\log(x+5)+\log(3x+5)\) as a single logarithm.
Solution
We apply the power rule first:
\(2\log x−4\log(x+5)+\\log(3x+5)=\log(x^2)−\log{(x+5)}^4+\log({(3x+5)}^{x^{−1}})\)
Next we rearrange and apply the product rule to the sum:
\[\begin{align*} \log(x^2)-\log{(x+5)}^4+\log({(3x+5)})&= \log(x^2)+\log({(3x+5)}-\log{(x+5)}^4\\[4pt] &= \log(x^2{(3x+5)})-\log{(x+5)}^4 \qquad \text{Apply the product rule}\\[4pt] &= \log\dfrac{x^2{(3x+5)}}{{(x+5)}^4} \qquad \text{Apply the quotient rule} \end{align*}\]
Rewrite \(\log(5)+0.5\log(x)−\log(7x−1)+3\log(x−1)\) as a single logarithm.
- Answer
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\(\log \dfrac{5\sqrt{x}{(x−1)}^3}{(7x−1)}\)
Condense \(4(3\log(x)+\log(x+5)−\log(2x+3))\).
- Answer
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\(\log\dfrac{x^{12}{(x+5)}^4}{{(2x+3)}^4}\); this answer could also be written \(\log{ \left (\dfrac{x^3(x+5)}{(2x+3)} \right )}^4\)
In chemistry, pH is defined as \({pH}=−\log[H+]\) where \(H+\) is the concentration of hydrogen ions in a liquid. 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
\(pH=P−\log(2)≈P−0.301\)
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
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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 e, we use the change-of-base formula to rewrite the logarithm as the quotient of logarithms of any other base;. When using a calculator to approximate a logarithm, 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 \(M\), \(b\), and \(n\), where \(n≠1\) and \(b≠1\),we show
\({\log}_bM=\dfrac{{\log}_nM}{{\log}_nb}\)
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
\[\begin{align*} {\log}_n(b^y)&= {\log}_nM \qquad \text{Apply the one-to-one property}\\[4pt] y{\log}_nb&= {\log}_nM \qquad \text{Apply the power rule for logarithms}\\[4pt] y&= \dfrac{{\log}_nM}{{\log}_nb} \qquad \text{Isolate y}\\[4pt] {\log}_bM&= \dfrac{{\log}_nM}{{\log}_nb} \qquad \text{Substitute for y} \end{align*}\]
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.
\[\begin{align*} {\log}_536&= \dfrac{\log(36)}{\log(5)} \qquad \text{Apply the change of base formula using base 10}\\[4pt] &\approx 2.2266 \qquad \text{Use a calculator to evaluate to 4 decimal places} \end{align*}\]
The change-of-base formula can be used to evaluate a logarithm with any base.
For any positive real numbers \(M\), \(b\), and \(n\), where \(n≠1\) and \(b≠1\),
\[{\log}_bM=\dfrac{{\log}_nM}{{\log}_nb}\nonumber\]
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.
\[{\log}_bM=\dfrac{\ln M}{\ln b}\nonumber\]
and
\[{\log}_bM=\dfrac{\log M}{\log b}\nonumber\]
Change \({\log}_53\) to a quotient of natural logarithms.
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 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.
\({\log}_bM=\dfrac{\ln M}{\ln b}\)
\({\log}_53=\dfrac{\ln3}{\ln5}\)
Evaluate \({\log}_2(10)\) using the change-of-base formula with a calculator and round to four decimal places.
Solution
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 \(e\).
\[\begin{align*} {\log}_210&= \dfrac{\ln10}{\ln2} \qquad \text{Apply the change of base formula using base } e\\[4pt] &\approx 3.3219 \qquad \text{Use a calculator to evaluate to 4 decimal places} \end{align*}\]
Evaluate \({\log}_5(100)\) using the change-of-base formula. and round to three decimal places.
- Answer
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\(\dfrac{\ln100}{\ln5}≈\dfrac{4.6051}{1.6094}=2.861\)