Learning Objectives
- Use the properties of logarithms
- Use the Change of Base Formula
Be Prepared 10.4
Before you get started, take this readiness quiz.
- Evaluate: ⓐ ${a}^{0}$ ⓑ ${a}^{1}.$
If you missed this problem, review Example 5.14. - Write with a rational exponent: $\sqrt[3]{{x}^{2}y}.$
If you missed this problem, review Example 8.27. - Round to three decimal places: 2.5646415.
If you missed this problem, review Example 1.34.
Use the Properties of Logarithms
Now that we have learned about exponential and logarithmic functions, we can introduce some of the properties of logarithms. These will be very helpful as we continue to solve both exponential and logarithmic equations.
The first two properties derive from the definition of logarithms. Since ${a}^{0}=1,$ we can convert this to logarithmic form and get ${\text{log}}_{a}1=0.$ Also, since ${a}^{1}=a,$ we get ${\text{log}}_{a}a=1.$
Properties of Logarithms
In the next example we could evaluate the logarithm by converting to exponential form, as we have done previously, but recognizing and then applying the properties saves time.
Example 10.28
Evaluate using the properties of logarithms: ⓐ ${\text{log}}_{8}1$ and ⓑ ${\text{log}}_{6}6.$
Solution
ⓐ
$\begin{array}{cccccc}& & & \phantom{\rule{2em}{0ex}}{\text{log}}_{8}1\hfill & & \\ \text{Use the property,}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}1=0.\hfill & & & \phantom{\rule{3em}{0ex}}0\hfill & & \phantom{\rule{2em}{0ex}}{\text{log}}_{8}1=0\hfill \end{array}$
ⓑ
$\begin{array}{cccccc}& & & \phantom{\rule{2em}{0ex}}{\text{log}}_{6}6\hfill & & \\ \text{Use the property,}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}a=1.\hfill & & & \phantom{\rule{3em}{0ex}}1\hfill & & \phantom{\rule{2em}{0ex}}{\text{log}}_{6}6=1\hfill \end{array}$
Try It 10.55
Evaluate using the properties of logarithms: ⓐ ${\text{log}}_{13}1$ ⓑ ${\text{log}}_{9}9.$
Try It 10.56
Evaluate using the properties of logarithms: ⓐ ${\text{log}}_{5}1$ ⓑ ${\text{log}}_{7}7.$
The next two properties can also be verified by converting them from exponential form to logarithmic form, or the reverse.
The exponential equation ${a}^{{\text{log}}_{a}x}=x$ converts to the logarithmic equation ${\text{log}}_{a}x={\text{log}}_{a}x,$ which is a true statement for positive values for x only.
The logarithmic equation ${\text{log}}_{a}{a}^{x}=x$ converts to the exponential equation ${a}^{x}={a}^{x},$ which is also a true statement.
These two properties are called inverse properties because, when we have the same base, raising to a power “undoes” the log and taking the log “undoes” raising to a power. These two properties show the composition of functions. Both ended up with the identity function which shows again that the exponential and logarithmic functions are inverse functions.
Inverse Properties of Logarithms
For $a>0,$$x>0$ and $a\ne 1,$
In the next example, apply the inverse properties of logarithms.
Example 10.29
Evaluate using the properties of logarithms: ⓐ ${4}^{{\text{log}}_{4}9}$ and ⓑ ${\text{log}}_{3}{3}^{5}.$
Solution
ⓐ
$\begin{array}{cccc}& & & \phantom{\rule{2em}{0ex}}{4}^{{\text{log}}_{4}9}\hfill \\ \text{Use the property,}\phantom{\rule{0.2em}{0ex}}{a}^{{\text{log}}_{a}x}=x.\hfill & & & \phantom{\rule{3em}{0ex}}9\hfill & \phantom{\rule{2em}{0ex}}{4}^{{\text{log}}_{4}9}=9\hfill \end{array}$
ⓑ
$\begin{array}{cccc}& & & \phantom{\rule{2em}{0ex}}{\text{log}}_{3}{3}^{5}\hfill \\ \text{Use the property,}\phantom{\rule{0.2em}{0ex}}{a}^{{\text{log}}_{a}x}=x.\hfill & & & \phantom{\rule{3em}{0ex}}5\hfill & \phantom{\rule{2em}{0ex}}{\text{log}}_{3}{3}^{5}=5\hfill \end{array}$
Try It 10.57
Evaluate using the properties of logarithms: ⓐ ${5}^{{\text{log}}_{5}15}$ ⓑ ${\text{log}}_{7}{7}^{4}.$
Try It 10.58
Evaluate using the properties of logarithms: ⓐ ${2}^{{\text{log}}_{2}8}$ ⓑ ${\text{log}}_{2}{2}^{15}.$
There are three more properties of logarithms that will be useful in our work. We know exponential functions and logarithmic function are very interrelated. Our definition of logarithm shows us that a logarithm is the exponent of the equivalent exponential. The properties of exponents have related properties for exponents.
In the Product Property of Exponents, ${a}^{m}\xb7{a}^{n}={a}^{m+n},$ we see that to multiply the same base, we add the exponents. The Product Property of Logarithms, ${\text{log}}_{a}M\xb7N={\text{log}}_{a}M+{\text{log}}_{a}N$ tells us to take the log of a product, we add the log of the factors.
Product Property of Logarithms
If $M>0,N>0\text{,}\phantom{\rule{0.2em}{0ex}}\text{a}>0$ and $\text{a}\ne 1,$ then,
The logarithm of a product is the sum of the logarithms.
We use this property to write the log of a product as a sum of the logs of each factor.
Example 10.30
Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible: ⓐ ${\text{log}}_{3}7x$ and ⓑ ${\text{log}}_{4}64xy.$
Solution
ⓐ
$\begin{array}{cccc}& & & \phantom{\rule{2em}{0ex}}{\text{log}}_{3}7x\hfill \\ \text{Use the Product Property,}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}\left(M\xb7N\right)={\text{log}}_{a}M+{\text{log}}_{a}N.\hfill & & & \phantom{\rule{2em}{0ex}}{\text{log}}_{3}7+{\text{log}}_{3}x\hfill \\ \\ \\ & & & \phantom{\rule{2em}{0ex}}{\text{log}}_{3}7x={\text{log}}_{3}7+{\text{log}}_{3}x\hfill \end{array}$
ⓑ
$\begin{array}{cccc}& & & \phantom{\rule{2em}{0ex}}{\text{log}}_{4}64xy\hfill \\ \text{Use the Product Property,}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}\left(M\xb7N\right)={\text{log}}_{a}M+{\text{log}}_{a}N.\hfill & & & \phantom{\rule{2em}{0ex}}{\text{log}}_{4}64+{\text{log}}_{4}x+{\text{log}}_{4}y\hfill \\ \\ \\ \\ \text{Simplify by evaluating}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{4}64.\hfill & & & \phantom{\rule{2em}{0ex}}3+{\mathrm{log}}_{4}x+{\mathrm{log}}_{4}y\hfill \\ & & & \phantom{\rule{2em}{0ex}}{\text{log}}_{4}64xy=3+{\text{log}}_{4}x+{\text{log}}_{4}y\hfill \end{array}$
Try It 10.59
Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible.
ⓐ ${\text{log}}_{3}3x$ ⓑ ${\text{log}}_{2}8xy$
Try It 10.60
Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible.
ⓐ ${\text{log}}_{9}9x$ ⓑ ${\text{log}}_{3}27xy$
Similarly, in the Quotient Property of Exponents, $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n},$ we see that to divide the same base, we subtract the exponents. The Quotient Property of Logarithms, ${\text{log}}_{a}\frac{M}{N}={\text{log}}_{a}M-{\text{log}}_{a}N$ tells us to take the log of a quotient, we subtract the log of the numerator and denominator.
Quotient Property of Logarithms
If $M>0,N>0\text{,}\text{a}>0\phantom{\rule{0.2em}{0ex}}$ and $\text{a}\ne 1,$ then,
The logarithm of a quotient is the difference of the logarithms.
Note that ${\text{log}}_{a}M-{\text{log}}_{a}N\ne {\text{log}}_{a}(M-N).$
We use this property to write the log of a quotient as a difference of the logs of each factor.
Example 10.31
Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.
ⓐ ${\text{log}}_{5}\frac{5}{7}$ and ⓑ $\text{log}\frac{x}{100}$
Solution
ⓐ
$\begin{array}{cccc}& & & \phantom{\rule{2em}{0ex}}{\text{log}}_{5}\frac{5}{7}\hfill \\ \text{Use the Quotient Property,}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}\frac{M}{N}={\text{log}}_{a}M-{\text{log}}_{a}N.\hfill & & & \phantom{\rule{2em}{0ex}}{\text{log}}_{5}5-{\text{log}}_{5}7\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{2em}{0ex}}1-{\text{log}}_{5}7\hfill \\ \\ \\ & & & \phantom{\rule{2em}{0ex}}{\text{log}}_{5}\frac{5}{7}=1-{\text{log}}_{5}7\hfill \end{array}$
ⓑ
$\begin{array}{cccc}& & & \phantom{\rule{2em}{0ex}}\text{log}\frac{x}{100}\hfill \\ \text{Use the Quotient Property,}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}\frac{M}{N}={\text{log}}_{a}M-{\text{log}}_{a}N.\hfill & & & \phantom{\rule{2em}{0ex}}\text{log}\phantom{\rule{0.2em}{0ex}}x-\text{log}100\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{2em}{0ex}}\text{log}\phantom{\rule{0.2em}{0ex}}x-2\hfill \\ \\ \\ & & & \phantom{\rule{2em}{0ex}}\text{log}\frac{x}{100}=\text{log}\phantom{\rule{0.2em}{0ex}}x-2\hfill \end{array}$
Try It 10.61
Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.
ⓐ ${\text{log}}_{4}\frac{3}{4}$ ⓑ $\text{log}\frac{x}{1000}$
Try It 10.62
Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.
ⓐ ${\text{log}}_{2}\frac{5}{4}$ ⓑ $\text{log}\frac{10}{y}$
The third property of logarithms is related to the Power Property of Exponents, ${\left({a}^{m}\right)}^{n}={a}^{m\xb7n},$ we see that to raise a power to a power, we multiply the exponents. The Power Property of Logarithms, ${\text{log}}_{a}{M}^{p}=p\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}M$ tells us to take the log of a number raised to a power, we multiply the power times the log of the number.
Power Property of Logarithms
If $M>0,\phantom{\rule{0.2em}{0ex}}\text{a}>0,\phantom{\rule{0.2em}{0ex}}\text{a}\ne 1$ and $p$ is any real number then,
The log of a number raised to a power as the product product of the power times the log of the number.
We use this property to write the log of a number raised to a power as the product of the power times the log of the number. We essentially take the exponent and throw it in front of the logarithm.
Example 10.32
Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.
ⓐ ${\text{log}}_{5}{4}^{3}$ and ⓑ $\text{log}{x}^{10}$
Solution
ⓐ
$\begin{array}{cccc}& & & \phantom{\rule{2em}{0ex}}{\text{log}}_{5}{4}^{3}\hfill \\ \text{Use the Power Property,}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}{M}^{p}=p\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}M.\hfill & & & \phantom{\rule{2em}{0ex}}3{\text{log}}_{5}4\hfill \\ \\ \\ & & & \phantom{\rule{2em}{0ex}}{\text{log}}_{5}{4}^{3}=3{\text{log}}_{5}4\hfill \end{array}$
ⓑ
$\begin{array}{cccc}& & & \phantom{\rule{2em}{0ex}}\text{log}{x}^{10}\hfill \\ \text{Use the Power Property,}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}{M}^{p}=p\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}M.\hfill & & & \phantom{\rule{2em}{0ex}}10\phantom{\rule{0.2em}{0ex}}\text{log}\phantom{\rule{0.2em}{0ex}}x\hfill \\ \\ \\ & & & \phantom{\rule{2em}{0ex}}\text{log}{x}^{10}=10\text{log}\phantom{\rule{0.2em}{0ex}}x\hfill \end{array}$
Try It 10.63
Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.
ⓐ ${\text{log}}_{7}{5}^{4}$ ⓑ $\text{log}{x}^{100}$
Try It 10.64
Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.
ⓐ ${\text{log}}_{2}{3}^{7}$ ⓑ $\text{log}{x}^{20}$
We summarize the Properties of Logarithms here for easy reference. While the natural logarithms are a special case of these properties, it is often helpful to also show the natural logarithm version of each property.
Properties of Logarithms
If $M>0,\phantom{\rule{0.2em}{0ex}}N>0,\text{a}>0,\phantom{\rule{0.2em}{0ex}}\text{a}\ne 1$ and $p$ is any real number then,
Property | Base $a$ | Base $e$ |
---|---|---|
${\text{log}}_{a}1=0$ | $\text{ln}\phantom{\rule{0.2em}{0ex}}1=0$ | |
${\text{log}}_{a}a=1$ | $\text{ln}\phantom{\rule{0.2em}{0ex}}e=1$ | |
Inverse Properties | $\begin{array}{c}{a}^{{\text{log}}_{a}x}=x\hfill \\ {\text{log}}_{a}{a}^{x}=x\hfill \end{array}$ | $\begin{array}{c}{e}^{\text{ln}\phantom{\rule{0.2em}{0ex}}x}=x\hfill \\ \text{ln}\phantom{\rule{0.2em}{0ex}}{e}^{x}=x\hfill \end{array}$ |
Product Property of Logarithms | ${\text{log}}_{a}\left(M\xb7N\right)={\text{log}}_{a}M+{\text{log}}_{a}N$ | $\text{ln}\left(M\phantom{\rule{0.2em}{0ex}}\xb7\phantom{\rule{0.2em}{0ex}}N\right)=\text{ln}\phantom{\rule{0.2em}{0ex}}M+\text{ln}\phantom{\rule{0.2em}{0ex}}N$ |
Quotient Property of Logarithms | ${\text{log}}_{a}\frac{M}{N}={\text{log}}_{a}M-{\text{log}}_{a}N$ | $\text{ln}\phantom{\rule{0.2em}{0ex}}\frac{M}{N}=\text{ln}\phantom{\rule{0.2em}{0ex}}M-\text{ln}\phantom{\rule{0.2em}{0ex}}N$ |
Power Property of Logarithms | ${\text{log}}_{a}{M}^{p}=p\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}M$ | $\text{ln}\phantom{\rule{0.2em}{0ex}}{M}^{p}=p\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}M$ |
Now that we have the properties we can use them to “expand” a logarithmic expression. This means to write the logarithm as a sum or difference and without any powers.
We generally apply the Product and Quotient Properties before we apply the Power Property.
Example 10.33
Use the Properties of Logarithms to expand the logarithm ${\text{log}}_{4}\left(2{x}^{3}{y}^{2}\right)$. Simplify, if possible.
Solution
$\begin{array}{cccc}& & & \phantom{\rule{2em}{0ex}}{\text{log}}_{4}\left(2{x}^{3}{y}^{2}\right)\hfill \\ \text{Use the Product Property,}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}M\xb7N={\text{log}}_{a}M+{\text{log}}_{a}N.\hfill & & & \phantom{\rule{2em}{0ex}}{\text{log}}_{4}2+{\text{log}}_{4}{x}^{3}+{\text{log}}_{4}{y}^{2}\hfill \\ \text{Use the Power Property,}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}{M}^{p}=p\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}M,\phantom{\rule{0.2em}{0ex}}\text{on the last two terms.}\hfill & & & \phantom{\rule{2em}{0ex}}{\text{log}}_{4}2+3{\text{log}}_{4}x+2{\text{log}}_{4}y\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{2em}{0ex}}\frac{1}{2}+3{\text{log}}_{4}x+2{\text{log}}_{4}y\hfill \\ \\ \\ & & & \phantom{\rule{2em}{0ex}}{\text{log}}_{4}\left(2{x}^{3}{y}^{2}\right)=\frac{1}{2}+3{\text{log}}_{4}x+2{\text{log}}_{4}y\hfill \end{array}$
Try It 10.65
Use the Properties of Logarithms to expand the logarithm ${\text{log}}_{2}\left(5{x}^{4}{y}^{2}\right)$. Simplify, if possible.
Try It 10.66
Use the Properties of Logarithms to expand the logarithm ${\text{log}}_{3}\left(7{x}^{5}{y}^{3}\right)$. Simplify, if possible.
When we have a radical in the logarithmic expression, it is helpful to first write its radicand as a rational exponent.
Example 10.34
Use the Properties of Logarithms to expand the logarithm ${\text{log}}_{2}\sqrt[4]{\frac{{x}^{3}}{3{y}^{2}z}}$. Simplify, if possible.
Solution
$\begin{array}{cccc}& & & \phantom{\rule{2em}{0ex}}{\text{log}}_{2}\sqrt[4]{\frac{{x}^{3}}{3{y}^{2}z}}\hfill \\ \text{Rewrite the radical with a rational exponent.}\hfill & & & \phantom{\rule{2em}{0ex}}{\text{log}}_{2}{\left(\frac{{x}^{3}}{3{y}^{2}z}\right)}^{\frac{1}{4}}\hfill \\ \\ \\ \text{Use the Power Property,}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}{M}^{p}=p\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}M.\hfill & & & \phantom{\rule{2em}{0ex}}\frac{1}{4}{\text{log}}_{2}\left(\frac{{x}^{3}}{3{y}^{2}z}\right)\hfill \\ \\ \\ \text{Use the Quotient Property,}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}M\xb7N={\text{log}}_{a}M-{\text{log}}_{a}N.\hfill & & & \phantom{\rule{2em}{0ex}}\frac{1}{4}\left({\text{log}}_{2}\left({x}^{3}\right)-{\text{log}}_{2}\left(3{y}^{2}z\right)\right)\hfill \\ \\ \\ \begin{array}{c}\text{Use the Product Property,}\hfill \\ {\text{log}}_{a}M\xb7N={\text{log}}_{a}M+{\text{log}}_{a}N,\phantom{\rule{0.2em}{0ex}}\text{in the second term.}\hfill \end{array}\hfill & & & \phantom{\rule{2em}{0ex}}\frac{1}{4}\left({\text{log}}_{2}\left({x}^{3}\right)-\left({\text{log}}_{2}3+{\text{log}}_{2}{y}^{2}+{\text{log}}_{2}z\right)\right)\hfill \\ \\ \\ \begin{array}{c}\text{Use the Power Property,}\hfill \\ {\text{log}}_{a}{M}^{p}=p\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}M,\phantom{\rule{0.2em}{0ex}}\text{inside the parentheses.}\hfill \end{array}\hfill & & & \phantom{\rule{2em}{0ex}}\frac{1}{4}\left(3{\text{log}}_{2}x-\left({\text{log}}_{2}3+2{\text{log}}_{2}y+{\text{log}}_{2}z\right)\right)\hfill \\ \\ \\ \text{Simplify by distributing.}\hfill & & & \phantom{\rule{2em}{0ex}}\frac{1}{4}\left(3{\text{log}}_{2}x-{\text{log}}_{2}3-2{\text{log}}_{2}y-{\text{log}}_{2}z\right)\hfill \\ \\ \\ & & & {\text{log}}_{2}\sqrt[4]{\frac{{x}^{3}}{3{y}^{2}z}}=\frac{1}{4}\left(3{\text{log}}_{2}x-{\text{log}}_{2}3-2{\text{log}}_{2}y-{\text{log}}_{2}z\right)\hfill \end{array}$
Try It 10.67
Use the Properties of Logarithms to expand the logarithm ${\text{log}}_{4}\sqrt[5]{\frac{{x}^{4}}{2{y}^{3}{z}^{2}}}$. Simplify, if possible.
Try It 10.68
Use the Properties of Logarithms to expand the logarithm ${\text{log}}_{3}\sqrt[3]{\frac{{x}^{2}}{5{y}^{}z}}$. Simplify, if possible.
The opposite of expanding a logarithm is to condense a sum or difference of logarithms that have the same base into a single logarithm. We again use the properties of logarithms to help us, but in reverse.
To condense logarithmic expressions with the same base into one logarithm, we start by using the Power Property to get the coefficients of the log terms to be one and then the Product and Quotient Properties as needed.
Example 10.35
Use the Properties of Logarithms to condense the logarithm ${\text{log}}_{4}3+{\text{log}}_{4}x-{\text{log}}_{4}y$. Simplify, if possible.
Solution
$\begin{array}{cccc}\text{The log expressions all have the same base, 4.}\hfill & & & \phantom{\rule{6em}{0ex}}{\text{log}}_{4}3+{\text{log}}_{4}x-{\text{log}}_{4}y\hfill \\ \text{The first two terms are added, so we use the Product Property,}\hfill & & & \\ {\text{log}}_{a}M+{\text{log}}_{a}N={\text{log}}_{a}M\xb7N.\hfill & & & \phantom{\rule{6em}{0ex}}{\text{log}}_{4}3x-{\text{log}}_{4}y\hfill \\ \begin{array}{c}\text{Since the logs are subtracted, we use the Quotient Property,}\hfill \\ {\text{log}}_{a}M-{\text{log}}_{a}N={\text{log}}_{a}\frac{M}{N}.\hfill \end{array}\hfill & & & \phantom{\rule{6em}{0ex}}{\text{log}}_{4}\frac{3x}{y}\hfill \\ & & & \phantom{\rule{6em}{0ex}}{\text{log}}_{4}3+{\text{log}}_{4}x-{\text{log}}_{4}y={\text{log}}_{4}\frac{3x}{y}\hfill \end{array}$
Try It 10.69
Use the Properties of Logarithms to condense the logarithm ${\text{log}}_{2}5+{\text{log}}_{2}x-{\text{log}}_{2}y$. Simplify, if possible.
Try It 10.70
Use the Properties of Logarithms to condense the logarithm ${\text{log}}_{3}6-{\text{log}}_{3}x-{\text{log}}_{3}y$. Simplify, if possible.
Example 10.36
Use the Properties of Logarithms to condense the logarithm $2{\text{log}}_{3}x+4{\text{log}}_{3}\left(x+1\right)$. Simplify, if possible.
Solution
$\begin{array}{cccc}\text{The log expressions have the same base, 3.}\hfill & & & \phantom{\rule{2em}{0ex}}2{\text{log}}_{3}x+4{\text{log}}_{3}\left(x+1\right)\hfill \\ \text{Use the Power Property,}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}M+{\text{log}}_{a}N={\text{log}}_{a}M\xb7N.\hfill & & & \phantom{\rule{2em}{0ex}}{\text{log}}_{3}{x}^{2}+{\text{log}}_{3}{\left(x+1\right)}^{4}\hfill \\ \begin{array}{c}\text{The terms are added, so we use the Product}\hfill \\ \text{Property,}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}M+{\text{log}}_{a}N={\text{log}}_{a}M\xb7N.\hfill \end{array}\hfill & & & \phantom{\rule{2em}{0ex}}{\text{log}}_{3}{x}^{2}{\left(x+1\right)}^{4}\hfill \\ & & & \phantom{\rule{2em}{0ex}}2{\text{log}}_{3}x+4{\text{log}}_{3}\left(x+1\right)={\text{log}}_{3}{x}^{2}{\left(x+1\right)}^{4}\hfill \end{array}$
Try It 10.71
Use the Properties of Logarithms to condense the logarithm $3{\text{log}}_{2}x+2{\text{log}}_{2}\left(x-1\right)$. Simplify, if possible.
Try It 10.72
Use the Properties of Logarithms to condense the logarithm $2\text{log}\phantom{\rule{0.2em}{0ex}}x+2\text{log}\left(x+1\right)$. Simplify, if possible.
Use the Change-of-Base Formula
To evaluate a logarithm with any other base, we can use the Change-of-Base Formula. We will show how this is derived.
$\begin{array}{cccccc}\text{Suppose we want to evaluate}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}M.\hfill & & & \phantom{\rule{2em}{0ex}}{\text{log}}_{a}M\hfill & & \\ \text{Let}\phantom{\rule{0.2em}{0ex}}y={\text{log}}_{a}M.\hfill & & & \hfill \phantom{\rule{2em}{0ex}}y& =\hfill & {\text{log}}_{a}M\hfill \\ \text{Rewrite the expression in exponential form.}\hfill & & & \hfill \phantom{\rule{2em}{0ex}}{a}^{y}& =\hfill & M\hfill \\ \text{Take the}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{b}\phantom{\rule{0.2em}{0ex}}\text{of each side.}\hfill & & & \hfill \phantom{\rule{2em}{0ex}}{\text{log}}_{b}{a}^{y}& =\hfill & {\text{log}}_{b}M\hfill \\ \text{Use the Power Property.}\hfill & & & \hfill \phantom{\rule{2em}{0ex}}y{\text{log}}_{b}a& =\hfill & {\text{log}}_{b}M\hfill \\ \text{Solve for}\phantom{\rule{0.2em}{0ex}}y.\hfill & & & \hfill \phantom{\rule{2em}{0ex}}y& =\hfill & \frac{{\text{log}}_{b}M}{{\text{log}}_{b}a}\hfill \\ \text{Substitute}\phantom{\rule{0.2em}{0ex}}y={\text{log}}_{a}M.\hfill & & & \hfill \phantom{\rule{2em}{0ex}}{\text{log}}_{a}M& =\hfill & \frac{{\text{log}}_{b}M}{{\text{log}}_{b}a}\hfill \end{array}$
The Change-of-Base Formula introduces a new base $b.$ This can be any base b we want where $b>0,b\ne 1.$ Because our calculators have keys for logarithms base 10 and base e, we will rewrite the Change-of-Base Formula with the new base as 10 or e.
Change-of-Base Formula
For any logarithmic bases $a,b$ and $M>0,$
When we use a calculator to find the logarithm value, we usually round to three decimal places. This gives us an approximate value and so we use the approximately equal symbol $\text{(\u2248)}$.
Example 10.37
Rounding to three decimal places, approximate ${\text{log}}_{4}35.$
Solution
Use the Change-of-Base Formula. | |
Identify a and M. Choose 10 for b. | |
Enter the expression $\frac{\text{log}35}{\text{log}4}$ in the calculator using the log button for base 10. Round to three decimal places. |
Try It 10.73
Rounding to three decimal places, approximate ${\text{log}}_{3}42.$
Try It 10.74
Rounding to three decimal places, approximate ${\text{log}}_{5}46.$
Media
Access these online resources for additional instruction and practice with using the properties of logarithms.
Section 10.4 Exercises
Practice Makes Perfect
Use the Properties of Logarithms
In the following exercises, use the properties of logarithms to evaluate.
ⓐ ${\text{log}}_{4}1$ ⓑ ${\text{log}}_{8}8$
ⓐ ${3}^{{\text{log}}_{3}6}$ ⓑ ${\text{log}}_{2}{2}^{7}$
ⓐ ${8}^{{\text{log}}_{8}7}$ ⓑ ${\text{log}}_{6}{6}^{\mathrm{-2}}$
ⓐ ${10}^{\text{log}\sqrt{5}}$ ⓑ $\text{log}{10}^{\mathrm{-2}}$
ⓐ ${e}^{\text{ln}4}$ ⓑ $\text{ln}\phantom{\rule{0.2em}{0ex}}{e}^{2}$
In the following exercises, use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify if possible.
${\text{log}}_{4}6x$
${\text{log}}_{2}32xy$
$\text{log}100x$
In the following exercises, use the Quotient Property of Logarithms to write each logarithm as a sum of logarithms. Simplify if possible.
${\text{log}}_{3}\frac{3}{8}$
${\text{log}}_{4}\frac{16}{y}$
$\text{log}\frac{x}{10}$
$\text{ln}\frac{{e}^{3}}{3}$
In the following exercises, use the Power Property of Logarithms to expand each. Simplify if possible.
${\text{log}}_{3}{x}^{2}$
$\text{log}{x}^{\mathrm{-2}}$
${\text{log}}_{4}\sqrt{x}$
$\text{ln}\phantom{\rule{0.2em}{0ex}}{x}^{\sqrt{3}}$
In the following exercises, use the Properties of Logarithms to expand the logarithm. Simplify if possible.
${\text{log}}_{5}\left(4{x}^{6}{y}^{4}\right)$
${\text{log}}_{3}\left(\sqrt{2}{x}^{2}\right)$
${\text{log}}_{3}\frac{x{y}^{2}}{{z}^{2}}$
${\text{log}}_{4}\frac{\sqrt{x}}{16{y}^{4}}$
${\text{log}}_{2}\frac{\sqrt{2x+{y}^{2}}}{{z}^{2}}$
${\text{log}}_{2}\sqrt[4]{\frac{5{x}^{3}}{2{y}^{2}{z}^{4}}}$
In the following exercises, use the Properties of Logarithms to condense the logarithm. Simplify if possible.
${\text{log}}_{6}4+{\text{log}}_{6}9$
${\text{log}}_{2}80-{\text{log}}_{2}5$
${\text{log}}_{3}4+{\text{log}}_{3}\left(x+1\right)$
${\text{log}}_{7}3+{\text{log}}_{7}x-{\text{log}}_{7}y$
$4{\text{log}}_{2}x+6{\text{log}}_{2}y$
${\text{log}}_{3}\left({x}^{2}-1\right)-2{\text{log}}_{3}\left(x-1\right)$
$4\text{log}\phantom{\rule{0.2em}{0ex}}x-2\text{log}y-3\text{log}z$
$3\text{ln}\phantom{\rule{0.2em}{0ex}}x+4\text{ln}\phantom{\rule{0.2em}{0ex}}y-2\text{ln}\phantom{\rule{0.2em}{0ex}}z$
$\frac{1}{3}\text{log}\phantom{\rule{0.2em}{0ex}}x-3\text{log}\left(x+1\right)$
Use the Change-of-Base Formula
In the following exercises, use the Change-of-Base Formula, rounding to three decimal places, to approximate each logarithm.
${\text{log}}_{3}42$
${\text{log}}_{12}87$
${\text{log}}_{\sqrt{2}}17$
Writing Exercises
Write the Product Property in your own words. Does it apply to each of the following? ${\text{log}}_{a}5x,$${\text{log}}_{a}\left(5+x\right).$ Why or why not?
Write the Power Property in your own words. Does it apply to each of the following? ${\text{log}}_{a}{x}^{p},$${\left({\text{log}}_{a}x\right)}^{r}.$ Why or why not?
Use an example to show that
$\text{log}\left(a+b\right)\ne \text{log}a+\text{log}b.$
Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ On a scale of $1-10,$ how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?