# 4.5 Logarithmic properties  (Page 6/10)

 Page 6 / 10

Given a logarithm with the form $\text{\hspace{0.17em}}{\mathrm{log}}_{b}M,$ use the change-of-base formula to rewrite it as a quotient of logs with any positive base $\text{\hspace{0.17em}}n,$ where $\text{\hspace{0.17em}}n\ne 1.$

1. Determine the new base $\text{\hspace{0.17em}}n,$ remembering that the common log, $\text{\hspace{0.17em}}\mathrm{log}\left(x\right),$ has base 10, and the natural log, $\text{\hspace{0.17em}}\mathrm{ln}\left(x\right),$ has base $\text{\hspace{0.17em}}e.$
2. Rewrite the log as a quotient using the change-of-base formula
• The numerator of the quotient will be a logarithm with base $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ and argument $\text{\hspace{0.17em}}M.$
• The denominator of the quotient will be a logarithm with base $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ and argument $\text{\hspace{0.17em}}b.$

## Changing logarithmic expressions to expressions involving only natural logs

Change $\text{\hspace{0.17em}}{\mathrm{log}}_{5}3\text{\hspace{0.17em}}$ to a quotient of natural logarithms.

Because we will be expressing $\text{\hspace{0.17em}}{\mathrm{log}}_{5}3\text{\hspace{0.17em}}$ as a quotient of natural logarithms, the new base, $\text{\hspace{0.17em}}n=e.$

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.

$\begin{array}{ll}{\mathrm{log}}_{b}M\hfill & =\frac{\mathrm{ln}M}{\mathrm{ln}b}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\mathrm{log}}_{5}3\hfill & =\frac{\mathrm{ln}3}{\mathrm{ln}5}\hfill \end{array}$

Change $\text{\hspace{0.17em}}{\mathrm{log}}_{0.5}8\text{\hspace{0.17em}}$ to a quotient of natural logarithms.

$\frac{\mathrm{ln}8}{\mathrm{ln}0.5}$

Can we change common logarithms to natural logarithms?

Yes. Remember that $\text{\hspace{0.17em}}\mathrm{log}9\text{\hspace{0.17em}}$ means $\text{\hspace{0.17em}}{\text{log}}_{\text{10}}\text{9}.$ So, $\text{\hspace{0.17em}}\mathrm{log}9=\frac{\mathrm{ln}9}{\mathrm{ln}10}.$

## Using the change-of-base formula with a calculator

Evaluate $\text{\hspace{0.17em}}{\mathrm{log}}_{2}\left(10\right)\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}e.$

Evaluate $\text{\hspace{0.17em}}{\mathrm{log}}_{5}\left(100\right)\text{\hspace{0.17em}}$ using the change-of-base formula.

$\frac{\mathrm{ln}100}{\mathrm{ln}5}\approx \frac{4.6051}{1.6094}=2.861$

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

## Key equations

 The Product Rule for Logarithms ${\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)$ The Quotient Rule for Logarithms ${\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}M-{\mathrm{log}}_{b}N$ The Power Rule for Logarithms ${\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M$ The Change-of-Base Formula

## 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 $\text{\hspace{0.17em}}e\text{\hspace{0.17em}}$ as the quotient of natural or common logs. That way a calculator can be used to evaluate. See [link] .

## Verbal

How does the power rule for logarithms help when solving logarithms with the form $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(\sqrt[n]{x}\right)?$

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, $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left({x}^{\frac{1}{n}}\right)=\frac{1}{n}{\mathrm{log}}_{b}\left(x\right).$

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.

${\mathrm{log}}_{b}\left(7x\cdot 2y\right)$

${\mathrm{log}}_{b}\left(2\right)+{\mathrm{log}}_{b}\left(7\right)+{\mathrm{log}}_{b}\left(x\right)+{\mathrm{log}}_{b}\left(y\right)$

$\mathrm{ln}\left(3ab\cdot 5c\right)$

${\mathrm{log}}_{b}\left(\frac{13}{17}\right)$

${\mathrm{log}}_{b}\left(13\right)-{\mathrm{log}}_{b}\left(17\right)$

$\mathrm{ln}\left(\frac{1}{{4}^{k}}\right)$

$-k\mathrm{ln}\left(4\right)$

${\mathrm{log}}_{2}\left({y}^{x}\right)$

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

$\mathrm{ln}\left(7\right)+\mathrm{ln}\left(x\right)+\mathrm{ln}\left(y\right)$

$\mathrm{ln}\left(7xy\right)$

${\mathrm{log}}_{3}\left(2\right)+{\mathrm{log}}_{3}\left(a\right)+{\mathrm{log}}_{3}\left(11\right)+{\mathrm{log}}_{3}\left(b\right)$

${\mathrm{log}}_{b}\left(28\right)-{\mathrm{log}}_{b}\left(7\right)$

${\mathrm{log}}_{b}\left(4\right)$

$\mathrm{ln}\left(a\right)-\mathrm{ln}\left(d\right)-\mathrm{ln}\left(c\right)$

$-{\mathrm{log}}_{b}\left(\frac{1}{7}\right)$

${\text{log}}_{b}\left(7\right)$

$\frac{1}{3}\mathrm{ln}\left(8\right)$

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.

$\mathrm{log}\left(\frac{{x}^{15}{y}^{13}}{{z}^{19}}\right)$

$15\mathrm{log}\left(x\right)+13\mathrm{log}\left(y\right)-19\mathrm{log}\left(z\right)$

$\mathrm{ln}\left(\frac{{a}^{-2}}{{b}^{-4}{c}^{5}}\right)$

$\mathrm{log}\left(\sqrt{{x}^{3}{y}^{-4}}\right)$

$\frac{3}{2}\mathrm{log}\left(x\right)-2\mathrm{log}\left(y\right)$

$\mathrm{ln}\left(y\sqrt{\frac{y}{1-y}}\right)$

$\mathrm{log}\left({x}^{2}{y}^{3}\sqrt[3]{{x}^{2}{y}^{5}}\right)$

$\frac{8}{3}\mathrm{log}\left(x\right)+\frac{14}{3}\mathrm{log}\left(y\right)$

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

$\mathrm{log}\left(2{x}^{4}\right)+\mathrm{log}\left(3{x}^{5}\right)$

$\mathrm{ln}\left(6{x}^{9}\right)-\mathrm{ln}\left(3{x}^{2}\right)$

$\mathrm{ln}\left(2{x}^{7}\right)$

$2\mathrm{log}\left(x\right)+3\mathrm{log}\left(x+1\right)$

$\mathrm{log}\left(x\right)-\frac{1}{2}\mathrm{log}\left(y\right)+3\mathrm{log}\left(z\right)$

$\mathrm{log}\left(\frac{x{z}^{3}}{\sqrt{y}}\right)$

$4{\mathrm{log}}_{7}\left(c\right)+\frac{{\mathrm{log}}_{7}\left(a\right)}{3}+\frac{{\mathrm{log}}_{7}\left(b\right)}{3}$

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

${\mathrm{log}}_{7}\left(15\right)\text{\hspace{0.17em}}$ to base $\text{\hspace{0.17em}}e$

${\mathrm{log}}_{7}\left(15\right)=\frac{\mathrm{ln}\left(15\right)}{\mathrm{ln}\left(7\right)}$

${\mathrm{log}}_{14}\left(55.875\right)\text{\hspace{0.17em}}$ to base $\text{\hspace{0.17em}}10$

For the following exercises, suppose $\text{\hspace{0.17em}}{\mathrm{log}}_{5}\left(6\right)=a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{\mathrm{log}}_{5}\left(11\right)=b.\text{\hspace{0.17em}}$ Use the change-of-base formula along with properties of logarithms to rewrite each expression in terms of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b.\text{\hspace{0.17em}}$ Show the steps for solving.

${\mathrm{log}}_{11}\left(5\right)$

${\mathrm{log}}_{11}\left(5\right)=\frac{{\mathrm{log}}_{5}\left(5\right)}{{\mathrm{log}}_{5}\left(11\right)}=\frac{1}{b}$

${\mathrm{log}}_{6}\left(55\right)$

${\mathrm{log}}_{11}\left(\frac{6}{11}\right)$

${\mathrm{log}}_{11}\left(\frac{6}{11}\right)=\frac{{\mathrm{log}}_{5}\left(\frac{6}{11}\right)}{{\mathrm{log}}_{5}\left(11\right)}=\frac{{\mathrm{log}}_{5}\left(6\right)-{\mathrm{log}}_{5}\left(11\right)}{{\mathrm{log}}_{5}\left(11\right)}=\frac{a-b}{b}=\frac{a}{b}-1$

## Numeric

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

${\mathrm{log}}_{3}\left(\frac{1}{9}\right)-3{\mathrm{log}}_{3}\left(3\right)$

$6{\mathrm{log}}_{8}\left(2\right)+\frac{{\mathrm{log}}_{8}\left(64\right)}{3{\mathrm{log}}_{8}\left(4\right)}$

$3$

$2{\mathrm{log}}_{9}\left(3\right)-4{\mathrm{log}}_{9}\left(3\right)+{\mathrm{log}}_{9}\left(\frac{1}{729}\right)$

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.

${\mathrm{log}}_{3}\left(22\right)$

$2.81359$

${\mathrm{log}}_{8}\left(65\right)$

${\mathrm{log}}_{6}\left(5.38\right)$

$0.93913$

${\mathrm{log}}_{4}\left(\frac{15}{2}\right)$

${\mathrm{log}}_{\frac{1}{2}}\left(4.7\right)$

$-2.23266$

## Extensions

Use the product rule for logarithms to find all $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values such that $\text{\hspace{0.17em}}{\mathrm{log}}_{12}\left(2x+6\right)+{\mathrm{log}}_{12}\left(x+2\right)=2.\text{\hspace{0.17em}}$ Show the steps for solving.

Use the quotient rule for logarithms to find all $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values such that $\text{\hspace{0.17em}}{\mathrm{log}}_{6}\left(x+2\right)-{\mathrm{log}}_{6}\left(x-3\right)=1.\text{\hspace{0.17em}}$ Show the steps for solving.

$x=4;\text{\hspace{0.17em}}$ By the quotient rule: ${\mathrm{log}}_{6}\left(x+2\right)-{\mathrm{log}}_{6}\left(x-3\right)={\mathrm{log}}_{6}\left(\frac{x+2}{x-3}\right)=1.$

Rewriting as an exponential equation and solving for $\text{\hspace{0.17em}}x:$

$\begin{array}{ll}{6}^{1}\hfill & =\frac{x+2}{x-3}\hfill \\ \text{\hspace{0.17em}}0\hfill & =\frac{x+2}{x-3}-6\hfill \\ \text{\hspace{0.17em}}0\hfill & =\frac{x+2}{x-3}-\frac{6\left(x-3\right)}{\left(x-3\right)}\hfill \\ \text{\hspace{0.17em}}0\hfill & =\frac{x+2-6x+18}{x-3}\hfill \\ \text{\hspace{0.17em}}0\hfill & =\frac{x-4}{x-3}\hfill \\ \text{​}\text{\hspace{0.17em}}x\hfill & =4\hfill \end{array}$

Checking, we find that $\text{\hspace{0.17em}}{\mathrm{log}}_{6}\left(4+2\right)-{\mathrm{log}}_{6}\left(4-3\right)={\mathrm{log}}_{6}\left(6\right)-{\mathrm{log}}_{6}\left(1\right)\text{\hspace{0.17em}}$ is defined, so $\text{\hspace{0.17em}}x=4.$

Can the power property of logarithms be derived from the power property of exponents using the equation $\text{\hspace{0.17em}}{b}^{x}=m?\text{\hspace{0.17em}}$ If not, explain why. If so, show the derivation.

Prove that $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(n\right)=\frac{1}{{\mathrm{log}}_{n}\left(b\right)}\text{\hspace{0.17em}}$ for any positive integers $\text{\hspace{0.17em}}b>1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}n>1.$

Let $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ be positive integers greater than $\text{\hspace{0.17em}}1.\text{\hspace{0.17em}}$ Then, by the change-of-base formula, $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(n\right)=\frac{{\mathrm{log}}_{n}\left(n\right)}{{\mathrm{log}}_{n}\left(b\right)}=\frac{1}{{\mathrm{log}}_{n}\left(b\right)}.$

Does $\text{\hspace{0.17em}}{\mathrm{log}}_{81}\left(2401\right)={\mathrm{log}}_{3}\left(7\right)?\text{\hspace{0.17em}}$ Verify the claim algebraically.

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