# 4.3 Logarithmic functions  (Page 3/9)

 Page 3 / 9

Write the following exponential equations in logarithmic form.

1. ${3}^{2}=9$
2. ${5}^{3}=125$
3. ${2}^{-1}=\frac{1}{2}$
1. ${3}^{2}=9\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}{\mathrm{log}}_{3}\left(9\right)=2$
2. ${5}^{3}=125\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}{\mathrm{log}}_{5}\left(125\right)=3$
3. ${2}^{-1}=\frac{1}{2}\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}{\text{log}}_{2}\left(\frac{1}{2}\right)=-1$

## Evaluating logarithms

Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider $\text{\hspace{0.17em}}{\mathrm{log}}_{2}8.\text{\hspace{0.17em}}$ We ask, “To what exponent must $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ be raised in order to get 8?” Because we already know $\text{\hspace{0.17em}}{2}^{3}=8,$ it follows that $\text{\hspace{0.17em}}{\mathrm{log}}_{2}8=3.$

Now consider solving $\text{\hspace{0.17em}}{\mathrm{log}}_{7}49\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{\mathrm{log}}_{3}27\text{\hspace{0.17em}}$ mentally.

• We ask, “To what exponent must 7 be raised in order to get 49?” We know $\text{\hspace{0.17em}}{7}^{2}=49.\text{\hspace{0.17em}}$ Therefore, $\text{\hspace{0.17em}}{\mathrm{log}}_{7}49=2$
• We ask, “To what exponent must 3 be raised in order to get 27?” We know $\text{\hspace{0.17em}}{3}^{3}=27.\text{\hspace{0.17em}}$ Therefore, $\text{\hspace{0.17em}}{\mathrm{log}}_{3}27=3$

Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let’s evaluate $\text{\hspace{0.17em}}{\mathrm{log}}_{\frac{2}{3}}\frac{4}{9}\text{\hspace{0.17em}}$ mentally.

• We ask, “To what exponent must $\text{\hspace{0.17em}}\frac{2}{3}\text{\hspace{0.17em}}$ be raised in order to get $\text{\hspace{0.17em}}\frac{4}{9}?\text{\hspace{0.17em}}$ ” We know $\text{\hspace{0.17em}}{2}^{2}=4\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{3}^{2}=9,$ so $\text{\hspace{0.17em}}{\left(\frac{2}{3}\right)}^{2}=\frac{4}{9}.\text{\hspace{0.17em}}$ Therefore, $\text{\hspace{0.17em}}{\mathrm{log}}_{\frac{2}{3}}\left(\frac{4}{9}\right)=2.$

Given a logarithm of the form $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right),$ evaluate it mentally.

1. Rewrite the argument $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ as a power of $\text{\hspace{0.17em}}b:\text{\hspace{0.17em}}$ ${b}^{y}=x.\text{\hspace{0.17em}}$
2. Use previous knowledge of powers of $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ identify $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ by asking, “To what exponent should $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ be raised in order to get $\text{\hspace{0.17em}}x?$

## Solving logarithms mentally

Solve $\text{\hspace{0.17em}}y={\mathrm{log}}_{4}\left(64\right)\text{\hspace{0.17em}}$ without using a calculator.

First we rewrite the logarithm in exponential form: $\text{\hspace{0.17em}}{4}^{y}=64.\text{\hspace{0.17em}}$ Next, we ask, “To what exponent must 4 be raised in order to get 64?”

We know

${4}^{3}=64$

Therefore,

$\mathrm{log}{}_{4}\left(64\right)=3$

Solve $\text{\hspace{0.17em}}y={\mathrm{log}}_{121}\left(11\right)\text{\hspace{0.17em}}$ without using a calculator.

${\mathrm{log}}_{121}\left(11\right)=\frac{1}{2}\text{\hspace{0.17em}}$ (recalling that $\text{\hspace{0.17em}}\sqrt{121}={\left(121\right)}^{\frac{1}{2}}=11$ )

## Evaluating the logarithm of a reciprocal

Evaluate $\text{\hspace{0.17em}}y={\mathrm{log}}_{3}\left(\frac{1}{27}\right)\text{\hspace{0.17em}}$ without using a calculator.

First we rewrite the logarithm in exponential form: $\text{\hspace{0.17em}}{3}^{y}=\frac{1}{27}.\text{\hspace{0.17em}}$ Next, we ask, “To what exponent must 3 be raised in order to get $\text{\hspace{0.17em}}\frac{1}{27}?$

We know $\text{\hspace{0.17em}}{3}^{3}=27,$ but what must we do to get the reciprocal, $\text{\hspace{0.17em}}\frac{1}{27}?\text{\hspace{0.17em}}$ Recall from working with exponents that $\text{\hspace{0.17em}}{b}^{-a}=\frac{1}{{b}^{a}}.\text{\hspace{0.17em}}$ We use this information to write

$\begin{array}{l}{3}^{-3}=\frac{1}{{3}^{3}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{27}\hfill \end{array}$

Therefore, $\text{\hspace{0.17em}}{\mathrm{log}}_{3}\left(\frac{1}{27}\right)=-3.$

Evaluate $\text{\hspace{0.17em}}y={\mathrm{log}}_{2}\left(\frac{1}{32}\right)\text{\hspace{0.17em}}$ without using a calculator.

${\mathrm{log}}_{2}\left(\frac{1}{32}\right)=-5$

## Using common logarithms

Sometimes we may see a logarithm written without a base. In this case, we assume that the base is 10. In other words, the expression $\text{\hspace{0.17em}}\mathrm{log}\left(x\right)\text{\hspace{0.17em}}$ means $\text{\hspace{0.17em}}{\mathrm{log}}_{10}\left(x\right).\text{\hspace{0.17em}}$ We call a base-10 logarithm a common logarithm . Common logarithms are used to measure the Richter Scale mentioned at the beginning of the section. Scales for measuring the brightness of stars and the pH of acids and bases also use common logarithms.

## Definition of the common logarithm

A common logarithm    is a logarithm with base $\text{\hspace{0.17em}}10.\text{\hspace{0.17em}}$ We write $\text{\hspace{0.17em}}{\mathrm{log}}_{10}\left(x\right)\text{\hspace{0.17em}}$ simply as $\text{\hspace{0.17em}}\mathrm{log}\left(x\right).\text{\hspace{0.17em}}$ The common logarithm of a positive number $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ satisfies the following definition.

For $\text{\hspace{0.17em}}x>0,$

We read $\text{\hspace{0.17em}}\mathrm{log}\left(x\right)\text{\hspace{0.17em}}$ as, “the logarithm with base $\text{\hspace{0.17em}}10\text{\hspace{0.17em}}$ of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ ” or “log base 10 of $\text{\hspace{0.17em}}x.$

The logarithm $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ is the exponent to which $\text{\hspace{0.17em}}10\text{\hspace{0.17em}}$ must be raised to get $\text{\hspace{0.17em}}x.$

Given a common logarithm of the form $\text{\hspace{0.17em}}y=\mathrm{log}\left(x\right),$ evaluate it mentally.

1. Rewrite the argument $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ as a power of $\text{\hspace{0.17em}}10:\text{\hspace{0.17em}}$ ${10}^{y}=x.$
2. Use previous knowledge of powers of $\text{\hspace{0.17em}}10\text{\hspace{0.17em}}$ to identify $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ by asking, “To what exponent must $\text{\hspace{0.17em}}10\text{\hspace{0.17em}}$ be raised in order to get $\text{\hspace{0.17em}}x?$

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
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I'm not sure why it wrote it the other way
Salomon
I got X =-6
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ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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