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Write the following exponential equations in logarithmic form.
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.
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.
Given a logarithm of the form $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right),$ evaluate it 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
Therefore,
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 $\text{\hspace{0.17em}}\sqrt{121}={(121)}^{\frac{1}{2}}=11$ )
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
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.
${\mathrm{log}}_{2}\left(\frac{1}{32}\right)=-5$
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.
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.
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