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Q2.

Expert-verified
Found in: Page 351

Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508

The function $f\left(x\right)={\mathrm{log}}_{9}x$ is the logarithm function with base____. So, width="46" height="21" role="math" style="max-width: none; vertical-align: -5px;" localid="1648528874754" $f\left(9\right)=$ _____, $f\left(1\right)=$ ____, $f\left(\frac{1}{9}\right)=$ ____, $f\left(81\right)=$ ____, and $f\left(3\right)=$ ____.

The function $f\left(x\right)={\mathrm{log}}_{9}x$ is the logarithm function with base 9. So, $f\left(9\right)=1,\text{\hspace{0.17em}\hspace{0.17em}}f\left(1\right)=0,\text{\hspace{0.17em}\hspace{0.17em}}f\left(\frac{1}{9}\right)=-1,\text{\hspace{0.17em}\hspace{0.17em}}f\left(81\right)=2,\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}f\left(3\right)=\frac{1}{2}$.

See the step by step solution

Step 1. Given.

The given function is $f\left(x\right)={\mathrm{log}}_{9}x$.

Step 2. To determine.

We have to find $f\left(9\right),\text{\hspace{0.17em}\hspace{0.17em}}f\left(1\right),\text{\hspace{0.17em}\hspace{0.17em}}f\left(\frac{1}{9}\right),\text{\hspace{0.17em}\hspace{0.17em}}f\left(81\right),\text{\hspace{0.17em}\hspace{0.17em}}f\left(3\right)$.

Step 3. Calculation.

We compare $f\left(x\right)={\mathrm{log}}_{9}x$ with $f\left(x\right)={\mathrm{log}}_{a}x$.

Here, $a=9$.

So, this given function $f\left(x\right)={\mathrm{log}}_{9}x$ has base $=9$.

We’ll use the formula: $\mathrm{log}\left({9}^{k}\right)=k$.

width="92" height="80" role="math" style="max-width: none; vertical-align: -41px;" localid="1648529347686" $f\left(9\right)={\mathrm{log}}_{9}9\phantom{\rule{0ex}{0ex}}={\mathrm{log}}_{9}\left({9}^{1}\right)\phantom{\rule{0ex}{0ex}}=1$

$f\left(1\right)={\mathrm{log}}_{9}1\phantom{\rule{0ex}{0ex}}={\mathrm{log}}_{9}\left({9}^{0}\right)\phantom{\rule{0ex}{0ex}}=0$

$f\left(\frac{1}{9}\right)={\mathrm{log}}_{9}\left(\frac{1}{9}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{log}}_{9}\left({9}^{-1}\right)\phantom{\rule{0ex}{0ex}}=-1$

$f\left(81\right)={\mathrm{log}}_{9}\left(81\right)\phantom{\rule{0ex}{0ex}}={\mathrm{log}}_{9}\left({9}^{2}\right)\phantom{\rule{0ex}{0ex}}=2$

$f\left(3\right)={\mathrm{log}}_{9}\left(3\right)\phantom{\rule{0ex}{0ex}}={\mathrm{log}}_{9}\left({9}^{\frac{1}{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Hence, the required values are $f\left(9\right)=1,\text{\hspace{0.17em}\hspace{0.17em}}f\left(1\right)=0,\text{\hspace{0.17em}\hspace{0.17em}}f\left(\frac{1}{9}\right)=-1,\text{\hspace{0.17em}\hspace{0.17em}}f\left(81\right)=2,\text{\hspace{0.17em}\hspace{0.17em}}f\left(3\right)=\frac{1}{2}$.