Question

The function $f\left(x\right)={\log }_{9}x$is the logarithm function with base

____. So, width="46" height="21" role="math">f9=_____, $f\left(1\right)=$____, $f\left(\frac{1}{9}\right)=$____, $f\left(81\right)=$____, and $f\left(3\right)=$____.

Short answer

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

Step-by-step solution

Step 1. Given.

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

Step 2. To determine.

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

Step 3. Calculation.

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

Here, $a=9$.

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

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

width="92" height="80" role="math">f(9)=log99=log991=1

$$\begin{gathered} f\left(1\right)={\log }_{9}1 \\ ={\log }_9\left(9^0\right) \\ =0 \end{gathered}$$

$$\begin{gathered} f\left(\frac{1}{9}\right)={\log }_9\left(\frac{1}{9}\right) \\ ={\log }_9\left(9^{-1}\right) \\ =-1 \end{gathered}$$

$$\begin{gathered} f\left(81\right)={\log }_9\left(81\right) \\ ={\log }_9\left(9^2\right) \\ =2 \end{gathered}$$

$$\begin{gathered} f\left(3\right)={\log }_9\left(3\right) \\ ={\log }_9\left(9^{\frac{1}{2}}\right) \\ =\frac{1}{2} \end{gathered}$$

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