Question

Find$a$so that the graph of$f\left(x\right)={\log }_{a}x$contains the point$(2,2)$

Short answer

The value of$\text{'}a\text{'}$is$\sqrt{2}$

Step-by-step solution

Step 1. Given information

Given function is $y={\log }_{a}x$

Step 2. Omitting logaritm

Omitting logarithm, we have,

$x=a^y$

Now, the function holds the point $(2,2)$

Hence, putting the point to the function we have,

$2=a^2$

By square rooting both sides, we have,

$a=\sqrt{2}$

Step 3. Check

Now, for $a=\sqrt{2}$,

we have,

$$\begin{gathered} {\log }_{\sqrt{2}}2 \\ ={\log }_{\sqrt{2}}{\left(\sqrt{2}\right)}^2=2\times {\log }_{\sqrt{2}}\sqrt{2} \\ =2\times 1 \\ =2 \end{gathered}$$