Question

Find $\text{'}a\text{'}$so that the graph of$f\left(x\right)={\log }_{a}x$ contains the point $(\frac{1}{2},-4)$

Short answer

The value of$\text{'}a\text{'}$is$-4$

Step-by-step solution

Step 1. Given information 

Given function is$f\left(x\right)={\log }_a\left(x\right)$

Step 2. Omitting logaritm 

Omitting logarithm, we have,

$x=a^y$

Now, the function holds the point $(\frac{1}{2},-4)$

Hence, putting the point to the function we have,

$\frac{1}{2}=a^{-4}$

Simplifying, we have,

$a=2^{\frac{1}{4}}$

Step 3. Check

Now, for$a=2^{\frac{1}{4}}$

we have,

$$\begin{gathered} {\log }_{2^{\frac{1}{4}}}\left(\frac{1}{2}\right) \\ =-4 \end{gathered}$$