Answer
$a=\sqrt[4]2.$
Work Step by Step
We know that if the graph of $f(x)=\log_a{x}$ contains the point $\left(\frac{1}{2},-4\right)$, then $-4=\log_a{\frac{1}{2}}$.
We know that if $y=\log_a{b}$, then $a^y=b$.
Hence here: $a^{-4}=\frac{1}{2}\\a^4=2\\a=\pm\sqrt[4]2.$
But the base of a logarithm can only be positive, hence $a=\sqrt[4]2.$
