## Precalculus (10th Edition)

$a=\sqrt2.$
We know that if the graph of $f(x)=\log_a{x}$ contains the point $(2,2)$, then $2=\log_a{2}$. We know that if $y=\log_a{b}$, then $a^y=b$. Hence here: $a^2=2\\a=\pm\sqrt2.$ But the base of a logarithm can only be positive, hence $a=\sqrt2.$