Answer
$-\dfrac{1}{2}$
Work Step by Step
RECALL:
$\log_b{b^x} = x.$
Note that
$\dfrac{1}{\sqrt{2}}=\dfrac{1}{2^{\frac{1}{2}}}=2^{-\frac{1}{2}}.$
Thus, the given expression can be written as
$\log_2{2^{-\frac{1}{2}}}.$
Use the rule above to obtain
$-\dfrac{1}{2}.$
