#### Answer

$\frac{1}{4}$

#### Work Step by Step

Rewrite the expression using exponents using the definition of logarithm.
$x^{-2}=16$
Raise both sides of the equation to the -1 power (find the reciprocals of both sides).
$x^2=\frac{1}{16}$
Find the square root of both sides.
$x=\pm\frac{1}{4}$
Rule out the solution $-\frac{1}{4}$ because the base of a logarithm must be greater than 0 and not equal to 1.
$x=\frac{1}{4}$