## College Algebra (11th Edition)

$\frac{1}{4}$
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. $x=\frac{1}{4}$