## College Algebra (10th Edition)

$x=2$
RECALL: $\log_a{x} = y \longrightarrow a^y=x, a\gt0, a\ne 1$ Use the definition above to obtain: $\log_x{4}=2 \\ \longrightarrow x^2=4$ Take the square root of both sides to obtain: $x = \pm \sqrt{4} \\x = \pm \sqrt{2^2} \\x = \pm 2$ Since the base of a logarithm must be positive, $x$ cannot be $-2$. Thus, the only solution to the given equation is $x=2$.