## Intermediate Algebra for College Students (7th Edition)

$x=5$
RECALL: $a \cdot \log_b{x} = \log_b{(x^a)}$ Use the rule above to obtain: $\log{(x^2)}=\log{25}$ Use the rule "$\log_b{a} =\log_b{c} \longrightarrow a = c$" to obtain: $x^2=25$ Take the square root of both sides to obtain: $x=\pm \sqrt{25} \\x = \pm 5$ Note, however, that in $y=\log_b{x}$, $x \gt 0$. This means that the solution to the given equation cannot be $-5$. Thus, the solution to the given equation is $x=5$.