Answer
$2+\log_5{x}$
Work Step by Step
Recall:
(1) $\sqrt[m]{a}=a^{\frac{1}{m}}$
(2) $\log_a {x^n}=n\cdot \log_a {x}$.
(3) $\log_a{xy}=\log_a{x} +\log_a{y}$
(4) $\log_a{\frac{x}{y}}=\log_a{x} -\log_a{y}$
Use rule (3) above to obtain
$\log_5 {(25x)}=\log_5 {25}+\log_5 {x}.$
Use rule (2) to obtain
$\log_5 {25}+\log_5 {x}\\=\log_5 {5^2}+\log_5 {x}\\=2\cdot \log_5 {5}+\log_5 {x}\\=2+\log_5{x}$
