Precalculus (10th Edition)

$2+\log_5{x}$
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 $\log_5 {25}+\log_5 {x}\\=\log_5 {5^2}+\log_5 {x}\\=2\cdot \log_5 {5}+\log_5 {x}\\=2+\log_5{x}$