## Precalculus (6th Edition) Blitzer

$3+5 \log_4 x$
Recall the log rules: (a) $\log_b{p \cdot q} =\log p + \log q$ (b) $\log{a^n}=n \log x$ (c) $\log_a{a^x}=x$ Use the rule (a) with $p=64$ and $q=x^5$ to obtain $\log_4 {64x^5}=\log_4 64+\log_4 x^5$ Simplify the second term by applying rules (b) and $(c)$ to obtain: $\log_4 4^3+\log_4 x^5= 3+5 \log_4 x$