Chapter 5 - Exponential and Logarithmic Functions - 5.4 Logarithmic Functions - 5.4 Assess Your Understanding - Page 296: 100

$x=-\frac{1}{5}$

We know that by definition then $\log_a {y}=x$, then $y=a^x$. Hence, if $\log_6 (36)=5x+3$, then $6^{5x+3}=36 \\6^{5x+3}=6^2$. Solve the equation above using the rule $a^x=a^y \longrightarrow x=y$ to obtain: \begin{align*}2=5x+3\end{align*}\begin{align*}-1=5x\end{align*}\begin{align*}x=-\frac{1}{5}\end{align*}