## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\text{Exact: } \dfrac{\ln 1.2}{\ln 8}$ $\text{Approximately: } -0.088$
$\because a^y = b \text{ is equivalent to } y = \log_a b$ $\therefore 8^{-x}=1.2 \text{ is equivalent to } -x = \log_8 1.2$ Solve the equation above using the Change of Base Formula, which is $\hspace{20pt} \log_ a M = \dfrac{\log_b M}{\log_b a}$, to obtain: $-x=\log_8 1.2 \\\\ -x= \dfrac{\ln 1.2}{\ln 8}\\\\ x= -\dfrac{\ln 1.2}{\ln 8}$ Therefore, $x = \boxed{-\dfrac{\ln 1.2}{\ln 8} \approx -0.088}$