# How do you solve 3^(0.1x)-4=5?

Oct 30, 2016

$x = 20$

#### Explanation:

${3}^{0.1 x} - 4 = 5$

Add $4$ to both sides.

${3}^{0.1 x} = 9$

Take the log of both sides.

$\log {3}^{0.1 x} = \log 9$

Use the log rule $\log {x}^{a} = a \log x$

$0.1 x \log 3 = \log 9$

Divide both sides by $0.1 \log 3$

$x = \log \frac{9}{0.1 \log 3} = \frac{10 \log 9}{\log} 3$

Use the change of base formula ${\log}_{b} a = \log \frac{a}{\log} b$

$x = 10 \cdot {\log}_{3} 9$

${\log}_{3} 9 = \textcolor{red}{2}$ because ${3}^{\textcolor{red}{2}} = 9$

$x = 10 \cdot \textcolor{red}{2} = 20$