How do you solve log_4 8 - log_4(x+6)=1?

Oct 20, 2016

${\log}_{\textcolor{m a \ge n t a}{4}} \textcolor{red}{8} - {\log}_{\textcolor{m a \ge n t a}{4}} \textcolor{b l u e}{\left(x + 6\right)} = 1$

Condense the log by using the log property $\log \textcolor{red}{a} - \log \textcolor{b l u e}{b} = \log \left(\frac{\textcolor{red}{a}}{\textcolor{b l u e}{b}}\right)$

${\log}_{\textcolor{m a \ge n t a}{4}} \left(\frac{\textcolor{red}{8}}{\textcolor{b l u e}{x + 6}}\right) = 1$

${\log}_{\textcolor{m a \ge n t a}{b}} \textcolor{\mathmr{and} a n \ge}{x} = \textcolor{\lim e g r e e n}{a}$ becomes ${\textcolor{m a \ge n t a}{b}}^{\textcolor{\lim e g r e e n}{a}} = \textcolor{\mathmr{and} a n \ge}{x}$.

log_color(magenta)4 color(orange)((8/(x+6))color(red)=color(limegreen)1

${\textcolor{m a \ge n t a}{4}}^{\textcolor{\lim e g r e e n}{1}} = \textcolor{\mathmr{and} a n \ge}{\frac{8}{x + 6}}$

$4 = \frac{8}{x + 6}$

$4 \left(x + 6\right) = 8$

$4 x + 24 = 8$

$\frac{4 x}{4} = - \frac{16}{4}$

$x = - 4$