How do you solve log_4 (m + 2) - log_4(m - 5) = log_4 8?

Apr 28, 2016

$m = 6$

Explanation:

Given,

${\log}_{4} \left(m + 2\right) - {\log}_{4} \left(m - 5\right) = {\log}_{4} \left(8\right)$

We can simplify the left side of the equation using the logarithmic property, ${\log}_{\textcolor{p u r p \le}{b}} \left(\frac{\textcolor{red}{m}}{\textcolor{b l u e}{n}}\right) = {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m}\right) - {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{b l u e}{n}\right)$.

${\log}_{4} \left(\frac{m + 2}{m - 5}\right) = {\log}_{4} \left(8\right)$

Since the equation now follows a "$\log = \log$" situation where the bases are the same on both sides, rewrite the equation without the "$\log$" portion.

$\frac{m + 2}{m - 5} = 8$

Solve for $m$.

$m + 2 = 8 \left(m - 5\right)$

$m + 2 = 8 m - 40$

$7 m = 42$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} m = 6 \textcolor{w h i t e}{\frac{a}{a}} |}}}$