# How do you solve log_4 (x^2 - 9) - log_4 (x+3) = 3?

Dec 25, 2015

$x = 67$

#### Explanation:

$\textcolor{w h i t e}{\times} {\log}_{4} \left({x}^{2} - 9\right) - {\log}_{4} \left(x + 3\right) = 3$

The logarithm of the ratio of two numbers is the difference of the logarithms.
$\implies {\log}_{4} \left(\frac{{x}^{2} - 9}{x + 3}\right) = 3$

DOTS RULE:
An expression in the form of a squared number less another squared number is called the difference of two squares.
$\implies {\log}_{4} \left(\frac{\left(x - 3\right) \left(x + 3\right)}{x + 3}\right) = 3$

$\implies {\log}_{4} \left(x - 3\right) = 3$
$\implies {4}^{3} = x - 3$
$\implies x - 3 + 3 = 64 + 3$
$\implies x = 67$