# How do you solve log_4 (x + 2) - log_4 (x - 4) = log_4 3?

Feb 12, 2016

${\log}_{4} \left(x + 2\right) - {\log}_{4} \left(x - 4\right) = {\log}_{4} \left(3\right)$

First calculate the dominion:

$x + 2 > 0 \mathmr{and} x - 4 > 0 \to x > - 2 \mathmr{and} x > 4 \implies x > 4$

Now add log_4(x-4) to both sides of the equality:

${\log}_{4} \left(x + 2\right) = {\log}_{4} \left(x - 4\right) + {\log}_{4} \left(3\right)$

We know that log(a)+log(b)=log(ab), so:

${\log}_{4} \left(x + 2\right) = {\log}_{4} \left(3 \left(x - 4\right)\right) = {\log}_{4} \left(3 x - 12\right)$

If a>0, and b>0, log(a)=log(b) => a=b

$x + 2 = 3 x - 12$

$- 2 x = - 14$

$x = 7$ (x>4, so the solution is valid)