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

Dec 6, 2015

$x = 8$

#### Explanation:

To solve this kind of equations, the strategy is to manipulate the expressions in order to arrive at something like

$\log \left(X\right) = \log \left(Y\right)$

To deduce that $X = Y$, since the logarithm function is injective.

So, the only thing we need to do in this case it to remember that

$\log \left(a\right) - \log \left(b\right) = \log \left(\frac{a}{b}\right)$

So you have that

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

Which is true if and only if

$\frac{x + 4}{x - 4} = 3$

If $x \setminus \ne 4$, we can multiply by $x - 4$ both terms and get

$x + 4 = 3 x - 12 \setminus \to 2 x = 16 \setminus \to x = 8$