# How do you solve log_(4) (x^2 - 4) - log_(4) (x + 2) = 2?

Jun 18, 2015

$x = 18$

#### Explanation:

For starters, let's determine the set of real numbers where this equation makes sense. It's necessary because we might engage in non-invariant transformations of this equation, which might produce extra solutions or we might lose certain solutions in the course of transformation. Even if we will use only invariant transformations, it's still a good practice to determine what kind of solutions our equation allows.

Any logarithm with a base 4 (as well as with any other positive base) is defined only for positive arguments. That why we have restrictions:
${x}^{2} - 4 > 0$ and
$x + 2 > 0$

The first condition is equivalent to ${x}^{2} > 4$ or $| x | > 2$ or a combination of two inequalities:
(1) $x < - 2$ or
(2) $x > 2$

The second condition is equivalent to
(3) $x > - 2$

The only condition that satisfies condition (3) AND either (1) or (2) above is
(4) $x > 2$

Now let's examine the problem at hand.
Recall that
${\log}_{c} \left(a \cdot b\right) = {\log}_{c} \left(a\right) + {\log}_{c} \left(b\right)$

Applying this to our equation and using an identity ${x}^{2} - 4 = \left(x - 2\right) \cdot \left(x + 2\right)$, we can rewrite it as
${\log}_{4} \left(x - 2\right) + {\log}_{4} \left(x + 2\right) - {\log}_{4} \left(x + 2\right) = 2$
or
(5) ${\log}_{4} \left(x - 2\right) = 2$

By definition of a logarithm, ${c}^{{\log}_{c} \left(d\right)} = d$
Applied this to our problem, it means that, by definition of logarithms,
${4}^{{\log}_{4} \left(x - 2\right)} = x - 2$

Since equation (5) is true, we can write it as
${4}^{2} = x - 2$ or
$16 = x - 2$ or
$x = 18$