# How do you solve log_ 4 (x – 6) + log _ 4 x = 2?

Jan 13, 2016

Step by step explanation is given below.

#### Explanation:

${\log}_{4} \left(x - 6\right) + {\log}_{4} x = 2$

Step 1: Condense the left-hand side using rule
${\log}_{b} \left(P\right) + {\log}_{b} \left(Q\right) = {\log}_{b} \left(P Q\right)$

Step 2: Convert logarithms to exponent form using the rule
${\log}_{b} \left(a\right) = k \to a = {b}^{k}$

Now let us solve our problem

${\log}_{4} \left(x - 6\right) + {\log}_{4} \left(x\right) = 2$

Step 1: Condensing...

${\log}_{4} \left(x - 6\right) x = 2$
${\log}_{4} \left({x}^{2} - 6 x\right) = 2$

Step 2: Converting to exponent form.

${x}^{2} - 6 x = {2}^{4}$

$\implies {x}^{2} - 6 x = 16$

$\implies {x}^{2} - 6 x - 16 = 0$

$\implies \left(x - 8\right) \left(x + 2\right) = 0$

$x - 8 = 0$ or $x + 2 = 0$

$x = 8 \mathmr{and} x = - 2$

$x = - 2$ is not possible as ${\log}_{4} \left(- 2\right)$ is not defined.

The solution is $x = 8$