# How do you solve log_6(b^2+2) + log_6 2=2?

Jun 20, 2015

$b = \pm 4$

#### Explanation:

Start by usig the following property of logs:

${\log}_{a} b + {\log}_{a} c = {\log}_{a} \left(b \cdot c\right)$
so you get:

${\log}_{6} \left[2 \left({b}^{2} + 2\right)\right] = 2$

now use the definition of log:
${\log}_{a} x = b \to x = {a}^{b}$

$2 \left({b}^{2} + 2\right) = {6}^{2}$
${b}^{2} + 2 = 18$
${b}^{2} = 16$
$b = \pm \sqrt{16} = \pm 4$