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

Jul 30, 2015

I found: $b = \pm 4$

Explanation:

You can start using a property of logs:
$\log x + \log y = \log x y$
so you get:
${\log}_{6} \left[2 \left({b}^{2} + 2\right)\right] = 2$
using the definition of log:
${\log}_{a} x = b \to x = {a}^{b}$ you get:
$2 \left({b}^{2} + 2\right) = {6}^{2} = 36$
${b}^{2} + 2 = \frac{36}{2} = 18$
${b}^{2} = 18 - 2 = 16$
so that $b = \pm \sqrt{16} = \pm 4$