# How do you solve log_3 x + log_3 (x - 8) = 2?

May 8, 2016

x = 9

#### Explanation:

Use ${\log}_{b} m + {\log}_{b} n = {\log}_{b} \left(m n\right)$

and inverse of $y = {\log}_{b} x$ is $x = {b}^{y}$

Here, ${\log}_{3} x + {\log}_{b} \left(x - 8\right) = {\log}_{3} \left(x \left(x - 8\right)\right) = 2$.

Inversely, $x \left(x - 8\right) = {3}^{2} = 9$.

So, ${x}^{2} - 8 x - 9 = \left(x + 1\right) \left(x - 9\right) = 0$.

$x = - 1 \mathmr{and} 9$. Negative x is inadmissible for ${\log}_{3} x$.