# How do you solve log_3x+log_3(x-8)=2?

Mar 31, 2018

Thus, x=9 is the valid solution

#### Explanation:

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

$\log m + \log n = \log m n$

${\log}_{3} x + {\log}_{3} \left(x - 8\right) = {\log}_{3} x \left(x - 8\right)$

${\log}_{3} x \left(x - 8\right) = 2$

$x \left(x - 8\right) = {3}^{2}$

$x \left(x - 8\right) = 9$

${x}^{2} - 8 x = 9$

${x}^{2} - 8 x - 9 = 0$

${x}^{2} - 9 x + 1 x - 9 = 0$

$x \left(x - 9\right) + 1 \left(x - 9\right) = 0$

$\left(x + 1\right) \left(x - 9\right) = 0$

$x = - 1 , x = 9$

$x = - 1 , 9$

Here, x needs to be a positive number .

Thus, x=9 is the valid solution