# How do you solve log_6 3x+ log_6 (x-1) =3 and find any extraneous solutions?

May 21, 2016

x = 9 and the extraneous solution is$- 8$.

#### Explanation:

${\log}_{6} \left(3 x\right) + {\log}_{6} \left(x - 1\right) = {\log}_{6} \left(3 x \left(x - 1\right)\right) = 3$

Inverting,

$3 x \left(x - 1\right) = {6}^{3} = 216$.

The roots of this quadratic are $x = 9 , - 8$.

Negative root is inadmissible. So, x = 9.