Question

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

Answer 1

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

Explanation:

`log_6 (3x) + log_6 (x-1)=log_6(3x(x-1))=3`

Inverting,

`3x(x-1)=6^3=216`.

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

Negative root is inadmissible. So, x = 9.