# `log_6(3x)+log_6(x-1)=3` Solve the equation. Check for extraneous solutions.

To evaluate the given equation `log_6(3x)+log_6(x-1)=3` , we may apply the logarithm property: `log_b(x)+log_b(y)=log_b(x*y)` .

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

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

`log_6(3x^2-3x)=3`

To get rid of the "log" function, we may apply the logarithm property: `b^(log_b(x))=x` .

Raise both sides by base of `6` .

`6^(log_6(3x^2-3x))=6^3`

`3x^2-3x=216`

Subtract `216` from both sides of the equation to simplify in standard form: `ax^2+bx+c= 0` .

`3x^2-3x-216=216-216`

`3x^2-3x-216=0`

Apply factoring on the trinomial.

`3*(x + 8)*(x - 9)=0`

Apply zero-factor property to solve for `x` by equating each factor in terms of `x` to `0` .

`x+8-8=0-8`

`x=-8`

`x-9=0`

`x-9+9=0+9`

`x=9`

Checking: Plug-in each `x` on  `log_6(3x)+log_6(x-1)=3` .

Let `x=-8` on `log_6(3x)+log_6(x-1)=3` .

`log_6(3*(-8))+log_6(-8-1)=?3`

`log_6(-24)+log_6(-9)=?3`

undefined +undefined =?3     FALSE

Note that `log_b(x)` is undefined on `xlt=0` .

Let `x=9` on `log_6(3x)+log_6(x-1)=3` .

`log_6(3*9)+log_6(9-1)=?3`

`log_6(27)+log_6(8)=?3`

`log_6(27*8)=?3`

`log_6(216)=?3`

`log_6(6^3)=?3`

`3log_6(6)=?3`

`3*1=?3`

`3=3 `    TRUE

Thus, the `x=-8` is an extraneous solution.

The `x=9` is the only real solution of the equation `log_6(3x)+log_6(x-1)=3` .