First, use the rule of logs: log(a) + log(b) = log(a x b)

So we have `log((x+1)(x-3)) = log(6x^2-6)`

Now take exponentials of both sides, giving

`(x+1)(x-3) = 6x^2 - 6`

Expand the lefthand side giving

`x^2 -3x + (1)x - (1)(3) = 6x^2 -6` or

`x^2 - 2x -...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

First, use the rule of logs: log(a) + log(b) = log(a x b)

So we have `log((x+1)(x-3)) = log(6x^2-6)`

Now take exponentials of both sides, giving

`(x+1)(x-3) = 6x^2 - 6`

Expand the lefthand side giving

`x^2 -3x + (1)x - (1)(3) = 6x^2 -6` or

`x^2 - 2x - 3 = 6x^2 - 6`

Gather terms to the lefthand side giving

`x^2 - 6x^2 -2x -3 + 6 = 0` or

`-5x^2 -2x + 3 = 0`

Solve for the roots `x_0` using the quadratic formula `x_0 = (-b+- sqrt(b^2 -4ac))/(2a)`

for a quadratic `ax^2 - bx + c = 0`

Thus `x_0 = (2 +- sqrt(2^2 - 4(-5)(3)))/(2(-5)) = (2 +- sqrt(4+60))/(-10) = (2+-sqrt(64))/(-10)`

`= (2 + 8)/(-10)` or `(2-8)/(-10)` = `-1` or `6/10`

Check: ` ``-5(x+1)(x-6/10) = -5(x^2 -(6/10)x + x -6/10) = -5x^2 - 3x + x + 3`

**` `**HOWEVER! Though this solves the quadratic, when we put these values into the original equation we find that we have some undefined values:

log(0), log(-4), log(-2.4) are all logs of negative numbers.

The log function is undefined below zero, and it's domain is only in [0,`oo`). Therefore the lines given on each side of the equation do not cross. Graph them on a calculator or computer to see this.

**So, the equation has no solutions**