Question

How do you solve `log (x^2-9) = log (5x+5)`?

Answer 1

`color(red)(x=7)`

Explanation:

`log(x^2-9) = log (5x+5)`

Convert the logarithmic equation to an exponential equation.

`10^( log(x^2-9)) = 10^( log (5x+5))`

Remember that `10^logx =x`, so

`x^2-9 = 5x+5`

Move all terms to the left hand side.

`x^2-9-5x-5 = 0`

Combine like terms.

`x^2-5x-14 = 0`

Factor.

`(x-7)(x+2) = 0`

`x-7 = 0` and `x+2=0`

`x=7` and `x=-2`

Check:

`log(x^2-9) = log (5x+5)`

If `x=7`

`log(7^2-9) = log (5(7)+5)`

`log(49-9) = log (35+5)`

`log40 = log40`

`x=7` is a solution.

If `x=-2`,

`log((-2)^2-9) = log (5(-2)+5)`

`log(4-9) = log (-10+5)`

`log(-5) = log (-5)`

#log(-5) is not defined,

`x=-2` is a spurious solution.