# Math Help - Logarithm

1. ## Logarithm

What's wrong with the workings in the attachment? Why can't I factorise?

2. ## Re: Logarithm

You're lucky I can read upside down.
I would advise in the future that you become acquainted with the use of latex, even if it's only a rudimentary understanding like mine to avoid attachments and to make the problem legible. We have

$\log_3(4x-6) - \log_3(2) = \log_3(x^2-1)$
$\log_3(2x-3) = \log_3(x^2-1)$
$2x-3 = x^2-1$
$x^2-2x+2=0$

You have done nothing wrong, if I have read what you've written correctly. Here we can use the quadratic formula to find the solutions for x directly, no need to factorize.