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

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- Nov 24th 2012, 03:27 AMThorpeliztsLogarithm
What's wrong with the workings in the attachment? Why can't I factorise?

- Nov 26th 2012, 08:38 PMMacstersUndeadRe: 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

$\displaystyle \log_3(4x-6) - \log_3(2) = \log_3(x^2-1)$

$\displaystyle \log_3(2x-3) = \log_3(x^2-1)$

$\displaystyle 2x-3 = x^2-1$

$\displaystyle x^2-2x+2=0$

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.