# How do you solve log_3 (5x+5) - log_3 (x^2 - 1) = 0?

May 11, 2015

The answer is $x = 6$

I recall the rule:
$\log \left(a\right) - \log \left(b\right) = \log \left(\frac{a}{b}\right)$
and $\log \left(a\right) = 0 \iff a = 1$

so ${\log}_{3} \left(\frac{5 x + 5}{{x}^{2} - 1}\right) = 0 \iff$

$\iff \frac{5 x + 5}{{x}^{2} - 1} = 1 \iff 5 x + 5 = {x}^{2} - 1$

So we gotta solve ${x}^{2} - 5 x - 6 = 0$

$\frac{5 \pm \sqrt{25 + 24}}{2} = \frac{5 \pm 7}{2} \implies x = 6 \mathmr{and} x = - 1$

6 is an acceptable solution:

$5 \cdot 6 + 5 = 35 > 0$ and $36 - 1 = 35 > 0$

-1 is not an acceptable answer:
$- 5 + 5 = 0 \mathmr{and} \log \left(0\right)$ is not defined, so that solution is an extraneous solution.