How do you solve log_7(2x^2 - 1) = log_7 4x?

Nov 24, 2015

$2 + \sqrt{6}$

Explanation:

Note that if ${\log}_{a} b = {\log}_{a} c ,$ then $b = c$.

Because of this, we can say that $2 {x}^{2} - 1 = 4 x$.

$2 {x}^{2} - 4 x - 1 = 0$

$x = \frac{4 \pm \sqrt{16 + 8}}{2}$

$x = 2 \pm \sqrt{6}$

However, the only answer is the positive one, $2 + \sqrt{6}$, sincein ${\log}_{a} b$, $b > 0$.