# How do you solve log x = log (2x^2 - 2)?

Jan 21, 2016

$x = \frac{1 + \sqrt{17}}{4}$

#### Explanation:

If
$\textcolor{w h i t e}{\text{XXX}} \log \left(x\right) = \log \left(2 {x}^{2} - 2\right)$
then
$\textcolor{w h i t e}{\text{XXX}} x = 2 {x}^{2} - 2$

$\Rightarrow 2 {x}^{2} - x - 2 = 0$

Using the quadratic formula for roots: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
gives
$\textcolor{w h i t e}{\text{XXX}} x = \frac{1 \pm \sqrt{17}}{4}$

Since the argument of $\log$ can not be negative,
we can eliminate $\frac{1 - \sqrt{17}}{4}$ as extraneous,
leaving only the answer above.