# How do you solve 8x -1 = 3e^(ln(x^2))?

Jun 14, 2016

$x = 4 \pm \sqrt{15}$

#### Explanation:

Let us have $a = \ln b$ i.e. $b = {e}^{a}$ or $b = {e}^{\ln b}$

Hence ${e}^{\ln {x}^{2}} = {x}^{2}$

and $8 x - 1 = {e}^{\ln {x}^{2}} \Leftrightarrow 8 x - 1 = {x}^{2}$ or

${x}^{2} - 8 x + 1 = 0$

$x = \frac{- \left(- 8\right) \pm \sqrt{{\left(- 8\right)}^{2} - 4 \cdot 1 \cdot 1}}{2}$
or $x = \frac{8 \pm \sqrt{64 - 4}}{2} = \frac{8 \pm 2 \sqrt{15}}{2} = 4 \pm \sqrt{15}$