# How do you solve e^[2 ln x – ln (x^2 + x - 3)] = 1?

May 6, 2016

$x = 3$

#### Explanation:

Use $m \ln n = \ln {m}^{n} \mathmr{and} {e}^{\log} a = a$

The given equation is

${e}^{\ln {x}^{2} - \ln \left({x}^{2} + x - 3\right)} = 1$

${e}^{\ln {x}^{2}} / {e}^{\ln} \left({x}^{2} + x - 3\right) = 1$

${x}^{2} / \left({x}^{2} + x - 3\right) = 1$

${x}^{2} = {x}^{2} + x - 3$

$x = 3$