How do you solve (lnx)^2 -1 = 0?

Dec 22, 2015

Rearrange then take exponents to find:

$x = e$ or $x = \frac{1}{e}$

Explanation:

Add $1$ to both sides of the equation to get:

${\left(\ln x\right)}^{2} = 1$

Hence:

$\ln x = \pm 1$

By definition of $\ln$ we know ${e}^{\ln x} = x$. Hence:

$x = {e}^{\ln x} = {e}^{\pm 1}$

That is $x = e$ or $x = \frac{1}{e}$