# What is the quadratic formula of e^(2x) - 2e^x = 1?

Oct 31, 2015

Recognise this as quadratic in ${e}^{x}$ and hence solve using the quadratic formula to find:

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

#### Explanation:

This is an equation that is quadratic in ${e}^{x}$, rewritable as:

${\left({e}^{x}\right)}^{2} - 2 \left({e}^{x}\right) - 1 = 0$

If we substitute $t = {e}^{x}$, we get:

${t}^{2} - 2 t - 1 = 0$

which is in the form $a {t}^{2} + b t + c = 0$, with $a = 1$, $b = - 2$ and $c = - 1$.

This has roots given by the quadratic formula:

$t = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$

Now $1 - \sqrt{2} < 0$ is not a possible value of ${e}^{x}$ for Real values of $x$.

So ${e}^{x} = 1 + \sqrt{2}$ and $x = \ln \left(1 + \sqrt{2}\right)$