# How do I find a solution to e^(2x) = 8e^(x) + 20 ?

## I've worked the equation to get to e^(x) = 10 and e^(x) = -2, but I don't understand how to advance it further

Nov 26, 2016

$x = \ln 10$

#### Explanation:

${e}^{2 x} - 8 {e}^{x} - 20 = 0$

By properties of exponents, we have:

${\left({e}^{x}\right)}^{2} - 8 {e}^{x} - 20 = 0$

We let $t = {e}^{x}$.

${t}^{2} - 8 t - 20 = 0$

$\left(t - 10\right) \left(t + 2\right) = 0$

$t = 10 \mathmr{and} - 2$

${e}^{x} = 10 \mathmr{and} {e}^{x} = - 2$

Take the natural logarithm of both sides.

$\ln \left({e}^{x}\right) = \ln 10 \mathmr{and} \ln {e}^{x} = \ln - 2$

Use the rule $\log {a}^{n} = n \log a$:

$x \ln e = \ln 10 \mathmr{and} x \ln e = \ln - 2$

$\ln$ and $e$ are opposites, so equal $1$.

$x = \ln 10$ and $x = \emptyset$, since $y = \ln x$ has a domain of $x > 0$.

Hopefully this helps!