# How do you solve e^(2x)-6e^x+8=0?

Dec 10, 2015

Use ${e}^{2 x} = {\left({e}^{x}\right)}^{2}$ to see that this equation is quadratic in ${e}^{x}$.

#### Explanation:

${e}^{2 x} - 6 {e}^{x} + 8 = 0$

${\left({e}^{x}\right)}^{2} - 6 {e}^{x} + 8 = 0$

Factor without substituting

$\left({e}^{x} - 4\right) \left({e}^{x} - 2\right) = 0$

So ${e}^{x} - 4 = 0$ $\text{ or }$ ${e}^{x} - 2 = 0$

${e}^{x} = 4$ $\text{ or }$ ${e}^{x} = 2$

$x = \ln 4$ $\text{ or }$ $x = \ln 2$

Substitute, then factor

Let $u = {e}^{x}$, then we want

${u}^{2} - 6 u + 8 = 0$

$\left(u - 4\right) \left(u - 2\right) = 0$

$u = 4$ $\text{ or }$ $u = 2$ Recall $u = {e}^{x}$, so

${e}^{x} = 4$ $\text{ or }$ ${e}^{x} = 2$

$x = \ln 4$ $\text{ or }$ $x = \ln 2$