How do you solve e^(2x)-5e^x+6=0?

Apr 21, 2016

$x = \ln 3 \mathmr{and} x = \ln 2$

Explanation:

Let $u = {e}^{x}$

Then we have

${u}^{2} - 5 u + 6 = 0$

$\left(u - 3\right) \left(u - 2\right) = 0$

$u - 3 = 0 \mathmr{and} u - 2 = 0$

${e}^{x} - 3 = 0 \mathmr{and} {e}^{x} - 2 = 0$

${e}^{x} = 3 \mathmr{and} {e}^{x} = 2$

$\ln {e}^{x} = \ln 3 \mathmr{and} \ln {e}^{x} = \ln 2$

$x \ln e = \ln 3 \mathmr{and} x \ln e = \ln 2$

$x = \ln 3 \mathmr{and} x = \ln 2$