How do you solve e^(2x) - 7e^(x) + 8 = 0?

Jul 7, 2015

Notice the overall pattern; there's an ${e}^{x}$ and an ${e}^{2 x}$ so think of those as a new thing and its square: $u$ and ${u}^{2}$.

Explanation:

If $u = {e}^{x}$ then ${u}^{2} = {\left({e}^{x}\right)}^{2} = {e}^{2 x} .$ Our equation becomes:

${u}^{2} - 7 u + 8 = 0$ which can't be solved by factoring but the quadratic formula says:

$u = \frac{7 \pm \sqrt{49 - 32}}{2} = \frac{7 \pm \sqrt{17}}{2}$.

And since $u = {e}^{x}$, $x = \ln u$ so the solutions are:

$x = \ln \left(\frac{7 + \sqrt{17}}{2}\right)$ and $x = \ln \left(\frac{7 - \sqrt{17}}{2}\right)$.

Just check that we are doing the natural log of positive numbers, otherwise those values are undefined.

\dansmath strikes again!\

Remember, If you want approximate decimal answers, be sure to watch for the correct order of operations!