# How do you solve e^(2x)-6e^x-16=0?

Mar 10, 2018

$x = \ln 8$

#### Explanation:

this is a quadratic in ${e}^{x}$

$u = {e}^{x}$

${e}^{2} x - 6 {e}^{x} - 16 = 0$

$\implies {u}^{2} - 6 u - 16 = 0$

$\left(u - 8\right) \left(u + 2\right) = 0$

$\therefore u = 8 \text{ or } - 2$

${e}^{x} = 8 \implies x = \ln 8$

${e}^{x} = - 2 \implies \text{ no soln for } x \in \mathbb{R}$

;.x=ln8.

Mar 10, 2018

Real root: $x = \ln \left(8\right)$

Complex roots: $x = \ln \left(2\right) + \left(2 \pi k + 1\right) i \pi$ and $x = \ln \left(8\right) + i 2 \pi k$ where $k$ is any integer.

#### Explanation:

Real roots
This is a quadratic in ${e}^{x}$. It will be easier to see what we're doing if we introduce a substitution. I will let $t = {e}^{x}$:

${\left({e}^{x}\right)}^{2} - 6 {e}^{x} - 16 = 0$

${t}^{2} - 6 t - 16 = 0$

We solve the quadratic by grouping:
${t}^{2} - 8 t + 2 t - 16 = 0$

$t \left(t - 8\right) + 2 \left(t - 8\right) = 0$

$\left(t + 2\right) \left(t - 8\right) = 0$

By the zero factor principle, this has solutions $t = - 2$ and $t = 8$. We can undo the substitution to get these in terms of $x$:

${e}^{x} = 8$

$\ln \left({e}^{x}\right) = \ln \left(8\right)$

$x = \ln \left(8\right)$

Complex roots
For the other root, $t = - 2$, we can try doing the same procedure:

${e}^{x} = - 2$

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

$x = \ln \left(- 2\right)$

But we run into a problem. The natural log of negative numbers is not defined with real numbers, which means that this root is complex.

We can find this root by using Euler's identity, ${e}^{i \pi} = - 1$. This lets us rewrite the logarithm as such:

$\ln \left(- 2\right) = \ln \left(2 \cdot - 1\right) = \ln \left(2 {e}^{i \pi}\right) = \ln \left(2\right) + \ln \left({e}^{i \pi}\right) = \ln \left(2\right) + i \pi$

Note that $i \pi$ is not the only imaginary exponent that gives $- 1$. Euler's formula actually gives an infinite number of ways to express $- 1$ in this way due to the periodicity of the sine and cosine functions. We can capture this by adding $2 \pi k$ (remember, the periodicity of sine and cosine is $2 \pi$) in the exponent, which after the same calculations as before gives the solutions:

$x = \ln \left(2\right) + \left(2 k + 1\right) i \pi$

Using complex numbers, we can also find even more solutions!

Because Euler's identity gives that ${e}^{i 2 \pi k} = 1$, we can do the same thing we did with $\ln \left(- 2\right)$ with $\ln \left(8\right)$:

$\ln \left(8\right) = \ln \left(8 {e}^{i 2 \pi k}\right) = \ln \left(8\right) + \ln \left({e}^{i 2 \pi k}\right) = \ln \left(8\right) + i 2 \pi k$