# How do you solve e^x(2e^x-1) = 10?

Jun 2, 2018

Real solution: $x = \ln \left(\frac{5}{2}\right)$

Complex solutions:

$x = \ln \left(\frac{5}{2}\right) + 2 n \pi i$

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

#### Explanation:

Given:

${e}^{x} \left(2 {e}^{x} - 1\right) = 10$

We can treat this as a quadratic in ${e}^{x}$ and factor it using an AC method:

Transposing and subtracting $10$ from both sides, this becomes:

$0 = {e}^{x} \left(2 {e}^{x} - 1\right) - 10$

$\textcolor{w h i t e}{0} = 2 {\left({e}^{x}\right)}^{2} - \left({e}^{x}\right) - 10$

$\textcolor{w h i t e}{0} = \left(2 {\left({e}^{x}\right)}^{2} - 5 \left({e}^{x}\right)\right) + \left(4 \left({e}^{x}\right) - 10\right)$

$\textcolor{w h i t e}{0} = {e}^{x} \left(2 {e}^{x} - 5\right) + 2 \left(2 {e}^{x} - 5\right)$

$\textcolor{w h i t e}{0} = \left({e}^{x} + 2\right) \left(2 {e}^{x} - 5\right)$

So:

${e}^{x} = - 2 \text{ }$ or $\text{ } {e}^{x} = \frac{5}{2}$

If $x$ is Real then ${e}^{x} > 0$. So the only Real solution to the given equation is given by taking the natural logarithm of both sides of the second equation to find:

$x = \ln \left(\frac{5}{2}\right)$

Note however that ${e}^{i \pi} = - 1$, and hence there are complex solutions:

$x = \ln \left(\frac{5}{2}\right) + 2 n \pi i$

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

for any integer $n$