# How do you solve -3e^(2x) + 16=5 and find any extraneous solutions?

Apr 12, 2017

$x = \ln \frac{\frac{11}{3}}{2}$
$\textcolor{w h i t e}{\text{XXX}}$...maybe I missed something, but I did not find any extraneous solutions.

#### Explanation:

$- 3 {e}^{2 x} + 16 = 5$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow - 3 {e}^{2 x} = - 11$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow {e}^{2 x} = \frac{11}{3}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow 2 x = \ln \left(\frac{11}{3}\right)$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow x = \ln \frac{\frac{11}{3}}{2} \approx 0.64964$ (I used a calculator for this approximation)

Checking this solution back into the original equation demonstrates that the solution is not extraneous. (Again, I used a calculator for this).