# How do you solve e^(0.6x-5)+9=18?

Sep 16, 2016

I got $x = 12$

#### Explanation:

We can first take the $9$ to the right:
${e}^{0.6 x - 5} = 18 - 9$
${e}^{0.6 x - 5} = 9$

we then take the natural log, $\ln$ of both sides:
$\ln \left[{e}^{0.6 x - 5}\right] = \ln \left(9\right)$

On the left the $\ln$ and the exponential cancel each other and we end up with:

$0.6 x - 5 = \ln \left(9\right)$

rearrange to isolate $x$:

$x = \frac{\ln \left(9\right) + 5}{0.6} = 11.99 \approx 12$