# How do you solve -9*e^(3-6r)=-3?

##### 1 Answer
Oct 8, 2016

I found: $r = \frac{1}{6} \left\{3 - \ln \left[\frac{1}{3}\right]\right\} = 0 , 6831$

#### Explanation:

Let us try rearranging it:
${e}^{3 - 6 r} = \frac{- 3}{- 9}$
I took the $- 9$ to the right:
we get:
${e}^{3 - 6 r} = \frac{1}{3}$
Take the natural log of both sides:
$\ln \left[{e}^{3 - 6 r}\right] = \ln \left[\frac{1}{3}\right]$
we get:
$3 - 6 r = \ln \left[\frac{1}{3}\right]$
$6 r = 3 - \ln \left[\frac{1}{3}\right]$
$r = \frac{1}{6} \left\{3 - \ln \left[\frac{1}{3}\right]\right\}$
if you can use a scientific calculator you'd get:
$r = \frac{1}{6} \left\{3 - \ln \left[\frac{1}{3}\right]\right\} = 0 , 6831$