# How do you solve 6e^(-4k-10)-4=63?

Jul 25, 2018

I tried this:

#### Explanation:

Let us rearrange it a bit:

${e}^{- 4 k - 10} = \frac{63 + 4}{6}$

now take the natural log of both sides:

$\ln \left[{e}^{- 4 k - 10}\right] = \ln \left[\frac{63 + 4}{6}\right]$

we can get rid of $\ln$ and $e$ on the left and write:

$- 4 k - 10 = \ln \left[\frac{63 + 4}{6}\right]$

rearrange to isolate $k$:

$4 k = - 10 - \ln \left[\frac{63 + 4}{6}\right]$

and:

$k = \frac{1}{4} \left[- 10 - \ln \left(\frac{63 + 4}{6}\right)\right] = - 3.10323$

$k = - \frac{1}{4} \left(10 + \setminus \ln \left(\frac{67}{6}\right)\right) = - 3.1032$

#### Explanation:

Given that

$6 {e}^{- 4 k - 10} - 4 = 63$

$6 {e}^{- 4 k - 10} = 67$

${e}^{- 4 k - 10} = \frac{67}{6}$

taking $\setminus \ln$ on both the sides as follows

$\setminus \ln {e}^{- 4 k - 10} = \setminus \ln \left(\frac{67}{6}\right)$

$\left(- 4 k - 10\right) \setminus \ln e = \setminus \ln \left(\frac{67}{6}\right)$

$- 4 k - 10 = \setminus \ln \left(\frac{67}{6}\right) \setminus \quad \left(\setminus \because \setminus \ln e = 1\right)$

$4 k = - 10 - \setminus \ln \left(\frac{67}{6}\right)$

$k = - \frac{1}{4} \left(10 + \setminus \ln \left(\frac{67}{6}\right)\right)$

$= - 3.1032$