# How do you solve -7*e^(4v)=-79?

Jun 12, 2018

$v = 0.543$ to 3 significant figures

#### Explanation:

The method for solving any equation is always to get all of the variables on one side of the equation, and all of the numbers on the other. So, the first step is to divide through by -7:

$- 7 \cdot {e}^{4 v} = - 79$

${e}^{4 v} = \frac{- 79}{- 7} = \frac{79}{7}$ (let's leave it as a fraction for now)

So then you need to take the natural logarithm of each side of the equation. This is because 'e' is the inverse function of natural logs. It's the same reason we divided through by -7 before: dividing is the inverse function of multiplying.

${e}^{4 v} = \frac{79}{7}$
$4 v = \ln \left(\frac{79}{7}\right)$
$v = \frac{1}{4} \ln \left(\frac{79}{7}\right) = 0.54305 \ldots = 0.543$ (3 significant figures)

Jun 13, 2018

$n \approx 0.606$

#### Explanation:

Let's start by isolating the term with $v$ in it. We can start by dividing both sides by $- 7$ to get

${e}^{4 v} = \frac{79}{7}$

Since $\ln$ cancels with base-$e$, we can take the natural log of both sides. We now have

$\cancel{\ln} {\cancel{e}}^{4 v} = \ln \left(\frac{79}{7}\right)$

$\implies 4 v = \ln \left(\frac{79}{7}\right)$

Dividing both sides by $4$, we get

$n = \ln \frac{\frac{79}{7}}{4}$

$n \approx 0.606$

Hope this helps!