# How do you solve -e^(-3.9n-1)-1=-3?

Sep 17, 2017

$n = - \frac{\ln \left(2\right) + 1}{3.9}$

#### Explanation:

We have: $- {e}^{- 3.9 n - 1} - 1 = - 3$

Multiplying both sides of the equation by $- 1$:

$R i g h t a r r o w - 1 \left(- {e}^{- 3.9 n - 1} - 1\right) = - 1 \times - 3$

$R i g h t a r r o w {e}^{- 3.9 n - 1} + 1 = 3$

Subtracting $1$ from both sides:

$R i g h t a r r o w {e}^{- 3.9 n - 1} + 1 - 1 = 3 - 1$

$R i g h t a r r o w {e}^{- 3.9 n - 1} = 2$

Applying $\ln$ to both sides:

$R i g h t a r r o w \ln \left({e}^{- 3.9 n - 1}\right) = \ln \left(2\right)$

Using the laws of logarithms:

$R i g h t a r r o w \left(- 3.9 n - 1\right) \left(\ln \left(e\right)\right) = \ln \left(2\right)$

$R i g h t a r r o w \left(- 3.9 n - 1\right) \times 1 = \ln \left(2\right)$

$R i g h t a r r o w - 3.9 n - 1 = \ln \left(2\right)$

Adding $1$ to both sides:

$R i g h t a r r o w - 3.9 n - 1 + 1 = \ln \left(2\right) + 1$

$R i g h t a r r o w - 3.9 n = \ln \left(2\right) + 1$

Dividing both sides by $- 3.9$:

$R i g h t a r r o w \frac{- 3.9 n}{- 3.9} = \frac{\ln \left(2\right) + 1}{- 3.9}$

$\therefore n = - \frac{\ln \left(2\right) + 1}{3.9}$