How do you solve $- e^{- 3.9 n - 1} - 1 = - 3$ $-e^(-3.9n-1)-1=-3$ ?

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Answer 1 · 2017-09-17T16:31:41

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

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

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

$\Rightarrow - 1 (- e^{- 3.9 n - 1} - 1) = - 1 \times - 3$ $Rightarrow - 1 (- e^(- 3.9 n - 1) - 1) = - 1 times - 3$

$\Rightarrow e^{- 3.9 n - 1} + 1 = 3$ $Rightarrow e^(- 3.9 n - 1) + 1 = 3$

Subtracting $1$ $1$ from both sides:

$\Rightarrow e^{- 3.9 n - 1} + 1 - 1 = 3 - 1$ $Rightarrow e^(- 3.9 n - 1) + 1 - 1 = 3 - 1$

$\Rightarrow e^{- 3.9 n - 1} = 2$ $Rightarrow e^(- 3.9 n - 1) = 2$

Applying $ln$ $ln$ to both sides:

$\Rightarrow ln (e^{- 3.9 n - 1}) = ln (2)$ $Rightarrow ln(e^(- 3.9 n - 1)) = ln(2)$

Using the laws of logarithms:

$\Rightarrow (- 3.9 n - 1) (ln (e)) = ln (2)$ $Rightarrow (- 3.9 n - 1)(ln(e)) = ln(2)$

$\Rightarrow (- 3.9 n - 1) \times 1 = ln (2)$ $Rightarrow (- 3.9 n - 1) times 1 = ln(2)$

$\Rightarrow - 3.9 n - 1 = ln (2)$ $Rightarrow - 3.9 n - 1 = ln(2)$

Adding $1$ $1$ to both sides:

$\Rightarrow - 3.9 n - 1 + 1 = ln (2) + 1$ $Rightarrow - 3.9 n - 1 + 1 = ln(2) + 1$

$\Rightarrow - 3.9 n = ln (2) + 1$ $Rightarrow - 3.9 n = ln(2) + 1$

Dividing both sides by $- 3.9$ $- 3.9$ :

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

$therefore n = - frac(ln(2) + 1)(3.9)$

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