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

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Answer 1 · 2017-09-17T16:31:41

#n = - frac(ln(2) + 1)(3.9)#

We have: #- e^(- 3.9 n - 1) - 1 = - 3#

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

#Rightarrow - 1 (- e^(- 3.9 n - 1) - 1) = - 1 times - 3#

#Rightarrow e^(- 3.9 n - 1) + 1 = 3#

Subtracting #1# from both sides:

#Rightarrow e^(- 3.9 n - 1) + 1 - 1 = 3 - 1#

#Rightarrow e^(- 3.9 n - 1) = 2#

Applying #ln# to both sides:

#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) times 1 = ln(2)#

#Rightarrow - 3.9 n - 1 = ln(2)#

Adding #1# to both sides:

#Rightarrow - 3.9 n - 1 + 1 = ln(2) + 1#

#Rightarrow - 3.9 n = ln(2) + 1#

Dividing both sides by #- 3.9#:

#Rightarrow frac(- 3.9 n)(- 3.9) = frac(ln(2) + 1)(- 3.9)#

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