How do you solve #-5e^(-x)+9=6#?

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Answer 1 · 2016-09-06T15:46:30

#x = - ln((3) / (5))#

We have: #- 5 e^(- x) + 9 = 6#

Let's begin by subtracting #9# from both sides of the equation:

#=> - 5 e^(- x) = - 3#

Then, let's divide both sides by #- 5#:

#=> e^(- x) = (3) / (5)#

Now, let's apply the natural logarithm to both sides:

#=> ln(e^(- x)) = ln((3) / (5))#

Using the laws of logarithms:

#=> - x ln(e) = ln((3) / (5))#

#=> - x cdot 1 = ln((3) / (5))#

#=> - x = ln((3) / (5))#

Finally, to solve for #x#, let's divide both sides by #- 1#:

#=> x = - ln((3) / (5))#