How do you solve 7e^{x} = 9- e^{- x}?

1 Answer
Nov 13, 2016

Two solutions
x_1=ln((9-root2(53))/14)
x_2=ln((9+root2(53))/14)

Explanation:

Let's rewrite the equation
7e^x-9+1/e^x=0
We can multiply both sides for e^x so that we get

7e^(2x)-9e^x+1=0
this can be solved as an ordinary second degree eqaution

e^x=(9+-root2(81-28))/14

and consequently

x=ln((9+-root2(53))/14)