Question

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

Answer 1

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)`