Question

How do you solve `e^x+4=1/(e^(2x))`?

Answer 1

Express as a cubic in `e^x`, solve that, then take natural log.

Explanation:

Multiply both sides by `e^(2x)` to get:

`e^(3x)+4e^(2x)=1`

Subtract `1` from both sides to get:

`e^(3x)+4e^(2x)-1 = 0`

Let `t = e^x`.

`t^3+4t^2-1 = 0`

This cubic has three Real roots, which are all irrational, but only one is positive.

graph{x^3+4x^2-1 [-10.58, 9.42, -1.36, 8.64]}

Solve the cubic by your favourite method to find:

`t_1 ~~ 0.472833909`

Then `x = ln(t_1) ~~ -0.749011`

Answer 2

Let `t = e^-x` to get a cubic `t^3-4t-1=0` with `3` Real roots, one of which is positive, giving `t ~~ 2.11490754` and

`x = -ln(t) ~~ -0.749011`

Explanation:

Divide both sides of the equation by `e^x` to get:

`1 + 4/e^x = 1/(e^x)^3`

Subtract the left hand side from the right to get:

`(1/e^x)^3-4(1/e^x)-1 = 0`

Let `t = 1/e^x` to get:

`t^3-4t-1 = 0`

This cubic has three Real roots, one of which is positive.

graph{x^3-4x-1 [-10, 10, -5, 5]}

Find by your favourite method (*):

`t_1 ~~ 2.11490754`

Then `x = -ln(t_1) ~~ -0.749011`

(*) For example, for a cubic with `3` Real roots that is already in the form `t^3+pt+q = 0` you can use the trigonometric formula:

`t_k = 2sqrt(-p/3) cos(1/3 arccos((3q)/(2p)sqrt(-3/p))-(2pik)/3)`

with `k = 0, 1, 2`

choosing `k` to give you the positive root.