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

2 Answers
Nov 28, 2015

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

Nov 28, 2015

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.