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

##### 2 Answers

Express as a cubic in

#### Explanation:

Multiply both sides by

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

Subtract

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

Let

#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

Let

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

#### Explanation:

Divide both sides of the equation by

#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^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

(*) For example, for a cubic with

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

with

choosing