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.