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