How to Solve e^x+e^{3x}-1=0 for x?

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To solve the equation e^x + e^{3x} - 1 = 0, the substitution y = e^x transforms it into the cubic equation y^3 + y - 1 = 0. The discussion reveals that there is one real solution and two complex solutions to this cubic equation. Participants express confusion about the possibility of x equating to zero, as substituting x = 0 leads to a contradiction. The conversation emphasizes the complexity of finding the real solution and suggests consulting resources like Mathworld for further guidance on cubic equations. Ultimately, the cubic function's nature indicates that solutions may not be straightforward.
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how do i solve
e^x+e^{3x}-1=0
for x?
 
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to make it simple let y=e^x

thus giving y^3 + y - 1 = 0
y(y^2 + 1) = 1

i looked at this quickly and i don't think that you would get a solution for x.
 
Last edited:
Let y= ex and the equation becomes y3+ y- 1= 0. Can you solve that?
 
how can the answer be x=0?
e^{x}+e^{3x}-1=0
e^{0}+e^{3\times{0}}-1=0
1+1-1=0
1=0??
 
the answer cannot be zero, i edited my post (sorry about that)
 
no prob, but is there really no solution?
 
well it makes sense to me because this is a cubic function right? how did you go about solving for y?
 
i didn't, i don't know how to
 
There is 1 real answer and 2 initial complex answers, being non-simple roots of the cubic equation they all look quite nasty.
 
  • #10
could someone explain how to find the real answer?
 
  • #11
Check out Mathworld's page on the cubic equation.

Daniel.
 
  • #12
ok, thanks
 
  • #13
you can set
y=e^x
then the equation will become y+y^3-1=0
now you can creat the term y^2,then minus the same term.
 
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