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

  • Thread starter Guero
  • Start date
In summary, to solve e^x+e^{3x}-1=0 for x, you can first substitute y=e^x, which simplifies the equation to y^3 + y - 1 = 0. By manipulating the equation, you can get y(y^2 + 1) = 1. However, there is no real solution for x as this is a cubic equation and the real solution is quite complex. To understand the concept better, you can refer to Mathworld's page on the cubic equation.
  • #1
Guero
15
0
how do i solve
[tex]e^x+e^{3x}-1=0[/tex]
for x?
 
Physics news on Phys.org
  • #2
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:
  • #3
Let y= ex and the equation becomes y3+ y- 1= 0. Can you solve that?
 
  • #4
how can the answer be x=0?
[tex]e^{x}+e^{3x}-1=0[/tex]
[tex]e^{0}+e^{3\times{0}}-1=0[/tex]
[tex]1+1-1=0[/tex]
[tex]1=0[/tex]??
 
  • #5
the answer cannot be zero, i edited my post (sorry about that)
 
  • #6
no prob, but is there really no solution?
 
  • #7
well it makes sense to me because this is a cubic function right? how did you go about solving for y?
 
  • #8
i didn't, i don't know how to
 
  • #9
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
[tex] y=e^x[/tex]
then the equation will become [tex] y+y^3-1=0 [/tex]
now you can creat the term [tex] y^2 [/tex],then minus the same term.
 

1. What is the purpose of solving e^x+e^{3x}-1=0 for x?

The purpose of solving this equation is to find the value(s) of x that make the equation true. This can help in understanding the behavior of exponential functions and in solving other equations involving exponents.

2. Can this equation be solved algebraically?

Yes, this equation can be solved algebraically by using logarithmic functions and properties of exponents. However, the solution may involve complex numbers.

3. Is there only one solution to this equation?

No, there can be multiple solutions to this equation depending on the values of x. The number of solutions can be determined by analyzing the behavior of the exponential functions involved.

4. Are there any special techniques or methods for solving this type of equation?

Yes, there are various techniques that can be used to solve this type of equation, such as substitution, factoring, and using logarithmic functions. It is important to carefully analyze the equation and choose the most appropriate method for solving it.

5. How can solving e^x+e^{3x}-1=0 for x be applied in real-world situations?

This type of equation can arise in various fields such as engineering, physics, and finance, where exponential functions are commonly used to model real-world phenomena. Solving this equation can help in predicting and analyzing the behavior of these systems.

Similar threads

  • Introductory Physics Homework Help
Replies
19
Views
673
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
328
  • Introductory Physics Homework Help
Replies
2
Views
452
  • Introductory Physics Homework Help
Replies
2
Views
535
  • Introductory Physics Homework Help
Replies
24
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
687
  • Introductory Physics Homework Help
2
Replies
48
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
285
Back
Top