Solution to an exponential equation

  • Thread starter Kb1jij
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  • #1
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Today I came across a very strange exponential equation to which neither my teacher nor I could find a solution. It is as follows:
[itex]e^t=3t^2[/itex]
This could be easily solved graphically, but could anyone show me how to do this algebraically?
Thanks in advance!
 

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  • #2
Tom Mattson
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I don't think it can be done.
 
  • #3
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Kb1jij said:
Today I came across a very strange exponential equation to which neither my teacher nor I could find a solution. It is as follows:
[itex]e^t=3t^2[/itex]
This could be easily solved graphically, but could anyone show me how to do this algebraically?
Thanks in advance!

you would need a second equation describing t. then you could use substitution to solve for it.

I mean even if you used the natural logarithm you still get:

[itex]t = 2ln(3t)[/itex] and that does not help you much since you still have t in terms of itself and if you then did this:

[itex]0 = 2ln(3t) - t[/itex]
you can not factor out t.

that does seem to be an interesting solution set though (is there even a solution set?).
 
Last edited:
  • #4
HallsofIvy
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In general, an equation involving a a transcendental function (such as et) and an algebraic function (such as t2) can't be solved in terms of "elementary" functions. It could, I think, be solved in terms of the "Lambert W function", which is defined as the inverse function to f(x)= xex.

(I edited this- my f(x)= xe2 was a typo.)
 
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  • #5
hotvette
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This reminds me of trying to solve [itex]y = x^x[/itex] for a given y.:smile:
 
  • #6
saltydog
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Kb1jij said:
Today I came across a very strange exponential equation to which neither my teacher nor I could find a solution. It is as follows:
[itex]e^t=3t^2[/itex]
This could be easily solved graphically, but could anyone show me how to do this algebraically?
Thanks in advance!

The Lambert W-function is defined as the inverse of the following:

[tex]
f(x)=xe^x=y
[/tex]

then:

[tex]
f^{-1}(y)=x=W(y)
[/tex]

with W being the Lambert W-function for y>-e^{-1}

If:

[tex]
g(x)=x^2e^x=y
[/tex]

then:

[tex]
g^{-1}(y)=2W(\frac{\sqrt y}{2})
[/tex]

and in general if:

[tex]
h(x)=x^ne^x=y
[/tex]

then:
[tex]
h^{-1}(y)=nW(\frac{y^\frac{1}{n}}{n})
[/tex]

Kindly proceed to express the solution of your equation in terms of this generalized Lambda W function, that is:

[tex]t=\text{some function of W}[/tex]
 
  • #7
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Thanks for all for your input!

I thought I was just missing some easy step...
guess I was wrong!
 

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