- #1

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[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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- Thread starter Kb1jij
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- #1

- 19

- 0

[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!

- #2

Tom Mattson

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I don't think it can be done.

- #3

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Kb1jij said:

[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 e^{t}) and an algebraic function (such as t^{2}) 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)= xe^{x}.

(I edited this- my f(x)= xe^{2} was a typo.)

(I edited this- my f(x)= xe

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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.

- #6

saltydog

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Kb1jij said:

[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!

I thought I was just missing some easy step...

guess I was wrong!

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