# Solution to an exponential equation

1. Oct 14, 2005

### Kb1jij

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

2. Oct 14, 2005

### Tom Mattson

Staff Emeritus
I don't think it can be done.

3. Oct 14, 2005

### ComputerGeek

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:

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

$0 = 2ln(3t) - t$
you can not factor out t.

that does seem to be an interesting solution set though (is there even a solution set?).

Last edited: Oct 14, 2005
4. Oct 14, 2005

### HallsofIvy

Staff Emeritus
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.)

Last edited: Oct 15, 2005
5. Oct 14, 2005

### hotvette

This reminds me of trying to solve $y = x^x$ for a given y.

6. Oct 16, 2005

### saltydog

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

$$f(x)=xe^x=y$$

then:

$$f^{-1}(y)=x=W(y)$$

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

If:

$$g(x)=x^2e^x=y$$

then:

$$g^{-1}(y)=2W(\frac{\sqrt y}{2})$$

and in general if:

$$h(x)=x^ne^x=y$$

then:
$$h^{-1}(y)=nW(\frac{y^\frac{1}{n}}{n})$$

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

$$t=\text{some function of W}$$

7. Oct 16, 2005

### Kb1jij

Thanks for all for your input!

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

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