# Solution to an exponential equation

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!

## Answers and Replies

Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
I don't think it can be done.

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:
$e^t=3t^2$
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:

$t = 2ln(3t)$ 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:
HallsofIvy
Science Advisor
Homework Helper
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 by a moderator:
hotvette
Homework Helper
This reminds me of trying to solve $y = x^x$ for a given y. saltydog
Science Advisor
Homework Helper
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:
$e^t=3t^2$
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:

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

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:

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

Thanks for all for your input!

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