Solution to an exponential equation

[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!

 

[quantumdude]

I don't think it can be done.

 

[ComputerGeek]

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?).

 

[HallsofIvy]

In general, an equation involving a a transcendental function (such as e^t) 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)= xe^x.

(I edited this- my f(x)= xe² was a typo.)

 

[hotvette]

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

 

[saltydog]

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:

<br /> f(x)=xe^x=y<br />

then:

<br /> f^{-1}(y)=x=W(y) <br />

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

If:

<br /> g(x)=x^2e^x=y <br />

then:

<br /> g^{-1}(y)=2W(\frac{\sqrt y}{2}) <br />

and in general if:

<br /> h(x)=x^ne^x=y<br />

then:
<br /> h^{-1}(y)=nW(\frac{y^\frac{1}{n}}{n})<br />

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}

 

[Kb1jij]

Thanks for all for your input!

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