# Finding the Intersection Of 2 Equations (difficult)

1. Aug 29, 2010

1. The problem statement, all variables and given/known data
Solve the following equation:

$$e^{2x}=3x^2$$

2. Relevant equations

3. The attempt at a solution
I can find an approximate solution with a graphing calculator easily, but I am interested how you would find the exact solution.

I can take the natural log of both sides and wind up with.

$$2x=\ln{3x^2}$$

I'm not sure how I could proceed from here to get it into terms of x.

Would perhaps another method be better?

2. Aug 29, 2010

### Mentallic

There is no analytic solution to this problem. You can solve it in terms of the lambert W function though.

The lambert W function is as follows:
if $$a=ue^u$$ then $$u=W(a)$$ where u is some function of x and a is a constant.

$$3x^2=e^{2x}$$

$$3x^2e^{-2x}=1$$ divided through by e2x

$$xe^{-x}=\frac{1}{\sqrt{3}}$$ divided through by 3 then square root of both sides

$$-xe^{-x}=\frac{-1}{\sqrt{3}}$$ negative of both sides.

So now we have it in the lambert W form and thus the solution is

$$-x=W\left(\frac{-1}{\sqrt{3}}\right)$$

$$x=-W\left(\frac{-1}{\sqrt{3}}\right)$$

But again, we can't express the solution in terms of elementary functions so you will only get a good approximation at best for the solution.

3. Aug 30, 2010

### HallsofIvy

Staff Emeritus
Mentallic, I'm not sure why you would say using the Lambert W function (which you did very nicely, by the way) would not be an "analytic" solution. Surely, it is as analytic as, say, using the function ln(x) to solve $e^x= 2$.

4. Aug 30, 2010

### Mentallic

Ahh I think I've mixed up the meaning of analytic with 'expressible in terms of elementary functions', my mistake
Thanks!

5. Aug 30, 2010

hmm. Interesting.

I'll bet If you wanted to take this even further you could use the series expansion for the lambert w function.

$$w(x)=\sum^{\infty}_{n=1}\frac{(-1)^{n-1}n^{n-2}x^{n}}{(n-1)!}$$

$$-w(\frac{-1}{\sqrt{3}})=-\sum^{\infty}_{n=1}\frac{(-1)^{n-1}n^{n-2}(\frac{-1}{\sqrt{3}})^{n}}{(n-1)!}$$

Otherwise a computer would be required to calculate that inverse function. This would allow you to compute the solution to any degree of accuracy. Right?

6. Aug 30, 2010

### Mentallic

I really need to learn about series expansions, they seem to exist in numerous forms for every possible function!

You could even simplify the summation,

$$(-1)^{n-1}\left(\frac{-1}{\sqrt{3}}\right)^n=(-1)^{n-1}(-1)^n.3^{-n/2}=(-1)^{2n-1}.3^{-n/2}=-3^{-n/2}$$

therefore the sum is:

$$-W\left(\frac{-1}{\sqrt{3}}\right)=\sum^{\infty}_{n=1}\frac{n^{n-2}}{3^{n/2}(n-1)!}$$

And yes, you can solve it to any degree of accuracy you wish

7. Aug 31, 2010

### ehild

x can be obtained to any degree of accuracy also with an iterative process

$$x_{i+1}=-\sqrt{\frac{e^{2x_i}}{3}}$$

I got -0.3906....

ehild

Last edited: Aug 31, 2010
8. Aug 31, 2010

### Dickfore

It is better to say that the solution of the equation is negative real number $x = -y, \; y > 0$ and the iteration should be:

$$y_{i + 1} = \frac{e^{-y_{i}}}{\sqrt{3}}$$