Question

How do you solve `e ^ { 2x } = 3x ^ { 2}`?

Answer 1

`x = -W_n(+-1/sqrt(3))` where `W_n` is the Lambert W function.

Explanation:

This has no solution in terms of elementary functions, but is expressible in terms of the Lambert W function...

Suppose:

`(-x)e^(-x) = +-1/sqrt(3)`

Squaring both sides, we find:

`x^2e^(-2x) = 1/3`

Multiplying both sides by `3e^(2x)` we get:

`3x^2=e^(2x)`

The Lambert W function (actually a multi-valued function or family of functions) satisfies:

`ze^z = a" " <=> " " z = W_n(a)` for some integer `n`

The two real valued branches of `W_n` are for `n=0` and `n=-1`.

For our example, we have:

`-x = W_n(+-1/sqrt(3))`

That is:

`x = -W_n(+-1/sqrt(3))`

The real valued solution is given by:

`x = -W_0(1/sqrt(3)) ~~ -0.39`

Here are the graphs of the functions `e^(2x)` and `3x^2`, so you can see their point of intersection...
graph{(y-e^(2x))(y-3x^2) = 0 [-2.657, 2.343, -0.34, 2.16]}

Answer 2

See below.

Explanation:

We have

`e^(2x)=(e^x)^2=3x^2` or

`1/3=x^2(e^-x)^2 = (-x e^-x)^2` and then

`pm1/sqrt3 = -xe^-x`

Regarding the Lambert function

https://en.wikipedia.org/wiki/Lambert_W_function

we have

`y = x e^x hArr x = W(y)` then

`-x=W(pm1/sqrt3)` or `x=-W(pm1/sqrt3)`

with real value for

`x = -W(1/sqrt3)=-0.39064638080205444`