How do you solve e ^ { 2x } = 3x ^ { 2}e2x=3x2?

2 Answers
Jul 21, 2017

x = -W_n(+-1/sqrt(3))x=Wn(±13) where W_nWn 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)(x)ex=±13

Squaring both sides, we find:

x^2e^(-2x) = 1/3x2e2x=13

Multiplying both sides by 3e^(2x)3e2x we get:

3x^2=e^(2x)3x2=e2x

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

ze^z = a" " <=> " " z = W_n(a)zez=a z=Wn(a) for some integer nn

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

For our example, we have:

-x = W_n(+-1/sqrt(3))x=Wn(±13)

That is:

x = -W_n(+-1/sqrt(3))x=Wn(±13)

The real valued solution is given by:

x = -W_0(1/sqrt(3)) ~~ -0.39x=W0(13)0.39

Here are the graphs of the functions e^(2x)e2x and 3x^23x2, so you can see their point of intersection...
graph{(y-e^(2x))(y-3x^2) = 0 [-2.657, 2.343, -0.34, 2.16]}

Jul 21, 2017

See below.

Explanation:

We have

e^(2x)=(e^x)^2=3x^2e2x=(ex)2=3x2 or

1/3=x^2(e^-x)^2 = (-x e^-x)^213=x2(ex)2=(xex)2 and then

pm1/sqrt3 = -xe^-x±13=xex

Regarding the Lambert function

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

we have

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

-x=W(pm1/sqrt3)x=W(±13) or x=-W(pm1/sqrt3)x=W(±13)

with real value for

x = -W(1/sqrt3)=-0.39064638080205444x=W(13)=0.39064638080205444