Question

`5e^x = x^3`, find `x`?

Answer 1

`x = -3 W_n(-5^(1/3)/3 omega^m)" "` for any `n in ZZ` and `m in { 0, 1, 2 }`

where `W_n` is the Lambert W function and `omega = -1/2+sqrt(3)/2i` is the primitive complex cube root of `1`

Explanation:

This kind of equation has no solution expressible in terms of elementary functions, but we can use a family of functions called the Lambert `W` function to help.

An equation of the form:

`z e^z = c`

has solutions:

`z = W_n(c)" "` for `n in ZZ`

Given:

`5e^x = x^3`

Taking into account all cube roots, this means:

`5^(1/3) omega^m e^(x/3) = x" "` for `m in { 0, 1, 2 }`

where `omega = -1/2+sqrt(3)/2i` is the primitive complex cube root of `1`.

Then:

`5^(1/3)/3 omega^m e^(x/3) = x/3`

Then:

`-5^(1/3)/3 omega^m = (-x/3) e^((-x/3))`

Note that the right hand side is in the form `z e^z` with `z = -x/3`.

Hence we can apply the Lambert W function to find:

`-x/3 = W_n(-5^(1/3)/3 omega^m)" "` for any `n in ZZ` and `m in { 0, 1, 2 }`

So:

`x = -3 W_n(-5^(1/3)/3 omega^m)" "` for any `n in ZZ` and `m in { 0, 1, 2 }`

This only takes non-real complex values.