How do i solve x = 2 ln(x) for x?

[Ava.T]

I keep getting stuck when trying to solve for x in equations where it appears both inside and outside of a logarithm, like in x = 2 ln(x). I understand the properties individually, but combining them into a single algebraic solution has me completely stumped.

[HannahIJ]

I did the same thing a while back. I rewrote it as x/2 = ln x, then e^{x/2} = x. When I compare the two curves they don’t cross for any positive x. The function x - 2 ln x has a minimum at x = 2 of about 0.614, so it doesn’t hit zero.

[Jack_B]

I tried Newton’s method on f(x) = x - 2 ln x. With guesses like 1 or 3 the updates didn’t settle; the derivative blows up near 0 and the steps jump around. It felt like there was no real root hiding there.

[AubreyPH]

Lambert W path gives x = -2 W(-1/2). Real branches don’t exist for that argument, so there aren’t any real solutions; you’d end up with complex numbers.

[ElizabethM]

Maybe the issue is the domain or the problem wording; if you meant x = a ln x with a different constant, you could get a root. Or maybe a misprint. I can test a variant if you want.