2

Can $x=8\log_2x$ ($x$ real) be solved analytically (i.e., without a calculator)?

Start wearing purple
  • 53,234
  • 13
  • 164
  • 223
jack
  • 113

2 Answers2

3

No, you had to use special functions, like Lambert W-function.

Norbert
  • 56,803
0

Suppose $x=k\log x$. It is equivalent to $e^x=x^k$, which is further equivalent to $e^{-x/k}=x=(-1/k)(1/(-x/k))$. Now you can see this is just $e^{-x/k}(-x/k)=-1/k$.

Consider $w e^w$, this decreases on $(-\infty,-1)$ and increases on $(-1,+\infty)$, so takes value in $[-1/e,+\infty)$. $w$ approaching $-\infty$, it gets to $0$, so given $x=w e^w$, when $-1/e<x<0$, $w$ has two solutions; otherwise one.

If solution of $x=w e^w$ is denoted $w=W'(x)$ (possibly multivalued), then $-x/k=W'(-1/k)$ and $x=-kW'(-1/k)$. The Lambert function $W(x)$ is the inverse function of $x=w e^w$ in the complex ($\mathbb{C}$) sense, and our $W'$ is just the choice of all its real values.

Ash GX
  • 1,361