What are the closed-form roots of $$\log_2x=\frac {2^{x-1}}{x}$$ ?
My attempts:
Closed-form means, I assume that the Lambert W function can work.
I know that, at least $x>0$.
Wolfram Alpha gives the result $x=2, x=4$. But how? Honestly, I don't like this method: We will prove that $x=2,x=4$ are only solutions. I think, this is not a good way.
I want to know that, where do roots come from? What is the closed form?
I did some manipulations.
$$x=2^{\frac {2^{x-1}}{x}}$$
$$x^x=2^{2^{x-1}}$$
But, nothing comes from here.
Then, I tried the substitution $x=2^t$, so we have
$$t=\frac{2^{2^t-1}}{{2^t}}$$
$$t2^t=2^{2^{t-1}-1}$$
This seems more reasonable:
Because, $f(t)=t2^t$ looks like the well-known form $f(x)=xe^x$,which is solvable with Lambert W function. But, I understand that RHS is not a constant. That was a just observation. I am wondering, can Lambert-W solve the given equation?
I couldn't proceed anymore.
Your equations are equivalent to $$2x\log_2(x)=2^x$$ $$x^2\log_2(x^2)=x2^x$$ $$x^2\ln(x^2)=\ln(2)xe^{\ln(2)x}$$
Apply $W$ to each side: $$\ln(x^2)=\ln(2)x$$ $$\frac{\ln(x)}{x}=\frac{\ln(2)}{2}$$ $$\frac{1}{x}\ln\left(\frac{1}{x}\right)=-\frac{\ln(2)}{2}=\frac{1}{2}\ln\left(\frac12\right)$$
Apply $W$ to each side: $$\ln(1/x)=W\left(\frac{1}{2}\ln\left(\frac12\right)\right)$$
Can you take it from here? Following the $W_0$ branch leads you to $2$. Following the $W_{-1}$ branch leads you to $4$.