Question: Obtain the set of real numbers $c$ such that $$\frac{\log_2x}{x}=c$$ has only one real solution.
My approach: Firstly observe that for some $x>0$, $$\frac{\log_2x}{x}=c\iff \log_2x=xc\iff \log_2x-xc=0.$$
Now let $f(x):=\log_2x-xc, \forall x>0.$ Now clearly $f$ is differentiable $\forall x>0$ and $f'(x)=\frac{1}{x\ln2}-c, \forall x>0.$
Now $$f'(x)=0\iff \frac{1}{x\ln2}-c=0\iff x=\frac{1}{c\ln 2}.$$ We know that $x>0$, thus we must have $\frac{1}{c\ln 2}>0\iff c>0.$ Thus $\forall c>0$, $f'(x)=0$ iff $x=\frac{1}{c\ln 2}$.
Now $f''(x)=-\frac{1}{x^2\ln2}.$ Thus $f''(x)>0, \forall x>0$ and $\forall c\in\mathbb{R}$. Therefore $f''\left(\frac{1}{c\ln2}\right)<0, \forall c>0.$
Observe that $\lim_{x\to 0+}f(x)=-\infty, \forall c\in\mathbb{R}.$
Also $\lim_{x\to+\infty}f(x)=\lim_{x\to+\infty}(\log_2x-xc)=\lim_{x\to+\infty}x\left(\frac{\log_2x}{x}-c\right)=-\infty, \forall c>0.$
Thus $f\left(\frac{1}{c\ln 2}\right)$ is the maximum value of $f(x), \forall c>0$. From here, we can also conclude that $f$ is strictly increasing on the interval $\left(0,\frac{1}{c\ln 2}\right]$ and strictly decreasing on the interval $\left[\frac{1}{c\ln 2}, +\infty\right)$.
Thus $f$ has exactly one root for that $c>0,$ whose maximum value is $0$. Let us find that $c$. $$f\left(\frac{1}{c\ln 2}\right)=0\iff c=\frac{2^{-1/\ln 2}}{\ln 2}=\frac{1}{e\ln 2}.$$
Now when $c<0$, we have $f'(x)>0, \forall x>0.$ Thus $\forall c<0$, $f$ is a strictly increasing function.
Also observe that $\forall c<0$, $f(1)=-c>0$ and $\lim_{x\to 0+}f(x)=-\infty.$ Thus by IVT (since, $f$ is continuous), we can conclude that when $c<0$, $f$ has exactly one root.
Now when $c=0$, $$f(x)=0\iff \log_2x=0\iff x=1.$$ Thus when $c=0$, $f$ has exactly one root.
Thus the required set of real numbers $c$ is $$S:=(-\infty,0]\cup\left\{\frac{1}{e\ln 2}\right\}.$$
Can someone check if my solution is completely correct or not? And a better and shorter solution will be appreciated.
The function $$f(x)=\frac{\log x}{x},$$ which is only a positive constant multiple of yours, and so essentially equivalent to it with respect to this problem, is increasing for $0<x<e$ and decreasing for $x>e$ since the derivative is $$\frac{1-\log x}{x^2}.$$ Since $f(x)<0$ for $x<1,$ it follows that there is a local maximum at $x=e.$ Thus the function is concave for all $x>0.$ Hence the equation $f(x)=c$ has at most two solutions for each $c\le f(e)=1/e.$
PS. For more exact analysis, note that as $x\to+\infty,$ we have that $f(x)\to 0^+$ and as $x\to +0,$ we have that the limit is $-\infty.$ Thus we have that precisely two solutions exist for all $c$ satisfying $0<c<1/e$ and when $c=1/e$ or $c\le 0,$ we have just one solution to $f(x)=c.$