Obtain the set of real numbers $c$

Question

Show that there exists a positive real number $x \ne 2$ such that $\log_2 x ={x\over2}$ . Hence obtain the set of real numbers $c$ such that $\log_2 x\over x $$= c$ has only one real solution.

Answer 1

Hint: the function $$x\longmapsto\frac{\log_2 x}x$$ is increasing from 0 to $e$ and decreasing from $e$ to $+\infty$ (why?). Also, $$\lim_{x\to 0^+}\frac{\log_2 x}x=\cdots$$ $$\lim_{x\to+\infty}\frac{\log_2 x}x=\cdots$$

Answer 2

Hint

If you consider the function $$f(x)= \frac{ \log_2 x} {x} - c$$ and compute its derivative, you find $$f'(x)=\frac{1-\log (x)}{x^2 \log (2)}$$ which cancels for $x=e$. The second derivative test shows that this corresponds to a maximum. Now, you have $$f(e)=\frac{1}{e \log (2)}-c$$ So, for $1 < x <\infty$ there are two solutions is $c < \frac{1}{e \log (2)}$, the two roots are identical if $c=\frac{1}{e \log (2)}$ and there is no root for $c > \frac{1}{e \log (2)}$. So a single solution happens for $0<x<1$.

I am sure that you can take from here.