If $p,q\in N$ satisfy the equation $x^{\sqrt{x}}=\left(\sqrt{x}\right)^x$
(A) relatively prime
(B) twin prime
(C) coprime
(D) if $\log_q{p}$ is defined then $\log_p{q}$ is not and vice versa.
My attempt is as follows:-
$$x^\sqrt{x}=x^{\frac{x}{2}}$$ $$x^{\frac{x}{2}-\sqrt{x}}=x^{0}$$ $$x=2\sqrt{x}$$ $$x^2=4x$$ $$x=0,4$$
But there is only one root which is a natural number. So $p$ and $q$ may be equal here but then no given option will match. So what am I missing here?
On
HINT.-The number $0$ can be taken as solution because calculation gives $0^0=0^0$.
On the other hand we should have $x\gt0$ so $$x^{\sqrt{x}}=\left(\sqrt{x}\right)^x\Rightarrow \sqrt x\ln(x)=x\ln\sqrt x=\dfrac x2\ln (x)$$ It follows first if $\ln(x)\ne0$ then $\sqrt x=\dfrac x2$ so $4x=x^2$ then $x=4$ and second if $\ln(x)=0$ then $x=1$. Thus there are three solutions are $0,1$ and $4$.
You can answer now taking the three possibilities $(p.q)=(0,1),(0,4)$ and $(1,4)$ (not necessary to consider the couple inverted such as, for example $(4,1)$.)
As my comment noted, $p,\,q\in\{1,\,4\}$. So if we require $p\ne q$, without loss of generality $p=1,\,q=4$. This satisfies (A) and (C) (which as best I can tell are synonyms) and (D).