Question

If $p,q\in N$ satisfies the equation $x^{\sqrt{x}}={\left(\sqrt{x}\right)}^x$, then $p$ and $q$ are

• A. relatively prime
• B. twin prime
• C. coprime
• D. if ${\log }_{q}p$ is defined, then ${\log }_{p}q$ is not and vice versa

Answer 1

Answer: (A), (C), (D)

relatively prime
coprime
if ${\log }_{q}p$ is defined, then ${\log }_{p}q$ is not and vice versa

Taking $\log$ on both sides of the given equation

$\sqrt{x}\log x=x\log \sqrt{x}\cdots [\because \log a^m=m\log a]$

$\sqrt{x}\log x=\frac{x}{2}\log x$

$\Rightarrow \left(\log x\right)\cdot (\sqrt{x}-\frac{x}{2})=0$

$\Rightarrow \log x=0$ or $(\sqrt{x}-\frac{x}{2})=0$

$\Rightarrow x=1$ or $x^2-4x=0$

$\Rightarrow x=1$ or $x=0$ or $x=4$

But since it's given that the solutions are natural numbers, we can take $p$ and $q$ as $1,4$ or vice versa.

here $1,4$ are co-prime numbers

Also, ${\log }_{4}1$ is defined but ${\log }_{1}4$ is not defined

Hence, option A, C and D are correct.