Question

If $p,q\in N$ satisfy 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 & vice versa.

Answer 1

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

The correct options are
A relatively prime.
C coprime.
D if ${\log }_{q}p$ is defined then ${\log }_{p}q$ is not & vice versa.

$x^{\sqrt{x}}={\left(\sqrt{x}\right)}^x$
Take log both sides,
$\Rightarrow \sqrt{x}\log x=x\log \sqrt{x}[\because \log a^m=m\log a]$
$\Rightarrow \sqrt{x}\log x=\frac{x}{2}\log x$
$\Rightarrow (\sqrt{x}-\frac{x}{2})\log x=0$
$\Rightarrow \log x=0$ and $\sqrt{x}=\frac{x}{2}$
$\Rightarrow x=1$ and $4x=x^2$
$\Rightarrow x=1$ and $x=0\left(\text{not possible}\right),4$
$\Rightarrow x=1,4$
The numbers $1$ and $4$ have $GCD=1$
Hence, they are relatively prime (co-prime).
They are not twin primes as the difference between the two numbers is not 2.
Also, ${\log }_{1}4$ is not defined but ${\log }_{4}1$ is defined.