For positive integers p and q, with pq 1, (pq)p/q = p(pq - 1). Then,

$\sqrt{\frac{q}{p}} = \sqrt{\frac{p}{q}}$

GATE 2024 XH2 (English) – Question 3

General Aptitude · Quantitative Aptitude: Algebra, Geometry and Mensuration · 1 mark · Multiple choice

For positive integers $p$ and $q$, with $\frac{p}{q} \ne 1$, $\left(\frac{p}{q}\right)^{p/q} = p^{\left(\frac{p}{q} - 1\right)}$. Then,

$q^p = p^q$
$q^p = p^2 q$
$\sqrt{q} = \sqrt{p}$
$\sqrt{\frac{q}{p}} = \sqrt{\frac{p}{q}}$

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Correct answer: (A) $q^p = p^q$

Explanation

Let $t = p/q$. Then $t^t = p^{t-1}$ with $p = tq$, so $t^t = t^{t-1} q^{t-1}$, which gives $t = q^{t-1}$, that is $p = q^{t} = q^{p/q}$. Raising both sides to the power $q$ gives $p^q = q^p$.