GATE CS 2024 Set 1 — Question 5

Given the equation:$$\log (p^2 + q^2) = \log p + \log q + 2 \log 3$$Using the properties of logarithms $\log a + \log b = \log(ab)$ and $n \log a = \log(a^n)$:$$\log (p^2 + q^2) = \log(pq) + \log(3^2)$$$$\log (p^2 + q^2) = \log(pq) + \log(9)$$$$\log (p^2 + q^2) = \log(9pq)$$Removing the logarithm from both sides:$$p^2 + q^2 = 9pq$$We need to find the value of $\frac{p^4 + q^4}{p^2 q^2}$.
Let's rewrite the expression:$$\frac{p^4 + q^4}{p^2 q^2} = \frac{p^4}{p^2 q^2} + \frac{q^4}{p^2 q^2} = \frac{p^2}{q^2} + \frac{q^2}{p^2}$$Divide the equation $p^2 + q^2 = 9pq$ by $pq$:$$\frac{p^2}{pq} + \frac{q^2}{pq} = 9$$$$\frac{p}{q} + \frac{q}{p} = 9$$Squaring both sides:$$\left(\frac{p}{q} + \frac{q}{p}\right)^2 = 9^2$$$$\frac{p^2}{q^2} + \frac{q^2}{p^2} + 2\left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = 81$$$$\frac{p^2}{q^2} + \frac{q^2}{p^2} + 2 = 81$$$$\frac{p^2}{q^2} + \frac{q^2}{p^2} = 81 - 2$$$$\frac{p^2}{q^2} + \frac{q^2}{p^2} = 79$$Thus, $\frac{p^4 + q^4}{p^2 q^2} = 79$.