For positive non-zero real variables \(p\) and \(q\), if
\[\log\,(p^{2} + q^{2}) = \log p + \log q + 2\log 3,\]
then, the value of \(\frac{p^{4}+q^{4}}{p^{2}q^{2}}\) is
Answer
(A)
Explanation
Combine the right-hand side into one logarithm: \(\log p + \log q + 2\log 3 = \log(9pq)\). Since the logarithm is one-to-one, the two arguments must be equal:
\[p^{2} + q^{2} = 9pq.\]
Now use \(p^{4} + q^{4} = (p^{2} + q^{2})^{2} - 2p^{2}q^{2}\). Squaring the relation gives \((p^{2} + q^{2})^{2} = 81p^{2}q^{2}\), so \(p^{4} + q^{4} = 79p^{2}q^{2}\) and
\[\frac{p^{4} + q^{4}}{p^{2}q^{2}} = 79.\]
Option B, 81, comes from forgetting to subtract the cross term \(2p^{2}q^{2}\).
Answer: A