Question

For positive non-zero real variables \( p \) and \( q \), if \[ \log \left( p^2 + q^2 \right) = \log p + \log q + 2 \log 3, \] then the value of \( \frac{p^4 + q^4}{p^2 q^2} \) is:

• A. \( 79 \)
• B. \( 81 \)
• C. \( 9 \)
• D. \( 83 \)

Hint:

When solving logarithmic equations involving sums of powers, always try to express the terms in a factored form or relate them to known identities for simplification.

Answer 1

The Correct Option is A

Step 1: Simplify the given equation.
From the given equation: \[ \log \left( p^2 + q^2 \right) = \log p + \log q + 2 \log 3, \] we know that: \[ \log \left( p^2 + q^2 \right) = \log \left( p \cdot q \cdot 3^2 \right). \] Exponentiating both sides, we get: \[ p^2 + q^2 = 9pq. \quad \cdots (1) \] Step 2: Evaluate \( \frac{p^4 + q^4}{p^2 q^2} \).
We start with: \[ p^4 + q^4 = \left( p^2 + q^2 \right)^2 - 2p^2q^2. \] Using equation (1), substitute \( p^2 + q^2 = 9pq \): \[ p^4 + q^4 = (9pq)^2 - 2p^2q^2 = 81p^2q^2 - 2p^2q^2 = 79p^2q^2. \] Thus, \[ \frac{p^4 + q^4}{p^2q^2} = \frac{79p^2q^2}{p^2q^2} = 79. \] Conclusion: The value of \( \frac{p^4 + q^4}{p^2q^2} \) is \( 79 \).