Question

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 ?

• A. 81
• B. 79
• C. 9
• D. 83

Hint:

When dealing with logarithmic identities, always simplify using log properties and then eliminate the log. Often the problem reduces to a quadratic or identity relation like $p^2+q^2 = k pq$.

Answer 1

The Correct Option is B

Step 1: Simplify the given logarithmic condition. \[ \log(p^2 + q^2) = \log p + \log q + 2 \log 3 \] Using properties of logarithms: \[ \log p + \log q = \log(pq), \quad 2\log 3 = \log(3^2) = \log 9 \] So, \[ \log(p^2 + q^2) = \log(9pq) \] Step 2: Remove logarithms. Since $\log A = \log B \;\Rightarrow\; A = B$ (for $A, B > 0$), \[ p^2 + q^2 = 9pq \] Step 3: Expression to be found. We need \[ \frac{p^4 + q^4}{p^2 q^2} \] Step 4: Manipulate numerator. \[ p^4 + q^4 = (p^2 + q^2)^2 - 2p^2 q^2 \] So, \[ \frac{p^4 + q^4}{p^2 q^2} = \frac{(p^2 + q^2)^2 - 2p^2 q^2}{p^2 q^2} \] \[ = \frac{(p^2 + q^2)^2}{p^2 q^2} - 2 \] Step 5: Substitute relation $p^2 + q^2 = 9pq$. \[ \frac{(p^2 + q^2)^2}{p^2 q^2} = \frac{(9pq)^2}{p^2 q^2} = \frac{81 p^2 q^2}{p^2 q^2} = 81 \] Step 6: Final answer. \[ \frac{p^4 + q^4}{p^2 q^2} = 81 - 2 = 79 \] Oops, correction here: Let's carefully redo. Actually: \[ \frac{p^4 + q^4}{p^2 q^2} = \frac{(p^2 + q^2)^2}{p^2 q^2} - 2 \] Substituting $p^2 + q^2 = 9pq$, \[ = \frac{(9pq)^2}{p^2 q^2} - 2 = \frac{81 p^2 q^2}{p^2 q^2} - 2 = 81 - 2 = 79 \] So final value = 79. \[ \boxed{79} \]