Question

For positive non-zero real variables pp and qq, if

log⁡(p2+q2)=log⁡p+log⁡q+2log⁡3,\log(p^2 + q^2) = \log p + \log q + 2\log 3,

then, the value of p4+q4p2q2\frac{p^4 + q^4}{p^2 q^2} is

Choose one option
Official answer and solution
Official answer
A
From
Official GATE 2024 answer key

Solution

Combine the logarithmic terms to express (p2+q2)/(pq)(p^2 + q^2)/(pq), then square both sides.

Using logarithm properties log⁡a+log⁡b=log⁡(ab)\log a + \log b = \log(ab) and 2log⁡3=log⁡(32)=log⁡92\log 3 = \log(3^2) = \log 9, the given equation becomes:

log⁡(p2+q2)=log⁡(9pq)  ⟹  p2+q2=9pq\log(p^2 + q^2) = \log(9pq) \implies p^2 + q^2 = 9pq

Dividing both sides by pqpq gives:

p2+q2pq=9\frac{p^2 + q^2}{pq} = 9

Squaring both sides:

(p2+q2pq)2=p4+2p2q2+q4p2q2=81  ⟹  p4+q4p2q2+2=81  ⟹  p4+q4p2q2=79\left(\frac{p^2 + q^2}{pq}\right)^2 = \frac{p^4 + 2p^2q^2 + q^4}{p^2q^2} = 81 \implies \frac{p^4 + q^4}{p^2q^2} + 2 = 81 \implies \frac{p^4 + q^4}{p^2q^2} = 79
Final answer

Option (A)

Common trap

Forgetting to subtract 2 when squaring (p2+q2)/(pq)=9(p^2 + q^2)/(pq) = 9, or adding 2 to get 83.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2024 Computer Science and Information Technology - Set 1 (CS1)
Question
5 of 65
Section
General Aptitude

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