GATE BT 2024 | GA | Question: 5

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. $83$
   

   

Answer 1

$\log(p^2+q^2) = \log(9pq)$ which is equal to $p^2+q^2=9pq$ , square on both sides.

$p^4+q^4+2p^2q^2=81p^2q^2$  divide on both sides by $p^2q^2$

$\frac{p^4+q^4}{9pq} = 81-2 \Rightarrow 79$

Hence option $A.\space 79$ is correct.

Answer 2

Step 1: Simplify the logarithmic equation.

Using the property $\log a + \log b = \log(ab)$ and $n \log a = \log(a^n)$:

$$\log(p^2 + q^2) = \log(p \cdot q \cdot 3^2)$$

$$\log(p^2 + q^2) = \log(9pq)$$

Step 2: Remove logs from both sides.

Since $\log x = \log y \implies x = y$:

$$p^2 + q^2 = 9pq$$

We need to find the value of $\frac{p^4 + q^4}{p^2q^2}$. if we rewrite this as:

$$\frac{p^4}{p^2q^2} + \frac{q^4}{p^2q^2} = \frac{p^2}{q^2} + \frac{q^2}{p^2}$$

Step 3: Use the result from Step 2.

Divide the equation $p^2 + q^2 = 9pq$ by $pq$:

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

$$\frac{p}{q} + \frac{q}{p} = 9$$

Step 4: Square both sides we get:

$$\left(\frac{p}{q} + \frac{q}{p}\right)^2 = 9^2$$

$$\left(\frac{p}{q}\right)^2 + \left(\frac{q}{p}\right)^2 + 2\left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = 81$$

$$\frac{p^2}{q^2} + \frac{q^2}{p^2} + 2 = 81$$

$$\frac{p^2}{q^2} + \frac{q^2}{p^2} = 81 - 2 = 79$$

Correct Answer: (A)