Correct Answer: Option (B) Q² = PR

Option (B) Q² = PR is true. Let k = log P = (1/2) log Q = (1/3) log R, so P = 10ᵏ, Q = 10²ᵏ, and R = 10³ᵏ. Substituting yields Q² = (10²ᵏ)² = 10⁴ᵏ and PR = 10ᵏ · 10³ᵏ = 10⁴ᵏ, confirming equality.

Step-by-Step Solution

  1. Assign common value: log P = k, log Q = 2k, log R = 3k
  2. Exponentiate base 10: P = 10ᵏ, Q = 10²ᵏ, R = 10³ᵏ
  3. Test relations algebraically without assuming base

 Option Breakdown

  • (A) P² = Q³R²: Left: (10ᵏ)² = 10²ᵏ; Right: 10⁶ᵏ · 10⁶ᵏ = 10¹²ᵏ. Unequal.
  • (B) Q² = PR: Left: 10⁴ᵏ; Right: 10⁴ᵏ. ✅ Matches.
  • (C) Q² = R³P: Left: 10⁴ᵏ; Right: 10⁹ᵏ · 10ᵏ = 10¹⁰ᵏ. No.
  • (D) R = P²Q²: Left: 10³ᵏ; Right: 10⁴ᵏ. False.

 Detailed Algebraic Proof

Set k = log P. Then log Q = 2k so Q = P²; log R = 3k so R = P³.

Thus Q² = (P²)² = P⁴ and PR = P · P³ = P⁴. Proven!

Options Verified with Exponents
Option Left Side Right Side Exponents Equal?
(A) P² 10²ᵏ Q³R²: 10¹²ᵏ ❌ No
(B) Q² 10⁴ᵏ PR: 10⁴ᵏ ✅ Yes
(C) Q² 10⁴ᵏ R³P: 10¹⁰ᵏ ❌ No
(D) R 10³ᵏ P²Q²: 10⁴ᵏ ❌ No

 Exam Tips for Log Equations

  • Equalize via common k: Set all logs equal to one variable
  • Raise to powers: Match coefficients in exponents
  • Verify numerically: Test P=10, Q=100, R=1000 (satisfies B only)
  • Practice GATE numerical aptitude: Focus on exponent balancing

 Introduction: Crack Log(P) Equation for GATE Success

Log(P) = (1/2)Log(Q) = (1/3)Log(R) from GATE Q.60 tests logarithm manipulation. Q² = PR emerges correct via exponentiation. Master this with breakdowns and shortcuts!