GATE AE 2018 • Q5 MCQ • 1 Mark General Aptitude

General Aptitude

Quantitative Aptitude

Subtopic: Arithmetic & Numerical Estimation

Question Statement:

The value of the expression $\frac{1}{1+\log_u(vw)} + \frac{1}{1+\log_v(wu)} + \frac{1}{1+\log_w(uv)}$ is _________.

Options:

  • (A) -1
  • (B) 0
  • (C) 1
  • (D) 3

Official Answer Key:

C

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2018 Q.5)** **1. Intuitive Physical Setup & Given Parameters**: - Evaluate the symmetric logarithmic expression: S = \frac{1}{1+\log_u(vw)} + \frac{1}{1+\log_v(wu)} + \frac{1}{1+\log_w(uv)} - Using base-change and logarithmic product identities: $\log_b(a) = \frac{\ln a}{\ln b}$. **2. Step-by-Step Mathematical Derivation**: 1. **Transforming the First Term**: Notice that $1 = \log_u(u)$. Therefore: 1 + \log_u(vw) = \log_u(u) + \log_u(vw) = \log_u(uvw) Taking the reciprocal: \frac{1}{1+\log_u(vw)} = \frac{1}{\log_u(uvw)} = \log_{uvw}(u) 2. **Transforming the Second and Third Terms**: Similarly: 1 + \log_v(wu) = \log_v(v) + \log_v(wu) = \log_v(uvw) \implies \frac{1}{1+\log_v(wu)} = \log_{uvw}(v) 1 + \log_w(uv) = \log_w(w) + \log_w(uv) = \log_w(uvw) \implies \frac{1}{1+\log_w(uv)} = \log_{uvw}(w) 3. **Summing all three terms**: S = \log_{uvw}(u) + \log_{uvw}(v) + \log_{uvw}(w) = \log_{uvw}(u \cdot v \cdot w) = \log_{uvw}(uvw) = 1 **3. Comprehensive Options & Distractor Analysis**: - **(A)** -1: ❌ Incorrect. Logarithmic sum evaluates to positive unity. - **(B)** 0: ❌ Incorrect. Would require arguments to evaluate to $1^0$. - **(C)** 1: ✅ **CORRECT**. Elegantly collapses to $\log_{uvw}(uvw) = 1$. - **(D)** 3: ❌ Incorrect. Result if each term were 1, but here the sum equals 1. **4. Official Master Key & Exam Takeaway**: - **Official Answer**: **C** - **Exam Rule**: Use $1 = \log_b(b)$ to convert $1 + \log_b(x) = \log_b(bx)$, then apply the reciprocal base-swap identity $\frac{1}{\log_b(a)} = \log_a(b)$.
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