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 _________.
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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