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Basics & Logarithms: Value Expression

JEE Maths question with a full step-by-step solution.

Question
If x=∏n=12026nx = \prod_{n=1}^{2026} n, then the value of the expression
11log⁡2x+1log⁡3x+⋯+1log⁡2026x\frac{1}{\dfrac{1}{\log_2 x} + \dfrac{1}{\log_3 x} + \cdots + \dfrac{1}{\log_{2026} x}}
 is
Solution
Answer: 1
Step 1: Convert each term using the change-of-base identity.
1log⁡kx=log⁡xk\frac{1}{\log_k x} = \log_x k
 So the denominator becomes:
log⁡x2+log⁡x3+⋯+log⁡x2026=log⁡x(2⋅3⋯2026)\log_x 2 + \log_x 3 + \cdots + \log_x 2026 = \log_x(2 \cdot 3 \cdots 2026)
 Step 2: Since x=1⋅2⋅3⋯2026x = 1 \cdot 2 \cdot 3 \cdots 2026, we have 2⋅3⋯2026=x2 \cdot 3 \cdots 2026 = x. Therefore the denominator equals log⁡xx=1\log_x x = 1. Step 3: The entire expression equals 11=1\dfrac{1}{1} = 1. Answer: 1
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