Question

\(\log x + \log x^3 + \log x^5 + \dots + \log x^{2n-1}\) is equal to

• A. \(2n \log x\)
• B. \((2n-1)\log x\)
• C. \(n^2 \log x\)
• D. \((n^2+1)\log x\)

Hint:

Sum of first n odd numbers = \(n^2\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem uses properties of logarithms to transform the series into an arithmetic progression sum.
: Key Formula or Approach:
1. \( \log a^b = b \log a \).
2. Sum of first \( n \) odd natural numbers: \( 1 + 3 + 5 + \dots + (2n - 1) = n^2 \).
Step 2: Detailed Explanation:
The series is: \[ S = \log x + \log x^3 + \log x^5 + \dots + \log x^{2n-1} \] Apply the power property \( \log a^b = b \log a \): \[ S = 1\log x + 3\log x + 5\log x + \dots + (2n-1)\log x \] Factor out \( \log x \): \[ S = [1 + 3 + 5 + \dots + (2n-1)] \log x \] The term in the bracket is the sum of the first \( n \) odd numbers. We know that: \[ \sum_{k=1}^{n} (2k-1) = n^2 \] Therefore: \[ S = n^2 \log x \].
Step 3: Final Answer:
The sum is \( n^2 \log x \).