Question

The coefficient of \( x^n \) in the expansion of \( \log_a(1 + x) \) is:

• A. \( \frac{(-1)^{n-1}}{n} \)
• B. \( \frac{(-1)^{n-1}}{n} \log_a e \)
• C. \( \frac{(-1)^{n-1}}{n} \log_a a \)
• D. \( \frac{(-1)^n}{n} \log_a e \)

Hint:

The expansion of \( \log(1 + x) \) is a standard series. For logarithms in different bases, use the change of base formula to convert to natural logarithms.

Answer 1

The Correct Option is B

We are asked to find the coefficient of \( x^n \) in the expansion of \( \log_a(1 + x) \). Step 1: Using the expansion of \( \log(1 + x) \). We know that the expansion of \( \log(1 + x) \) for \( |x|<1 \) is given by: \[ \log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n}. \] This is the standard Taylor series expansion for \( \log(1 + x) \). Step 2: Applying the change of base formula. We are given \( \log_a(1 + x) \), which can be written in terms of the natural logarithm as: \[ \log_a(1 + x) = \frac{\log_e(1 + x)}{\log_e a}. \] Substitute the expansion of \( \log_e(1 + x) \) into this: \[ \log_a(1 + x) = \frac{1}{\log_e a} \left( x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \right). \] Step 3: Finding the coefficient of \( x^n \). The coefficient of \( x^n \) in the expansion of \( \log_a(1 + x) \) is: \[ \frac{(-1)^{n-1}}{n \log_e a}. \] Since \( \log_e a = \log_a e \), we have: \[ \frac{(-1)^{n-1}}{n \log_e a} = \frac{(-1)^{n-1}}{n} \log_a e. \] Thus, the coefficient of \( x^n \) in the expansion of \( \log_a(1 + x) \) is \( \frac{(-1)^{n-1}}{n} \log_a e \).
Step 4: Conclusion.
Therefore, the correct answer is (b).