Question

Which of the following is/are CORRECT about the value of \( y = \log_e(-e^x) \), where \( x \) is a real number.

• A. \( y \) is real valued
• B. \( y \) is complex valued
• C. \( y = x \pm i n\pi \), where \( n \) is an odd integer and \( i = \sqrt{-1} \)
• D. \( y = x \pm i n\pi \), where \( n \) is an even integer and \( i = \sqrt{-1} \)

Hint:

When taking the natural logarithm of a negative number, the result will be complex, and it can be written in terms of a multiple of \( i\pi \).

Answer 1

The Correct Option is B, C

We are asked to evaluate the logarithmic function \( y = \log_e(-e^x) \), where \( x \) is a real number.
Step 1: Understanding the domain.
The natural logarithm function \( \log_e(z) \) is only defined for complex numbers with a positive real part. However, since we are taking the logarithm of \( -e^x \), where \( e^x \) is always positive, the argument \( -e^x \) is negative for all real values of \( x \).
Step 2: Logarithmic property.
Using the logarithmic identity:
\[ \log_e(-e^x) = \log_e(e^x) + \log_e(-1) \] We know that:
\[ \log_e(e^x) = x \] and
\[ \log_e(-1) = i\pi \quad \text{(since \( \log_e(-1) = i\pi \) is a standard result in complex analysis)} \] Therefore, we have:
\[ y = x + i\pi \] Since the result involves an imaginary number, \( y \) is complex valued. This matches option (B).
Step 3: Conclusion for \( y \)
For \( y = \log_e(-e^x) \), we can have multiple solutions due to the periodic nature of the logarithm. The value of \( y \) can be expressed as \( y = x + i(2n + 1)\pi \), where \( n \) is an integer, which matches option (C).
Thus, the correct answers are (B) and (C).