Question

Which of the following when simplified, vanishes

• A. (A) $\frac{1}{\log_3 2} + \frac{2}{\log_9 4} - \frac{3}{\log_{27} 8}$
• B. (B) $\log_2 (\frac{2}{3}) + \log_4 (\frac{9}{4})$
• C. (C) -$\log_8 \log_4 \log_2 16$
• D. (D) $\log_{10}$ cot 1° + $\log_{10}$cot 2° + $\log_{10}$ cot 3° + ........ + $\log_{10}$ cot 89°

Answer 1

Answer: (A), (B), (C), (D)

A, B, C, D

Answer 2

(A) $\frac{1}{\log_3 2} + \frac{2}{\log_{3^2} 2^2} - \frac{3}{\log_{3^3} 2^3} = \frac{1}{\log_3 2} + \frac{2}{2\log_3 2} - \frac{3}{3\log_3 2} = \frac{1}{\log_3 2} + \frac{1}{\log_3 2} - \frac{1}{\log_3 2} = \frac{1}{\log_3 2} \neq 0$. Correction: $\frac{1}{\log_3 2} + \frac{2}{2\log_3 2} - \frac{3}{3\log_3 2} = \frac{1}{\log_3 2} + \frac{1}{\log_3 2} - \frac{1}{\log_3 2}$ is incorrect. Correct calculation for (A): $\frac{1}{\log_3 2} + \frac{2}{\log_9 4} - \frac{3}{\log_{27} 8} = \frac{1}{\log_3 2} + \frac{2}{2\log_3 2} - \frac{3}{3\log_3 2} = \frac{1}{\log_3 2} + \frac{1}{\log_3 2} - \frac{1}{\log_3 2}$. This is still incorrect. Let's re-evaluate (A) using $\log_{a^m} b^n = \frac{n}{m} \log_a b$: $\log_9 4 = \log_{3^2} 2^2 = \frac{2}{2} \log_3 2 = \log_3 2$. $\log_{27} 8 = \log_{3^3} 2^3 = \frac{3}{3} \log_3 2 = \log_3 2$. So, the expression becomes $\frac{1}{\log_3 2} + \frac{2}{\log_3 2} - \frac{3}{\log_3 2} = \frac{1+2-3}{\log_3 2} = \frac{0}{\log_3 2} = 0$.

(B) $\log_2 (2/3) + \log_4 (9/4) = (\log_2 2 - \log_2 3) + \log_{2^2} (3^2/2^2) = (1 - \log_2 3) + \frac{1}{2} \log_2 (3^2/2^2) = (1 - \log_2 3) + \frac{1}{2} (2\log_2 3 - 2\log_2 2) = (1 - \log_2 3) + (\log_2 3 - 1) = 0$.

(C) -$\log_8 \log_4 \log_2 16 = -\log_8 \log_4 4 = -\log_8 1 = 0$.

(D) $\log_{10}(\cot 1° \cdot \cot 2° \cdot \dots \cdot \cot 89°) = \log_{10}((\cot 1° \cot 89°) \dots (\cot 44° \cot 46°) \cot 45°) = \log_{10}((1) \dots (1) \cdot 1) = \log_{10} 1 = 0$.