Question

For a real number \( n>1 \), \( \frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} = 1 \). The value of n is

• A. \( 4 \)
• B. \( 12 \)
• C. \( 24 \)
• D. \( 36 \)

Hint:

Remember the fundamental logarithm identities: the reciprocal rule (\( \frac{1}{\log_b a} = \log_a b \)) and the product rule (\( \log_b x + \log_b y = \log_b(xy) \)). These are often the key to simplifying complex logarithmic equations.

Answer 1

The Correct Option is C

Step 1: Apply the change-of-base formula. The reciprocal rule, a form of the change-of-base formula, states: \[ \frac{1}{\log_b a} = \log_a b \] Applying this to each term: \[ \frac{1}{\log_2 n} = \log_n 2 \] \[ \frac{1}{\log_3 n} = \log_n 3 \] \[ \frac{1}{\log_4 n} = \log_n 4 \] The equation then becomes: \[ \log_n 2 + \log_n 3 + \log_n 4 = 1 \]
Step 2: Apply the product rule for logarithms. This rule states: \[ \log_b x + \log_b y = \log_b(xy) \] Applying this to the left side of the equation: \[ \log_n (2 \times 3 \times 4) = 1 \] \[ \log_n (24) = 1 \]
Step 3: Convert to exponential form. The logarithmic equation \( \log_b a = c \) is equivalent to \( b^c = a \). Applying this to \( \log_n (24) = 1 \): \[ n^1 = 24 \] \[ n = 24 \]