Question

If \( \log_4 x = a \) and \( \log_{25} x = b \), then \( \log_x 10 \) is:

• A. \( \frac{a + b}{2(a - b)} \)
• B. \( \frac{a + b}{a + b} \)
• C. \( \frac{a + b}{2ab} \)
• D. \( \frac{a - b}{2} \)

Hint:

When solving logarithmic equations, combine the terms by applying the logarithmic properties such as \( \log(ab) = \log a + \log b \) and \( \log(a^n) = n\log a \).

Answer 1

The Correct Option is C

Given:

1. \( \log_4 x = a \implies x = 4^a = (2^2)^a = 2^{2a} \)
2. \( \log_{25} x = b \implies x = 25^b = (5^2)^b = 5^{2b} \)

From (1) and (2), we have \( 2^{2a} = 5^{2b} \). Taking logarithms base 10 on both sides, we get: \[ \log_{10} (2^{2a}) = \log_{10} (5^{2b}) \] \[ 2a \log_{10} 2 = 2b \log_{10} 5 \] \[ a \log_{10} 2 = b \log_{10} 5 \] We want to find \( \log_x 10 \). We know that \( \log_x 10 = \frac{\log_{10} 10}{\log_{10} x} = \frac{1}{\log_{10} x} \). We need to find \( \log_{10} x \). Since \( x = 2^{2a} = 5^{2b} \), we have: \[ \log_{10} x = \log_{10} (2^{2a}) = 2a \log_{10} 2 \] Also, \[ \log_{10} x = \log_{10} (5^{2b}) = 2b \log_{10} 5 \] We know that \( \log_{10} 10 = \log_{10} (2 \times 5) = \log_{10} 2 + \log_{10} 5 = 1 \). From \( a \log_{10} 2 = b \log_{10} 5 \), we have \( \log_{10} 5 = \frac{a}{b} \log_{10} 2 \). Substituting this into \( \log_{10} 2 + \log_{10} 5 = 1 \), we get: \[ \log_{10} 2 + \frac{a}{b} \log_{10} 2 = 1 \] \[ \log_{10} 2 \left( 1 + \frac{a}{b} \right) = 1 \] \[ \log_{10} 2 \left( \frac{a+b}{b} \right) = 1 \] \[ \log_{10} 2 = \frac{b}{a+b} \] Then, \( \log_{10} 5 = \frac{a}{b} \log_{10} 2 = \frac{a}{b} \cdot \frac{b}{a+b} = \frac{a}{a+b} \). Now, \( \log_{10} x = 2a \log_{10} 2 = 2a \left( \frac{b}{a+b} \right) = \frac{2ab}{a+b} \). Then, \( \log_x 10 = \frac{1}{\log_{10} x} = \frac{1}{\frac{2ab}{a+b}} = \frac{a+b}{2ab} \). Answer: (c) \( \frac{a+b}{2ab} \)