Question

For a real number a, if \(\frac{log_{15}a+log_{32}a}{(log_{15}a)(log_{32}a)}= 4\), then a must lie in the range

• A. \(4 < a < 5\)
• B. \(3 < a < 4\)
• C. \(a > 5\)
• D. \(2 < a < 3\)

Answer 1

The Correct Option is A

Given:

\[ \frac{\log_{15} a + \log_{32} a}{(\log_{15} a)(\log_{32} a)} = 4 \] 

Step 1: Convert to base 10 logarithms

Using the change of base formula: \[ \log_{b} a = \frac{\log a}{\log b} \] So, \[ \frac{\log_{15} a + \log_{32} a}{(\log_{15} a)(\log_{32} a)} = \frac{\frac{\log a}{\log 15} + \frac{\log a}{\log 32}}{\left(\frac{\log a}{\log 15}\right) \left(\frac{\log a}{\log 32}\right)} \]

Step 2: Simplify the expression

Multiply numerator and denominator: \[ \frac{\log a \left( \frac{1}{\log 15} + \frac{1}{\log 32} \right)}{\frac{(\log a)^2}{\log 15 \cdot \log 32}} \] \[ = \frac{\log a \left( \frac{\log 32 + \log 15}{\log 15 \cdot \log 32} \right)}{\frac{(\log a)^2}{\log 15 \cdot \log 32}} \] \[ = \frac{\log a (\log 32 + \log 15)}{(\log a)^2} \] \[ = \frac{(\log 32 + \log 15)}{\log a} \]

Step 3: Use the given equation

\[ \frac{(\log 32 + \log 15)}{\log a} = 4 \Rightarrow \log 32 + \log 15 = 4 \log a \] Using the property \(\log m + \log n = \log (mn)\): \[ \log(32 \cdot 15) = 4 \log a \] \[ \log 480 = \log a^4 \Rightarrow a^4 = 480 \]

Step 4: Estimate the value of \( a \)

Try nearby values: 
\[ 4^4 = 256 \quad \text{and} \quad 5^4 = 625 \] So, \[ 256 < a^4 = 480 < 625 \Rightarrow 4 < a < 5 \]

Final Answer: Option (A): \( \boxed{4 < a < 5} \)