Question:medium

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

Updated On: Jan 15, 2026
  • \(4 < a < 5\)

  • \(3 < a < 4\)

  • \(a > 5\)

  • \(2 < a < 3\)

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The Correct Option is A

Solution and Explanation

Given:

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

Step 1: Change base to common logarithm

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

Step 2: Simplify the expression

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

Step 3: Solve for \( a \) using the original equation

\[ \frac{\log 32 + \log 15}{\log a} = 4 \] This implies \[ \log 32 + \log 15 = 4 \log a \]. Using the logarithm 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 \( a \)

Consider powers of 4:  
\[ 4^4 = 256 \quad \text{and} \quad 5^4 = 625 \] Since \( 256<480<625 \), it follows that \( 4<a<5 \).

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

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