Question:medium

If \(\log_3 2\), \(\log_3(2x-5)\) and \(\log_3(2x-\tfrac{7}{2})\) are in AP, then the value of \(x\) is

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Always check domain in logarithmic equations.
Updated On: Apr 18, 2026
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
For three terms \( a, b, c \) to be in Arithmetic Progression (AP), the condition is \( 2b = a + c \).
: Key Formula or Approach:
1. \( 2 \log_3(2^x - 5) = \log_3 2 + \log_3(2^x - 7/2) \).
2. Properties of logs: \( n \log a = \log a^n \) and \( \log a + \log b = \log(ab) \).
Step 2: Detailed Explanation:
Applying the AP condition: \[ \log_3 (2^x - 5)^2 = \log_3 \left[ 2 \cdot (2^x - 7/2) \right] \] Removing the log from both sides: \[ (2^x - 5)^2 = 2(2^x) - 7 \] Let \( y = 2^x \): \[ (y - 5)^2 = 2y - 7 \] \[ y^2 - 10y + 25 = 2y - 7 \] \[ y^2 - 12y + 32 = 0 \] Factorize the quadratic equation: \[ (y - 8)(y - 4) = 0 \] This gives \( y = 8 \) or \( y = 4 \).
Case 1: If \( y = 4 \), then \( 2^x = 4 \), so \( 2^x - 5 = 4 - 5 = -1 \). Logarithm of a negative number is undefined. So \( x = 2 \) is rejected.
Case 2: If \( y = 8 \), then \( 2^x = 8 \implies 2^x = 2^3 \implies x = 3 \).
Check: \( 2^3 - 5 = 3>0 \). Condition satisfied.
Step 3: Final Answer:
The value of x is 3.
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