Question

For a real number \(x\) , if \(\frac{1}{2},\frac{log_3(2^x-9)}{log_34}\), and \(\frac{log_5\bigg(2^x+\frac{17}{2}\bigg)}{log_54}\) are in an arithmetic progression, then the common difference is

• A. \(log_4\bigg(\frac{23}{2}\bigg)\)
• B. \(log_4\bigg(\frac{3}{2}\bigg)\)
• C. \(log_47\)
• D. \(log_4\bigg(\frac{7}{2}\bigg)\)

Answer 1

The Correct Option is D

Given expression: \[ \frac{\log_3(2^x - 9)}{\log_3 4} \] Using the base change formula: \[ \frac{\log_3(2^x - 9)}{\log_3 4} = \log_4(2^x - 9) \] Similarly, \[ \frac{\log_5\left(2^x + \frac{17}{2}\right)}{\log_5 4} = \log_4\left(2^x + \frac{17}{2}\right) \]

Given equation: \[ 2\log_4(2^x - 9) = \frac{1}{2} + \log_4\left(2^x + \frac{17}{2}\right) \] Note that: \[ \frac{1}{2} = \log_4 2 \] Therefore, the equation becomes: \[ 2\log_4(2^x - 9) = \log_4 2 + \log_4\left(2^x + \frac{17}{2}\right) \]

Using the logarithmic identity: \[ a\log_b m = \log_b m^a \quad \text{and} \quad \log_b m + \log_b n = \log_b (mn) \] Left side: \[ \log_4(2^x - 9)^2 \] Right side: \[ \log_4 \left[ 2 \cdot \left(2^x + \frac{17}{2} \right) \right] \] So we equate: \[ \log_4(2^x - 9)^2 = \log_4 \left[ 2 \cdot \left(2^x + \frac{17}{2} \right) \right] \] Removing logarithms: \[ (2^x - 9)^2 = 2 \cdot \left(2^x + \frac{17}{2} \right) \] 

Expanding both sides: 
Left: \[ (2^x)^2 - 2 \cdot 9 \cdot 2^x + 81 = 2^{2x} - 18 \cdot 2^x + 81 \] Right: \[ 2 \cdot 2^x + 2 \cdot \frac{17}{2} = 2 \cdot 2^x + 17 \] Equating: \[ 2^{2x} - 18 \cdot 2^x + 81 = 2 \cdot 2^x + 17 \] Bring all terms to one side: \[ 2^{2x} - 20 \cdot 2^x + 64 = 0 \]

Let \( y = 2^x \), then: \[ 2y^2 - 20y + 64 = 0 \] Factorizing: \[ 2y^2 - 16y - 4y + 64 = 0 \] \[ 2y(y - 8) - 4(y - 8) = 0 \] \[ (y - 4)(2y - 16) = 0 \Rightarrow y = 4 \text{ or } y = 8 \] So, \( 2^x = 4 \) or \( 2^x = 16 \), hence \( x = 2 \) or \( x = 4 \)

Check both values:

• For \( x = 2 \), \( 2^x - 9 = -5 \): Log is not defined.
• For \( x = 4 \), \( 2^x - 9 = 16 - 9 = 7 \): Valid.

So, only valid value: \( 2^x = 16 \)

 

Find the common difference: \[ \log_4(2^x - 9) - \log_4 2 = \log_4 \left( \frac{2^x - 9}{2} \right) \] Substituting \( 2^x = 16 \): \[ = \log_4 \left( \frac{16 - 9}{2} \right) = \log_4 \left( \frac{7}{2} \right) \]

Correct option: (D) \( \log_4 \left( \frac{7}{2} \right) \)