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

Expressed as a base-4 logarithm using the change of base formula, the term \( \frac{\log_3(2^x - 9)}{\log_3 4} \) becomes \( \log_4(2^x - 9) \). Similarly, \( \frac{\log_5\left(2^x + \frac{17}{2}\right)}{\log_5 4} \) simplifies to \( \log_4\left(2^x + \frac{17}{2}\right) \).

The given equation is \( 2\log_4(2^x - 9) = \frac{1}{2} + \log_4\left(2^x + \frac{17}{2}\right) \). Since \( \frac{1}{2} \) is equivalent to \( \log_4 2 \), the equation transforms into \( 2\log_4(2^x - 9) = \log_4 2 + \log_4\left(2^x + \frac{17}{2}\right) \).

Applying the logarithmic properties \( a\log_b m = \log_b m^a \) and \( \log_b m + \log_b n = \log_b (mn) \), the left side becomes \( \log_4(2^x - 9)^2 \) and the right side becomes \( \log_4 \left[ 2 \cdot \left(2^x + \frac{17}{2} \right) \right] \). Equating these yields \( \log_4(2^x - 9)^2 = \log_4 \left[ 2 \cdot \left(2^x + \frac{17}{2} \right) \right] \). Removing the logarithms gives \( (2^x - 9)^2 = 2 \cdot \left(2^x + \frac{17}{2} \right) \).

Expanding both sides: The left side is \( (2^x)^2 - 2 \cdot 9 \cdot 2^x + 81 = 2^{2x} - 18 \cdot 2^x + 81 \). The right side is \( 2 \cdot 2^x + 2 \cdot \frac{17}{2} = 2 \cdot 2^x + 17 \). Equating these expressions results in \( 2^{2x} - 18 \cdot 2^x + 81 = 2 \cdot 2^x + 17 \). Rearranging the terms to one side produces the quadratic equation \( 2^{2x} - 20 \cdot 2^x + 64 = 0 \).

Let \( y = 2^x \). The equation becomes \( 2y^2 - 20y + 64 = 0 \). Factoring the quadratic: \( 2y^2 - 16y - 4y + 64 = 0 \), which simplifies to \( 2y(y - 8) - 4(y - 8) = 0 \). This gives \( (y - 4)(2y - 16) = 0 \), yielding solutions \( y = 4 \) or \( y = 8 \). Substituting back \( y = 2^x \), we have \( 2^x = 4 \) or \( 2^x = 16 \), leading to \( x = 2 \) or \( x = 4 \).

Verification of solutions:

• For \( x = 2 \), \( 2^x - 9 = 2^2 - 9 = 4 - 9 = -5 \). The logarithm of a negative number is undefined.
• For \( x = 4 \), \( 2^x - 9 = 2^4 - 9 = 16 - 9 = 7 \). This value is valid.

Therefore, the only valid solution is \( 2^x = 16 \).

The common difference is calculated as \( \log_4(2^x - 9) - \log_4 2 = \log_4 \left( \frac{2^x - 9}{2} \right) \). Substituting \( 2^x = 16 \), the common difference is \( \log_4 \left( \frac{16 - 9}{2} \right) = \log_4 \left( \frac{7}{2} \right) \).

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