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

\(\frac{log_3(2^x-9)}{log_34}\) can be written as \(log_4(2^x-9)\), and \(\frac{log_5\bigg(2^x+\frac{17}{2}\bigg)}{log_54}\) can be written as \(log_4\bigg(2^x+\frac{17}{2}\bigg)\)
Hence, \(2log_4(2^x-9)=\frac12+log_4\bigg(2^x+\frac{17}{2}\bigg)\)
\(\frac12\) can be written as \(log_42\)
Therefore,
\(2log_4(2^x-9)=log_4\bigg(2^x+\frac{17}{2}\bigg)\)

\(log_4(2^x-9)^2=log_4\bigg(2^x+\frac{17}{2}\bigg)\)

\((2^x-9)^2=2(2^x+\frac{17}{2})\)
\(2^{2x}-18.2^x+81=2.2^x+17\)
\(2.2^{2x}-20.2^x+64=0\)
\(2.2^{2x}-16.2^x-4.2^x+64=0\)
\(2^x(2^x-16)-4(2^x-16)=0\)
\((2^x-4)(2^x-16)=0\)
The values of \(2^x\) cant be 4 (log will be undefined), which implies The value of \(2^x\) is 16.
 Therefore, the common difference,
\(=log_4 (2^x — 9) — log_42\)
  \(=log_4 7 — log_4 2 = log_4 (\frac72)\)

So, the correct option is (D): \(log_4\bigg(\frac{7}{2}\bigg)\).