\( 12 / 29 \) If \( \log _{3} 2, \log _{3}\left(2^{x}-5\right), \log _{3}\left(2^{x}-\frac{7}{2}\right) \) form an arithmetic sequence, then \( x=\ldots \ldots \ldots \ldots \) \( \begin{array}{llll}\text { (a) } 2 & \text { (b) } 3 & \text { (c) } 4 & \text { (d) } 5\end{array} \)

To solve this problem, we need to find the value of \( x \) such that the given logarithmic expressions form an arithmetic sequence. Given: \[ \log _{3} 2, \log _{3}\left(2^{x}-5\right), \log _{3}\left(2^{x}-\frac{7}{2}\right) \] Let's denote the common difference of the arithmetic sequence as \( d \). The common difference between consecutive terms in an arithmetic sequence is constant. Therefore, we can write: \[ \log _{3}\left(2^{x}-5\right) - \log _{3} 2 = d \] \[ \log _{3}\left(2^{x}-\frac{7}{2}\right) - \log _{3}\left(2^{x}-5\right) = d \] Now, we can simplify the expressions using logarithmic properties: \[ \log _{3}\left(\frac{2^{x}-5}{2}\right) = d \] \[ \log _{3}\left(\frac{2^{x}-\frac{7}{2}}{2^{x}-5}\right) = d \] Since the common difference is the same, we can equate the two expressions: \[ \log _{3}\left(\frac{2^{x}-5}{2}\right) = \log _{3}\left(\frac{2^{x}-\frac{7}{2}}{2^{x}-5}\right) \] Now, we can solve for \( x \) by equating the arguments of the logarithms: \[ \frac{2^{x}-5}{2} = \frac{2^{x}-\frac{7}{2}}{2^{x}-5} \] Solving this equation will give us the value of \( x \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{2^{x}-5}{2}=\frac{2^{x}-\frac{7}{2}}{2^{x}-5}\) - step1: Find the domain: \(\frac{2^{x}-5}{2}=\frac{2^{x}-\frac{7}{2}}{2^{x}-5},x\neq \log_{2}{\left(5\right)}\) - step2: Divide the terms: \(\frac{2^{x}-5}{2}=\frac{2^{x+1}-7}{2^{x+1}-10}\) - step3: Cross multiply: \(\left(2^{x}-5\right)\left(2^{x+1}-10\right)=2\left(2^{x+1}-7\right)\) - step4: Simplify the equation: \(2^{2x+1}-20\times 2^{x}+50=2^{x+2}-14\) - step5: Rewrite the expression: \(2\left(2^{2x}-20\times 2^{x-1}+25\right)=2\left(2^{x+1}-7\right)\) - step6: Evaluate: \(2^{2x}-20\times 2^{x-1}+25=2^{x+1}-7\) - step7: Move the expression to the left side: \(2^{2x}-20\times 2^{x-1}+25-\left(2^{x+1}-7\right)=0\) - step8: Calculate: \(2^{2x}-24\times 2^{x-1}+32=0\) - step9: Factor the expression: \(\left(2^{x}-8\right)\left(2^{x}-4\right)=0\) - step10: Separate into possible cases: \(\begin{align}&2^{x}-8=0\\&2^{x}-4=0\end{align}\) - step11: Solve the equation: \(\begin{align}&x=3\\&x=2\end{align}\) - step12: Check if the solution is in the defined range: \(\begin{align}&x=3\\&x=2\end{align},x\neq \log_{2}{\left(5\right)}\) - step13: Find the intersection: \(\begin{align}&x=3\\&x=2\end{align}\) - step14: Rewrite: \(x_{1}=2,x_{2}=3\) The solutions to the equation are \( x = 2 \) and \( x = 3 \). Therefore, the value of \( x \) that makes the given logarithmic expressions form an arithmetic sequence is \( x = 3 \). So, the correct answer is (b) 3.