Question:
If $\log_3 2$, $\log_3 (2^x - 5)$ and $\log_3 \left(2^x - \frac{7}{2}\right)$ are in AP, the value of $x$ is
If $\log_3 2$, $\log_3 (2^x - 5)$ and $\log_3 \left(2^x - \frac{7}{2}\right)$ are in AP, the value of $x$ is
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In AP: $2b = a + c$. Also, argument of log must be positive.
Updated On: May 2, 2026
- $2$
- $3$
- $0$
- $\frac{1}{3}$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
In AP, $2 \times \text{middle term} = \text{first term} + \text{third term}$.
Step 2: Detailed Explanation:
Given: $\log_3 2$, $\log_3 (2^x - 5)$, $\log_3 (2^x - 7/2)$ are in AP.
So $2\log_3 (2^x - 5) = \log_3 2 + \log_3 (2^x - 7/2)$
$\Rightarrow \log_3 (2^x - 5)^2 = \log_3 \left[2 \cdot (2^x - 7/2)\right]$
$\Rightarrow (2^x - 5)^2 = 2 \cdot (2^x - 7/2) = 2^{x+1} - 7$
Let $t = 2^x$. Then $(t-5)^2 = 2t - 7 \Rightarrow t^2 - 10t + 25 = 2t - 7 \Rightarrow t^2 - 12t + 32 = 0$
$\Rightarrow (t-4)(t-8) = 0 \Rightarrow t = 4$ or $t = 8$.
For $t=4$, $2^x = 4 \Rightarrow x=2$, but then $2^x - 5 = -1$ (log undefined). So $t=8 \Rightarrow x=3$ works.
Step 3: Final Answer:
$x = 3$.
In AP, $2 \times \text{middle term} = \text{first term} + \text{third term}$.
Step 2: Detailed Explanation:
Given: $\log_3 2$, $\log_3 (2^x - 5)$, $\log_3 (2^x - 7/2)$ are in AP.
So $2\log_3 (2^x - 5) = \log_3 2 + \log_3 (2^x - 7/2)$
$\Rightarrow \log_3 (2^x - 5)^2 = \log_3 \left[2 \cdot (2^x - 7/2)\right]$
$\Rightarrow (2^x - 5)^2 = 2 \cdot (2^x - 7/2) = 2^{x+1} - 7$
Let $t = 2^x$. Then $(t-5)^2 = 2t - 7 \Rightarrow t^2 - 10t + 25 = 2t - 7 \Rightarrow t^2 - 12t + 32 = 0$
$\Rightarrow (t-4)(t-8) = 0 \Rightarrow t = 4$ or $t = 8$.
For $t=4$, $2^x = 4 \Rightarrow x=2$, but then $2^x - 5 = -1$ (log undefined). So $t=8 \Rightarrow x=3$ works.
Step 3: Final Answer:
$x = 3$.
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