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Question:

For a real number x, if $\frac{1}{2}$, $\frac{{\log }_3(2^x-9)}{{\log }_{3}4}$, and $\frac{{\log }_5(2^x+{\displaystyle \frac{17}{2}})}{{\log }_{5}4}$ are in arithmetic progression, then the common difference is

Explanation:

Given, $\frac{1}{2}$, $\frac{{\log }_3(2^x-9)}{{\log }_{3}4}$ and $\frac{{\log }_5(2^x+{\displaystyle \frac{17}{2}})}{{\log }_{5}4}$ are in arithmetic progression,

⇒ 1/2, log₄ (2^x - 9) and log₄ (2^x + 17/2) are in AP

⇒ 2 × log₄ (2^x - 9) = 1/2 + log₄ (2^x + 17/2)

⇒ log₄ (2^x - 9)² = log₄ 2 + log₄ (2^x + 17/2)

⇒ log₄ (2^x - 9)² = log₄ 2 + log₄ (2^x + 17/2)

⇒ log₄ (2^x - 9)² = log₄ 2 × (2^x + 17/2)

⇒ (2^x - 9)² = 2 × (2^x + 17/2)

⇒ (a - 9)² = 2a + 17    [Assuming 2^x = a]

⇒ a² - 18a + 81 = 2a + 17

⇒ a² - 20a + 64 = 0

⇒ (a - 16)(a - 4) = 0

⇒ a = 2^x = 16   [4 is rejected as (a - 9) cannot be negative]

Now the first term of the AP = 1/2 and

second term of the AP = log₄ (2^x - 9) = log₄ (16 - 9) = log₄ 7

Common difference = log₄ 7 - 1/2 = log₄ 7 - log₄ 2 = log₄ 7/2

Hence, option (a).