Discussion

Explanation:

log₃ 2, log₃ (2^x − 5), log₃ (2^x − 7/2) are in A.P.

∴ 2 × log₃ (2^x − 5) = log₃ 2 + log₃ (2^x − 7/2)

∴ log₃ (2^x − 5)² = log₃ [2 × (2^x − 7/2)]

Let 2^x = a, then we have,

(a − 5)² = 2 × (a − 7/2)

∴ a² − 10a + 25 = 2a − 7

∴ a² − 12a + 32 = 0

∴ a² − 8a − 4a + 32 = 0

(a − 8)(a − 4) = 0

a = 8 or 4

2^x = 8 or 2^x = 4

x = 3 and x = 2

x = 2 cannot be the answer as (2^x − 5) would become negative and logarithms of negative numbers are not defined.

∴ x = 3

Hence, option (d).

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