Question

If log₃2, log₃ (2^x – 5) and log₃ (2^x – 7/2) are in A.P., then x is equal to

• A. 2
• B. 3
• C. 4
• D. (2, 3)

Answer 1

Answer: (B)

3

Answer 2

Given log₃2, log₃(2^x – 5), log₃$\left( 2 ^ { x } - \frac { 7 } { 2 } \right)$ are in A.P.

̃ 2 log₃(2^x – 5) = log₃2 + log₃ $\left( 2 ^ { x } - \frac { 7 } { 2 } \right)$

(2^x – 5)² = 2$\left( 2 ^ { x } - \frac { 7 } { 2 } \right)$

Put 2^x = t and get t² – 12t + 32 = 0

(t – 8)(t – 4)= 0 ̃ t = 8, 4

2^x = 8 or 2^x = 4

x = 3 x ¹ 2 Q 2^x – 5 < 0 for x = 2