Discussion

Explanation:

−16, 2x+3 − 22x−1 − 16, 22x−1 + 16 are in AP

If a, b and c are in AP ⇒ 2b = a + c

∴ 2 × (2x+3 − 22x−1 − 16) = -16 + 22x−1 + 16

⇒ 2x+4 − 22x − 32) = 22x−1

⇒ 16 × 2− 22x − 32 = 22x​​/2

⇒ 32 × 2− 2 × 22x − 64 = 22x​​

Take 2x = a and 22x = a2 

⇒ 32a − 2a− 64 = a2​​

⇒ 3a2​​ - 32a + 64 = 0

⇒ 3a2​​ - 8a - 24a + 64 = 0

⇒ (3a - 8)(a - 8) = 0

⇒ a = 8/3 or 8

⇒ 2x = 8/3 or 8 

[8/3 is rejected as 2x cannot be a fraction]

 

⇒ 2x = 8 = 23 

⇒ x = 3 

Hence, 3.

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