Question

If log₁₀2, log₁₀ (2^x – 1), and log₁₀ (2^x + 3) be three consecutive terms of an A.P., then find the value of  x.

Answer 1

 Given,

 log₁₀ 2, log₁₀ (2^x – 1), log₁₀ (2^x + 3) are in A.P. Then,

log₁₀ (2^x – 1) – log₁₀ 2 = log₁₀ (2^x + 3) – log₁₀ (2^x – 1)

⇒ \(log_{10}(\frac{2^x-1}{2})\) = \(log_{10}(\frac{2^x+3}{2^x-1})\)

⇒ \(\frac{2^x-1}{2}\)  =  \(\frac{2^x+3}{2^x-1}\) 

⇒ (2^x – 1)² = 2 (2^x + 3)

⇒ 2^2x – 2.2^x + 1 = 2.2^x + 6 

⇒ 2^2x – 4.2^x – 5 = 0 

⇒ 2^2x – 5.2^x + 2^x – 5 = 0 

⇒ 2^x (2^x – 5) + 1 (2^x – 5) = 0 

⇒ (2^x – 5) (2^x + 1) = 0 

⇒ 2^x = 5 (\(\because\) 2x ≠ –1) 

⇒ x = log₂ 5.