Find the value of x, if log2 (5.2x + 1), log4(21–x + 1) and 1 are

1.	Find the value of x, if log2 (5.2x + 1), log4(21–x + 1) and 1 are in A.P.(a) 1 + log52 (b) 1 – log25 (c) log210 (d) log25 + 1
Answer» (b) 1 – log₂5 Given, log₂ (5.2^x + 1), log₄ (2^{1– x} + 1), 1 are in A.P. ⇒ log₂ (5.2^x + 1) + 1 = 2 log₄ (2^{1 – x} + 1) ⇒ log₂ (5.2^x + 1) + log₂2 = 2 log_2²(2^1-x+ 1) ⇒ log₂ (5.2^x + 1).2 = 2 x \(\frac12\)log₂ (2^1-x+ 1) \(\big(\because\,\text{log}_{a^n}x=\frac{1}{n}\text{log}_ax\big)\) ⇒ log₂ (10.2^x + 2) = log₂(2^1-x+ 1) ⇒ 10.2^x + 2 = 2^{1 – x}+ 1 ⇒ 10.2^x + 2 = \(\frac{2}{2^x}\) + 1 Let 2^x= a, then 10. a + 2 = \(\frac{2}{a}\) + 1 ⇒ 10a + 1 = \(\frac{2}{a}\) ⇒ 10 a² + a – 2 = 0 ⇒ (5 a – 2) (2a + 1) = 0 ⇒ a = \(\frac{2}{5}\) ⇒ 2^x = \(\frac{2}{5}\) \(\big(\because\,2^x>0,\text{reject}\,a=-\frac{1}{2}\big)\) ⇒ log 2^x = log \(\frac{2}{5}\) ⇒ x log₂ 2 = log₂ 2 – log₂ 5 ⇒ \(x\) = 1 – log₂ 5.

Find the value of x, if log2 (5.2x + 1), log4(21–x + 1) and 1 are in A.P.(a) 1 + log52 (b) 1 – log25 (c) log210 (d) log25 + 1

Answer»

(b) 1 – log₂5

Given, log₂ (5.2^x + 1), log₄ (2^{1– x} + 1), 1 are in A.P.

⇒ log₂ (5.2^x + 1) + 1 = 2 log₄ (2^{1 – x} + 1)

⇒ log₂ (5.2^x + 1) + log₂2 = 2 log_2²(2^1-x+ 1)

⇒ log₂ (5.2^x + 1).2 = 2 x \(\frac12\)log₂ (2^1-x+ 1)

\(\big(\because\,\text{log}_{a^n}x=\frac{1}{n}\text{log}_ax\big)\)

⇒ log₂ (10.2^x + 2) = log₂(2^1-x+ 1)

⇒ 10.2^x + 2 = 2^{1 – x}+ 1 ⇒ 10.2^x + 2 = \(\frac{2}{2^x}\) + 1

Let 2^x= a, then

10. a + 2 = \(\frac{2}{a}\) + 1 ⇒ 10a + 1 = \(\frac{2}{a}\) ⇒ 10 a² + a – 2 = 0

⇒ (5 a – 2) (2a + 1) = 0 ⇒ a = \(\frac{2}{5}\) ⇒ 2^x = \(\frac{2}{5}\)

\(\big(\because\,2^x>0,\text{reject}\,a=-\frac{1}{2}\big)\)

⇒ log 2^x = log \(\frac{2}{5}\)

⇒ x log₂ 2 = log₂ 2 – log₂ 5 ⇒ \(x\) = 1 – log₂ 5.