(b) 1 – log25
Given, log2 (5.2x + 1), log4 (21– x + 1), 1 are in A.P.
⇒ log2 (5.2x + 1) + 1 = 2 log4 (21 – x + 1)
⇒ log2 (5.2x + 1) + log22 = 2 log22 (21-x + 1)
⇒ log2 (5.2x + 1).2 = 2 x \(\frac12\)log2 (21-x + 1)
\(\big(\because\,\text{log}_{a^n}x=\frac{1}{n}\text{log}_ax\big)\)
⇒ log2 (10.2x + 2) = log2 (21-x + 1)
⇒ 10.2x + 2 = 21 – x + 1 ⇒ 10.2x + 2 = \(\frac{2}{2^x}\) + 1
Let 2x = 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 = \(\frac{2}{5}\) ⇒ 2x = \(\frac{2}{5}\)
\(\big(\because\,2^x>0,\text{reject}\,a=-\frac{1}{2}\big)\)
⇒ log 2x = log \(\frac{2}{5}\)
⇒ x log2 2 = log2 2 – log2 5 ⇒ \(x\) = 1 – log2 5.