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

(a) 1 + log5

(b) 1 – log2

(c) log210 

(d) log25 + 1

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(b) 1 – log2

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) = log(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.

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