Gujarati
8. Sequences and Series
medium

${\log _{\sqrt 3 }}x + {\log _{\sqrt[4]{3}}}x + {\log _{\sqrt[6]{3}}}x + ..... + {\log _{\sqrt[{16}]{3}}}x = 36$ का हल है

A

$x = 3$

B

$x = 4\sqrt 3 $

C

$x = 9$

D

$x = \sqrt 3 $

Solution

(d) ${\log _{\sqrt 3 }}x + {\log _{\sqrt[4]{3}}}x + {\log _{\sqrt[6]{3}}}x + …… + {\log _{\sqrt[{16}]{3}}}x = 36$

$ \Rightarrow $$\frac{1}{{{{\log }_x}\sqrt 3 \,}} + \frac{1}{{{{\log }_x}\sqrt[4]{3}}} + \frac{1}{{{{\log }_x}\sqrt[6]{3}}} + … + \frac{1}{{{{\log }_x}\sqrt[{16}]{3}}} = 36$

$ \Rightarrow $ $\frac{1}{{(1/2){{\log }_x}3}} + \frac{1}{{(1/4){{\log }_x}3}} + \frac{1}{{(1/6){{\log }_x}3}} + ….. + \frac{1}{{(1/16){{\log }_x}3}} = 36$

  $ \Rightarrow $ $({\log _3}x)(2 + 4 + 6 + ….. + 16) = 36$

 $ \Rightarrow $ $({\log _3}x)\frac{8}{2}[2 + 16] = 36$

$ \Rightarrow $${\log _3}x = \frac{1}{2}$

 $ \Rightarrow $$x = {3^{1/2}}$

$ \Rightarrow x = \sqrt 3 $.

Standard 11
Mathematics

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