The solution of log_{sqrt{3}} x + log_{sqrt[4 | vedclass

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(D) Given equation: $\log_{\sqrt{3}} x + \log_{\sqrt[4]{3}} x + \log_{\sqrt[6]{3}} x + \dots + \log_{\sqrt[16]{3}} x = 36$Using the property $\log_{a^

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$x = \sqrt{3}$

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(D) Given equation: $\log_{\sqrt{3}} x + \log_{\sqrt[4]{3}} x + \log_{\sqrt[6]{3}} x + \dots + \log_{\sqrt[16]{3}} x = 36$Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$,we can write $\log_{3^{1/n}} x = n \log_3 x$.Substituting this into the equation:$2 \log_3 x + 4 \log_3 x + 6 \log_3 x + \dots + 16 \log_3 x = 36$Factor out $\log_3 x$:$(\log_3 x) (2 + 4 + 6 + \dots + 16) = 36$The sum of the arithmetic progression $2 + 4 + 6 + \dots + 16$ is given by $\frac{n}{2}(a + l)$,where $n=8$,$a=2$,and $l=16$:Sum $= \frac{8}{2}(2 + 16) = 4 	imes 18 = 72$.So,$(\log_3 x) 	imes 72 = 36$$\log_3 x = \frac{36}{72} = \frac{1}{2}$$x = 3^{1/2} = \sqrt{3}$.

The solution of $\log_{\sqrt{3}} x + \log_{\sqrt[4]{3}} x + \log_{\sqrt[6]{3}} x + \dots + \log_{\sqrt[16]{3}} x = 36$ is

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