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Question

Mathematics Question on Series

If x>0x > 0 and log3x+log3(x)+log3(x4)+log3x8+log3(x16)+....=4,\log_{3} x+\log_{3}\left(\sqrt{x}\right)+\log_{3}\left(\sqrt[4]{x}\right)+\log_{3}\sqrt[8]{x}+\log_{3}\left(\sqrt[16]{x}\right)+....=4, then x equals

A

9

B

81

C

1

D

27

Answer

9

Explanation

Solution

Given : log3x+log3(x)+log3(x4)+log3x8+log3(x16)+=4\log_{3} x+\log_{3}\left(\sqrt{x}\right)+\log_{3}\left(\sqrt[4]{x}\right)+\log_{3}\sqrt[8]{x}+\log_{3}\left(\sqrt[16]{x}\right)+--=4
log3x1+12+14+18+116+=4\Rightarrow\log_{3} x^{^{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+---\infty}}=4
log3x1112=4[S=a1r]\Rightarrow \log_{3} x^{^{1-\frac{1}{\frac{1}{2}}}}=4\quad\quad\left[\because S_{\infty}=\frac{a}{1-r}\right]
log3x2=4x2=34x=9\Rightarrow \log_{3} x^{2}=4\Rightarrow x^{2}=3^{4}\Rightarrow x=9