Question

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

• A. 9
• B. 81
• C. 1
• D. 27

Answer 1

The Correct Option is A

Given : $\log_{3} x+\log_{3}\left(\sqrt{x}\right)+\log_{3}\left(\sqrt[4]{x}\right)+\log_{3}\sqrt[8]{x}+\log_{3}\left(\sqrt[16]{x}\right)+--=4$
$\Rightarrow\log_{3} x^{^{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+---\infty}}=4$
$\Rightarrow \log_{3} x^{^{1-\frac{1}{\frac{1}{2}}}}=4\quad\quad\left[\because S_{\infty}=\frac{a}{1-r}\right]$
$\Rightarrow \log_{3} x^{2}=4\Rightarrow x^{2}=3^{4}\Rightarrow x=9$