Question

The solution of ${\log }_{\sqrt{3}}x+{\log }_{\sqrt[4]{43}}x+{\log }_{\sqrt[6]{63}}x+......+{\log }_{\sqrt[16]{163}}x=36$ is

• A. $x=3$
• B. $x=4\sqrt{3}$
• C. $x=9$
• D. $x=\sqrt{3}$

Answer 1

The correct option is D $x=\sqrt{3}$

We have, ${\log }_{\sqrt{3}}x+{\log }_{\sqrt[4]{43}}x+{\log }_{\sqrt[6]{63}}x+...+{\log }_{\sqrt[16]{163}}x=36.$
$\Rightarrow \frac{1}{{\log }_x\sqrt{3}}+\frac{1}{{\log }_x\sqrt[4]{43}}+\frac{1}{{\log }_x\sqrt[6]{63}}+...+\frac{1}{{\log }_x\sqrt[16]{163}}=36$
$\Rightarrow \frac{1}{\frac{1}{2}{\log }_{x}3}+\frac{1}{\frac{1}{4}{\log }_{x}3}+\frac{1}{\frac{1}{6}{\log }_{x}3}+...+\frac{1}{\frac{1}{16}{\log }_{x}3}=36$
$\Rightarrow \frac{1}{{\log }_{x}3}[2+4+6+...+16]=36$
$\Rightarrow ({\log }_{3}x)72=36$
$\Rightarrow {\log }_{3}x=\frac{1}{2}$
$\Rightarrow x=3^{\frac{1}{2}}$
$\Rightarrow x=\sqrt{3}$