Question

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

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

Answer 1

The correct option is D

$x=\sqrt{3}$

Explanation for the correct option:

Given: ${\log }_{\sqrt{3}}x+{\log }_{\sqrt[4]{3}}x+{\log }_{\sqrt[6]{3}}x+.........+{\log }_{\sqrt[16]{3}}x=36$

$$\begin{gathered} {\log }_{\sqrt{3}}x+{\log }_{\sqrt[4]{3}}x+{\log }_{\sqrt[6]{3}}x+.........+{\log }_{\sqrt[16]{3}}x=36 \\ \Rightarrow \frac{\log x}{\log {\left(3\right)}^{{\displaystyle \frac{1}{2}}}}+\frac{\log x}{\log {\left(3\right)}^{{\displaystyle \frac{1}{4}}}}+\frac{\log x}{\log {\left(3\right)}^{{\displaystyle \frac{1}{6}}}}+...........+\frac{\log x}{\log {\left(3\right)}^{{\displaystyle \frac{1}{16}}}}=36\left[\mathrm{as}{\log }_{a}b=\frac{\mathrm{logb}}{\mathrm{loga}}\right] \\ \Rightarrow \frac{\log x}{{\displaystyle \frac{1}{2}}\log 3}+\frac{\log x}{{\displaystyle \frac{1}{4}}\log 3}+\frac{\log x}{{\displaystyle \frac{1}{6}}\log 3}+...........+\frac{\log x}{{\displaystyle \frac{1}{16}}\log 3}=36\left[{\mathrm{loga}}^b=\mathrm{bloga}\right] \\ \Rightarrow \frac{\log x}{\log 3}\left(2+4+6+........+16\right)=36 \\ \Rightarrow 2\frac{\log x}{\log 3}\left(1+2+3+........+8\right)=36 \\ \Rightarrow \frac{\log x}{\log 3}\left(36\right)=18 \\ \Rightarrow \log x=\frac{1}{2}\log 3 \\ \Rightarrow \log x=\log 3^{\frac{1}{2}} \end{gathered}$$

comparing both sides

$x=\sqrt{3}$

Hence, option D is correct.