Question

The solution set of the equation $\left[4{\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+.....\right)}^{{\log }_{2}x}\right]={\left[54\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+.....\right)\right]}^{{\log }_{x}2}$ is

• A. $\left\{4,\frac{1}{4}\right\}$
• B. $\left\{2,\frac{1}{2}\right\}$
• C. $\left\{1,2\right\}$
• D. $\left\{8,\frac{1}{8}\right\}$

Answer 1

The correct option is A

$\left\{4,\frac{1}{4}\right\}$

Explanation for the correct option:

Find the solution set of the given equation.

In the question, an equation $\left[4{\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+.....\right)}^{{\log }_{2}x}\right]={\left[54\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+.....\right)\right]}^{{\log }_{x}2}$ is given.

We know that the sum of infinite GP can be given by $S_{\infty }=\frac{a}{1-r}$. where $a$ is the first term and $r$ is the common ratio,

Therefore,

$$\begin{gathered} \left[4{\left(\frac{1}{1+{\displaystyle \frac{1}{3}}}\right)}^{{\log }_{2}x}\right]={\left[54\left(\frac{1}{1-{\displaystyle \frac{1}{3}}}\right)\right]}^{{\log }_{x}2} \\ \Rightarrow \left[4{\left(\frac{1}{1+{\displaystyle \frac{1}{3}}}\right)}^{{\log }_{2}x}\right]={\left[54\left(\frac{1}{1-{\displaystyle \frac{1}{3}}}\right)\right]}^{{\log }_{x}2} \\ \Rightarrow \left[4{\left(\frac{3}{4}\right)}^{{\log }_{2}x}\right]={\left[54\left(\frac{3}{2}\right)\right]}^{{\log }_{x}2} \\ \Rightarrow 3^{{\log }_{2}x}={81}^{{\log }_{x}2} \\ \Rightarrow 3^{{\log }_{2}x}=3^{4{\log }_{x}2} \\ \Rightarrow {\log }_{2}x=4{\log }_{x}2 \\ \Rightarrow {\left({\log }_{2}x\right)}^2=4 \\ \Rightarrow \left({\log }_{2}x\right)=\pm 2 \\ \Rightarrow x=2^{-2},2^2 \end{gathered}$${Since, ${\log }_{a}b=\frac{1}{{\log }_{b}a}$ and $a^{{\log }_{a}b}=b$}

Therefore, the solution set of the given equation is $\left\{4,\frac{1}{4}\right\}$.

Hence, option $A$ is the correct answer.