Question

The solution set of the equation $\left({\log }_{x}2\right)\left({\log }_{2x}2\right)=\left({\log }_{4x}2\right)$ is

• A. $\{2^{-\sqrt{2}},2^{\sqrt{2}}\}$
• B. $\left\{\frac{1}{3},3\right\}$
• C. $\left\{\frac{1}{4},4\right\}$
• D. None of these

Answer 1

The correct option is A

$\{2^{-\sqrt{2}},2^{\sqrt{2}}\}$

Explanation for the correct option:

Find the solution set of the given equation.

In the question, an equation $\left({\log }_{x}2\right)\left({\log }_{2x}2\right)=\left({\log }_{4x}2\right)$ is given.

We know that, ${\log }_{a}b=\frac{{\log }_{c}b}{{\log }_{c}a}$.

So,

$$\begin{gathered} \frac{1}{\left({\log }_{2}x\right)}\frac{1}{\left({\log }_{2}2x\right)}=\frac{1}{\left({\log }_{2}4x\right)} \\ \Rightarrow \frac{1}{\left({\log }_{2}x\right)}\frac{1}{\left[{\log }_{2}2+\left({\log }_{2}x\right)\right]}=\frac{1}{\left(2\mathrm{lo}{g}_{2}2\right)+\left({\log }_{2}x\right)} \\ \Rightarrow \frac{1}{\left({\log }_{2}x\right)}\frac{1}{\left[1+\left({\log }_{2}x\right)\right]}=\frac{1}{2+\left({\log }_{2}x\right)} \\ \Rightarrow \left({\log }_{2}x\right)\left[1+\left({\log }_{2}x\right)\right]=2+\left({\log }_{2}x\right) \\ \Rightarrow \left({\log }_{2}x\right)+{\left({\log }_{2}x\right)}^2=2+\left({\log }_{2}x\right) \\ \Rightarrow {\left({\log }_{2}x\right)}^2=2 \\ \Rightarrow {\log }_{2}x=\pm \sqrt{2} \\ \Rightarrow x=2^{-\sqrt{2}},2^{\sqrt{2}} \end{gathered}$${Since,${\log }_{a}b=\frac{1}{{\log }_{b}a}$ and $\log \left(ab\right)=\log a+\log b$.}

Therefore, the solution set of the given equation is $\{2^{-\sqrt{2}},2^{\sqrt{2}}\}$.

Hence, option $A$ is the correct option.