Question

If $(x,y)$ is the solution of the following equations ${\left(2x\right)}^{\log 2}={\left(3y\right)}^{\log 3}\mathrm{and}{3}^{\log x}=2^{\log y}$, then $x$ is equal to

• A. $\frac{1}{6}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $6$

Answer 1

The correct option is C

$\frac{1}{2}$

Explanation for the correct option.

The first equation is ${\left(2x\right)}^{\log 2}={\left(3y\right)}^{\log 3}$

By taking log on the both sides we get

$\log 2\left(\log 2+\log x\right)=\log 3\left(\log 3+\log y\right)....\left(1\right)$ (by product and power property of log)

The second equation is $3^{\log x}=2^{\log y}$

By taking log on both sides we get;

$\begin{array}{rcl}\log x\log 3& =& \log y\log 2\left[(bypowerproperty)\right]\\ & \Rightarrow & \frac{\log x\log 3}{\log 2}=\log y\end{array}$

By substituting the value of $\log y$ in $\left(1\right)$, we get

$\begin{array}{rcl}\log 2\left(\log 2+\log x\right)& =& \log 3\left(\log 3+\frac{\log x\log 3}{\log 2}\right)\end{array}$

$\begin{array}{rcl}& & \\ & \Rightarrow & {\left(\log 2\right)}^2+\log 2\log x={\left(\log 3\right)}^2+\frac{{\left(\log 3\right)}^2\log x}{\log 2}\\ \Rightarrow {\left(\log 2\right)}^2-{\left(\log 3\right)}^2& =& \frac{{\left(\log 3\right)}^2\log x}{\log 2}-\log 2\log x\\ \Rightarrow {\left(\log 2\right)}^2-{\left(\log 3\right)}^2& =& \log x\left(\frac{{\left(\log 3\right)}^2-{\left(\log 2\right)}^2}{\log 2}\right)\\ \Rightarrow \frac{\log x}{\log 2}& =& -1\\ \Rightarrow \log x& =& -\log 2\\ \Rightarrow x& =& \frac{1}{2}\\ & & \end{array}$

Hence, option C is correct.