Question

If $2^{{\left({\log }_{2}3\right)}^x}=3^{{\left({\log }_{3}2\right)}^x}$ then the value of $x$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{3}$
• D. $\frac{1}{6}$

Answer 1

Answer: (A)

$2^{{\left({\log }_{2}3\right)}^x}=3^{{\left({\log }_{3}2\right)}^x}$ Taking $\log$ to the base 2 on both the sides, we get ${\left({\log }_{2}3\right)}^x\cdot {\log }_{2}2={\left({\log }_{3}2\right)}^x{\log }_{2}3$ ${\left({\log }_{2}3\right)}^{x-1}={\left({\log }_{3}2\right)}^x\Rightarrow \frac{{\left({\log }_{2}3\right)}^{x-1}}{{\left({\log }_{3}2\right)}^x}=1$ ${\left({\log }_{2}3\right)}^{2x-1}=1={\left({\log }_{2}3\right)}^0$ $\Rightarrow 2x-1=0\Rightarrow x=\frac{1}{2}$