Question

If $x=x_0$ is solution of the equation $(2x{)}^{{\log }_{5}2}-(3x{)}^{{\log }_{5}3}=0$, then the value of $\left(x_0+\frac{1}{x_0}\right)$ is equal to

• A. $\frac{37}{6}$
• B. $\frac{{\log }_{5}2+{\log }_{5}3}{{\log }_{5}3-{\log }_{5}2}$
• C. ${\log }_{2}3$
• D. 2

Answer 1

Answer: (A)

We have $(2x{)}^{{\log }_{5}2}-(3x{)}^{{\log }_{5}3}$
$\therefore \text{ Taking logarithm to the base }5\text{ on both sides, we get }$
$\left({\log }_{5}2\right)\cdot \left({\log }_{5}2+{\log }_{5}x\right)=\left({\log }_{5}3\right)\cdot \left({\log }_{5}3+{\log }_{5}x\right)$
$\Rightarrow -\left({\log }_{5}3-{\log }_{5}2\right)\cdot {\log }_{5}x=\left({\log }_{5}3-{\log }_{5}2\right)\cdot \left({\log }_{5}3+{\log }_{5}2\right)$
$\Rightarrow {\log }_5\left(\frac{1}{x}\right)={\log }_{5}6\Rightarrow x=\frac{1}{6}\equiv x_0\text{ (Given) }$
Hence, $\left(x_0+\frac{1}{x_0}\right)=\frac{37}{6}$.