Question

If $x=x_0$ is solution of the equation $(2 x)^{\log _5 2}-(3 x)^{\log _5 3}=0$, 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 $(2 x)^{\log _5 2}-(3 x)^{\log _5 3}$
$\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, $\left(x_0+\frac{1}{x_0}\right)=\frac{37}{6}$.