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If $x$ is a positive quantity such that $2^x=3^{\log_52}$, then $x$ is equal to

1. $1+\log_3\dfrac{5}{3}$
2. $\log_58$
3. $1+\log_5\dfrac{3}{5}$
4. $\log_59$

Given that, $2^{x} = 3^{\log_{5}{2}}$

Taking $\log_{2}$ on both sides.

$\log_{2}{2}^{x} = \log_{2} \left( 3^{\log_{5}{2}} \right)$

$\Rightarrow x = \log_{5}{2} \log_{2}{3}$ $\quad [\because{ \log_{a}{a}} = 1, \log_{b}{a}^{x} = x \log_{b}{a}$]

$\Rightarrow x = \frac{\log_{2}{3}}{\log_{2}{5}}$ $\quad \left [ \therefore \log_{a}{b} = \frac{1}{\log_{b}{a}} \right]$

$\Rightarrow \boxed{x = \log_{5}{3}}$ $\quad \left[ \therefore \frac{\log_{c}{a}}{\log_{c}{b}} = \log_{b}{a} \right]$

Now, we can check all the options.

1. $1+\log_{3}{\frac{5}{3}} = 1 + \log_{3}{5} – \log_{3}{3}$

$\qquad \qquad \quad =1 + \log_{3}{5} – 1$

$\qquad \qquad \quad = \log_{3}{5}$

Option $\text {(B)}$ and $\text{(D)}$ are not possible.

1. $1+\log_{5}{\frac{3}{5}} = 1 + \log_{5}{3} – \log_{5}{5}$

$\qquad \qquad \quad=1 + \log_{5}{3} – 1$

$\qquad \qquad \quad = \log_{5}{3}$

Correct Answer $: \text{C}$

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