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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. $\log_59$
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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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