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**Hint:**${\log _3}5 = x$and ${\log _{25}}11 = y$ we will some property of logarithmic

${\log _a}b = \dfrac{{\log b}}{{\log a}}$ and ${\log _{25}}11 = \dfrac{{\log 11}}{{\log 25}}$ .

$\log {a^b} = b\log a$then ${\log _{25}}11 = \dfrac{{\log 11}}{{2\log 5}}$. Then we will divide it with $\log 3$ then we get ${\log _{25}}11 = \dfrac{{{{\log }_3}11}}{{2{{\log }_3}5}}$ from this we will get the value of ${\log _3}11$

And $\log \dfrac{a}{b} = \log a - \log b$ then ${\log _3}\dfrac{{11}}{3} = {\log _3}11 - {\log _3}3$ and ${\log _a}a = 1$ then substituting all the values we will get the answer.

**Complete step-by-step answer:**given ${\log _3}5 = x$ and ${\log _{25}}11 = y$

it is known that ${\log _a}b = \dfrac{{\log b}}{{\log a}}$ then

${\log _{25}}11 = y$

$ \Rightarrow {\log _{25}}11 = \dfrac{{\log 11}}{{\log 25}} = y$

We know that $\log {a^b} = b\log a$ so,

$y = \dfrac{{\log 11}}{{2\log 5}}$

Dividing denominator and numerator with $\log 3$ then we get

\[y = \dfrac{{\dfrac{{\log 11}}{{\log 3}}}}{{\dfrac{{2\log 5}}{{\log 3}}}} = y = \dfrac{{{{\log }_3}11}}{{2{{\log }_3}5}}\]

According to question ${\log _3}5 = x$

Then ${\log _3}11 = 2yx$ …. (1)

Now, ${\log _3}\left( {\dfrac{{11}}{3}} \right) = {\log _3}11 - {\log _3}3$ as $\left[ {\log \dfrac{a}{b} = \log a - \log b} \right]$

And we know that ${\log _a}a = 1$ then ${\log _3}3 = 1$ and substituting the value (1) we get

${\log _3}\left( {\dfrac{{11}}{3}} \right) = 2xy - 1$

**Note:**Properties used in question are

${\log _a}b = \dfrac{{\log b}}{{\log a}}$

$\log {a^b} = b\log a$

$\log \dfrac{a}{b} = \log a - \log b$

${\log _a}a = 1$

If there is nothing is written is base then it has a default 10

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