Question

If ${\log _3}5 = x$ and ${\log _{25}}11 = y$ then the value of ${\log _3}\left( {\dfrac{{11}}{3}} \right)$ in terms of x and y is

Answer 1

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