# If a and b are different positive primes such that$\left(\frac{a^{-1} b^{2}}{a^{2} b^{-4}}\right)^{7} \p\left(\frac{a^{3} b^{-5}}{a^{-2} b^{3}}\right)=a^{x} b^{y}$, find $x$ and $y .$

Given:

a and b are different positive primes such that

$\left(\frac{a^{-1} b^{2}}{a^{2} b^{-4}}\right)^{7} \div\left(\frac{a^{3} b^{-5}}{a^{-2} b^{3}}\right)=a^{x} b^{y}$.

To do:

We have to find $x$ and $y$.

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

$(\frac{a^{-1} b^{2}}{a^{2} b^{-4}})^{7} \div(\frac{a^{3} b^{-5}}{a^{-2} b^{3}})=a^{x} b^{y}$

$\frac{a^{-7} b^{14}}{a^{14}b^{-28}} \div \frac{a^{3} b^{-5}}{a^{-2} b^{3}}=a^{x} b^{y}$

$\frac{a^{-7} b^{14}}{a^{14}b^{-28}} \times \frac{a^{-2} b^{3}}{a^{3} b^{-5}}=a^{x} b^{y}$

$a^{-7-14-2-3} \times b^{14+28+3+5}=a^{x} b^{y}$

$a^{-26} \times b^{50}=a^{x} b^{y}$

Comparing both sides, we get,

$x=-26$ and $y=50$

The values of $x$ and $y$ are $-26$ and $50$ respectively.

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Updated on: 10-Oct-2022

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