Question

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Answer 1

Solution is given in the image

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Answer 2

QUESTION:

$\sf If \ \left( \dfrac{a^{-1}b^2}{ {a}^{2} {b}^{ - 4} } \right)^{7} \div \left( \dfrac{a^{3}b^{ - 5}}{ {a}^{ - 2} {b}^{3} } \right)^{ - 5} = {a}^{p} . {b}^{q}$

$\sf find\ p\ and\ q$

ANSWER:

• The value of p = 4, q = 2

GIVEN:

$\sf \left( \dfrac{a^{-1}b^2}{ {a}^{2} {b}^{ - 4} } \right)^{7} \div \left( \dfrac{a^{3}b^{ - 5}}{ {a}^{ - 2} {b}^{3} } \right)^{ - 5} = {a}^{p} . {b}^{q}$

EXPLANATION:

$\sf \left( \dfrac{a^{-1}b^2}{ {a}^{2} {b}^{ - 4} } \right)^{7} = \left( \dfrac{a^{-7}b^{14}}{ {a}^{14} {b}^{ - 28} } \right)$

$\boxed{\bold{\large{\gray{\dfrac{x^m}{x^n}= x^{m - n}}}}}$

$\sf \left( \dfrac{a^{-1}b^2}{ {a}^{2} {b}^{ - 4} } \right)^{7} = (a^{-7 - 14})(b^{14 + 28})$

$\sf \left( \dfrac{a^{-1}b^2}{ {a}^{2} {b}^{ - 4} } \right)^{7} = (a^{-21})(b^{ 42})$

$\sf \left( \dfrac{a^{3}b^{ - 5}}{ {a}^{ - 2} {b}^{3} } \right)^{ - 5} = \left( \dfrac{a^{ - 15}b^{ 25}}{ {a}^{ 10} {b}^{ - 15} } \right)$

$\boxed{\bold{\large{\gray{\dfrac{x^m}{x^n}= x^{m - n}}}}}$

$\sf \left( \dfrac{a^{3}b^{ - 5}}{ {a}^{ - 2} {b}^{3} } \right)^{ - 5} = (a^{ - 15 - 10})(b^{ 25 + 15})$

$\sf \left( \dfrac{a^{3}b^{ - 5}}{ {a}^{ - 2} {b}^{3} } \right)^{ - 5} = (a^{ - 25})(b^{40})$

$\sf \left( \dfrac{a^{-1}b^2}{ {a}^{2} {b}^{ - 4} } \right)^{7} \div \left( \dfrac{a^{3}b^{ - 5}}{ {a}^{ - 2} {b}^{3} } \right)^{ - 5} = {a}^{p} . {b}^{q}$

$\sf \dfrac{(a^{-21})(b^{42})}{ (a^{ - 25})(b^{40})}= {a}^{p} . {b}^{q}$

$\boxed{\bold{\large{\gray{\dfrac{x^m}{x^n}= x^{m - n}}}}}$

$\sf (a^{-21 + 25})(b^{ 42 - 40})= {a}^{p} . {b}^{q}$

$\sf (a^{4})(b^{ 2})= {a}^{p} . {b}^{q}$

Bases are equal equate the powers

p = 4, q = 2

HENCE THE VALUE OF p = 4, q = 2.