Question

Find
$( { \frac{81}{16} })^{ \frac{ - 3}{4} } \times ( { \frac{25}{9} })^{ \frac{ - 3}{2} } \div ({ \frac{5}{2} })^{ - 3}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\: \bigg( { \dfrac{81}{16} } \bigg)^{ \dfrac{ - 3}{4} } \times \bigg( { \dfrac{25}{9} } \bigg)^{ \dfrac{ - 3}{2} } \div \bigg({ \dfrac{5}{2} } \bigg)^{ - 3}$

$\rm \: \: = \: \: \bigg( { \dfrac{16}{81} } \bigg)^{ \dfrac{3}{4} } \times \bigg( { \dfrac{9}{25} } \bigg)^{ \dfrac{3}{2} } \div \bigg({ \dfrac{2}{5} } \bigg)^{3}$

$\: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {\bigg(\dfrac{a}{b} \bigg) }^{ - m} = {\bigg(\dfrac{b}{a} \bigg) }^{m}\bigg \}}$

$\rm \: \: = \: \: \bigg( { \dfrac{2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3} } \bigg)^{ \dfrac{3}{4} } \times \bigg( { \dfrac{3 \times 3}{5 \times 5} } \bigg)^{ \dfrac{3}{2} } \times \bigg({ \dfrac{5}{2} } \bigg)^{3}$

$\rm \: \: = \: \: \bigg( { \dfrac{2}{3} } \bigg)^{4 \times \dfrac{3}{4} } \times \bigg( { \dfrac{3}{5} } \bigg)^{2 \times \dfrac{3}{2} } \times \bigg({ \dfrac{5}{2} } \bigg)^{3}$

$\rm \: \: = \: \: \bigg( { \dfrac{2}{3} } \bigg)^{3} \times \bigg( { \dfrac{3}{5} } \bigg)^{3} \times \bigg({ \dfrac{5}{2} } \bigg)^{3}$

$\rm \: \: = \: {\bigg(\dfrac{2}{3} \times \dfrac{3}{5} \times \dfrac{5}{2} \bigg) }^{3}$

$\: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {x}^{m} \times {y}^{m} = {(xy)}^{m} \bigg \}}$

$\rm \: \: = \: {(1)}^{3}$

$\rm \: \: = \: 1$

Hence,

$\bf :\longmapsto\:\: \bigg( { \dfrac{81}{16} } \bigg)^{ \dfrac{ - 3}{4} } \times \bigg( { \dfrac{25}{9} } \bigg)^{ \dfrac{ - 3}{2} } \div \bigg({ \dfrac{5}{2} } \bigg)^{ - 3} = 1$

Additional Information :-

$\green{\boxed{ \tt \: {x}^{m} \times {x}^{n} = {x}^{m + n}}}$

$\green{\boxed{ \tt \: {x}^{m} \div {x}^{n} = {x}^{m - n}}}$

$\green{\boxed{ \tt \: {( {x}^{m}) }^{n} = {x}^{mn} }}$

$\green{\boxed{ \tt \: {x}^{0} = 1 }}$

$\green{\boxed{ \tt \: {x}^{ - y} = \dfrac{1}{ {x}^{y} } }}$

$\green{\boxed{ \tt \: {x}^{m} \times {y}^{m} = {(xy)}^{m} }}$

Answer 2

Correct Question:-

$</u><u> \bigg( \frac{81}{16} \bigg)^{ \frac{ - 3}{4} } \times \left\{{ \bigg( \frac{25}{9} \bigg) ^{ \frac{ - 3}{2} } \div \bigg( \frac{5}{2} \bigg)^{ - 3} }\right\}$

$⟹ </u><u> \bigg( \frac{16}{81} \bigg)^{ \frac{ 3}{4} } \times \left\{{ \bigg( \frac{9}{25} \bigg) ^{ \frac{3}{2} } \div \bigg( \frac{2}{5} \bigg)^{ 3} }\right\}$

$⟹ \bigg( \frac{2}{3} \bigg) ^{4 \times \frac{3}{4} } \times \left \{ \bigg( \frac{3}{5} \bigg)^{3} \times \bigg( \frac{5}{2} \bigg)^{3} \right \}$

$⟹ \bigg( \frac{2}{3} \bigg) ^{ \cancel{4 }\times \frac{3}{ \cancel{4}} } \times \left \{ \bigg( \frac{3}{5} \bigg)^{3} \times \bigg( \frac{5}{2} \bigg)^{3} \right \}$

$⟹ \bigg( \frac{2}{3} \bigg) ^{3 } \times \left \{ \bigg( \frac{3}{5} \bigg)^{3} \times \bigg( \frac{5}{2} \bigg)^{3} \right \}$

$⟹ \bigg( \frac{2}{3} \bigg) ^{3} \times \bigg( \frac{ {3}^{3} }{ {5}^{3} } \times \frac{ {5}^{3} }{ {2}^{3} } \bigg)$

$⟹ \bigg( \frac{2}{3} \bigg) ^{3} \times \bigg( \frac{ {3}^{3} }{ \cancel{{5}^{3}} } \times \frac{ \cancel{{5}^{3}} }{ {2}^{3} } \bigg)$

$⟹ \bigg( \frac{ {2}^{3} }{ {3}^{3} } \bigg) \times \bigg( \frac{ {3}^{3} }{ {2}^{3} } \bigg)$

$⟹ \frac{ {2}^{3} }{ {3}^{ {3} } } \times \frac{ {3}^{3} }{ {2}^{3} }$

$⟹ \frac{\cancel{ {2}^{3} }}{ \cancel{{3}^{ {3} }} } \times \frac{ \cancel{{3}^{3}} }{ \cancel{{2}^{3}} }$

$⟹ \: 1 \: \: Ans.$

Know more laws of Integral Exponents

For any two real numbers a and b, a, b ≠ 0, and for any two positive integers, m and n

➲ If a be any non - zero rational number, then

a^0 = 1

➲ If a be any non - zero rational number and m,n be integer, then

(a^m)^n = a^mn

➲ If a be any non - zero rational number and m be any positive integer, then

a^-m = 1/a^m

➲ If a/b is a rational number and m is a positive integer, then

(a/b)^m = a^m/b^m

➲ For any Integers m and n and any rational number a, a ≠ 0

a^m × a^n = a^m+n

➲ For any Integers m and n for non - zero rational number a,

a^m ÷ a^n = a^m-n

➲ If a and b are non - zero rational numbers and m is any integer, then

(a+b)^m = a^m × b^m

• I hope it's help you...☺