Question

simplify with positive exponents: [(9/11)-³×(9/11)power is -7] ÷(9/11)-³​

Answer 1

$\bf \underline{ \underline{\maltese\:Solution } }$

$\sf \implies \bigg[ \bigg( \dfrac{9}{11} \bigg)^{ - 3} \times \bigg( \dfrac{9}{11} \bigg)^{ - 7} \bigg] \div \bigg( \dfrac{9}{11} \bigg)^{ - 3} \bigg]$

$\sf \implies \bigg( \dfrac{9}{11} \bigg)^{ - 3 + ( - 7)} \div \bigg( \dfrac{11}{9} \bigg)^{ 3}$

$\sf \implies \bigg( \dfrac{9}{11} \bigg)^{ - 10} \div \bigg( \dfrac{11}{9} \bigg)^{ 3}$

$\sf \implies \bigg( \dfrac{11}{9} \bigg)^{10} \div \bigg( \dfrac{11}{9} \bigg)^{ 3}$

$\sf \implies \bigg( \dfrac{11}{9} \bigg)^{10} \times \bigg( \dfrac{9}{11} \bigg)^{ 3}$

$\sf \implies\dfrac{(11) ^{10} }{(9)^{10} } \times \dfrac{(9)^{3} }{(11)^{3} }$

$\sf \implies\dfrac{11^{7} \times \cancel{11^{3}} }{9^{7} \times \cancel{9^{3}} } \times \dfrac{ \cancel{9^{3}} }{ \cancel{11^{3} } }$

$\sf \implies \dfrac{11^{7} }{9^{7}} = \bigg( \dfrac{11}{9} \bigg)^{7}$

$\boxed{ \bf \red{ Answer = \bigg( \dfrac{11}{9} \bigg)^{7} }}$

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$\bf \underline{ \underline{\maltese\:Laws \: of \: exponents } }$

$\sf \: \bf{ 1^{st} \: Law} = \bigg( \dfrac{m}{n} \bigg)^{a} \times \bigg( \dfrac{m}{n} \bigg)^{b} = \bigg( \dfrac{m}{n} \bigg)^{a + b}$

$\sf \bf{2^{nd} \: Law} =$

$\sf Case : (i) \: if \: a > b \: then, \bigg( \dfrac{m}{n}\bigg) ^{a} \div \bigg( \dfrac{m}{n}\bigg)^{b} = \bigg( \dfrac{m}{n}\bigg)^{a - b}$

$\sf Case : (ii) \: if \: a < b \: then, \bigg( \dfrac{m}{n}\bigg) ^{a} \div \bigg( \dfrac{m}{n}\bigg)^{b} = \dfrac{1}{\bigg( \dfrac{m}{n}\bigg)^{b - a}}$

$\sf \: \bf{3^{rd} \: Law }= \bigg\{ \bigg( \dfrac{m}{n} \bigg)^{a} \bigg\}^{b} = \bigg( \dfrac{m}{n} \bigg)^{a \times b} =\bigg( \dfrac{m}{n} \bigg)^{ab}$

$\sf \: \bf{4^{th} \: Law }= \bigg( \dfrac{m}{n} \bigg)^{ - 1} = \bigg( \dfrac{n}{m} \bigg) =\dfrac{n}{m}$

$\sf \: \bf{5^{th} \: Law }= \bigg( \dfrac{m}{n} \bigg)^{0} = 1$