Question :- Simplify the following :
[tex]\sf \: \dfrac{ {(15)}^{m + n} \times {(10)}^{n + p} \times {(6)}^{n} }{ {(10)}^{m + n} \times {(6)}^{n + p} } \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given expression is
We know,
[tex]\sf \: \boxed{ \sf{ \: \frac{ {x}^{m} }{ {x}^{n} } = {x}^{m - n} \: }} \\ \\ [/tex]
So, using this law of exponents, we get
[tex]\sf \: = \: {(15)}^{m + n} \times {(10)}^{n + p - (m + n)} \times {(6)}^{n - (n + p)} \\ \\ [/tex]
[tex]\sf \: = \: {(15)}^{m + n} \times {(10)}^{n + p - m - n} \times {(6)}^{n - n - p} \\ \\ [/tex]
[tex]\sf \: = \: {(15)}^{m + n} \times {(10)}^{ p - m} \times {(6)}^{ - p} \\ \\ [/tex]
[tex]\sf \: = \: {(5 \times 3)}^{m + n} \times {(5 \times 2)}^{ p - m} \times {(3 \times 2)}^{ - p} \\ \\ [/tex]
[tex]\sf \: = \: {(5)}^{m + n} \times {(3)}^{m + n} \times {(5)}^{ p - m} \times {(2)}^{p - m} \times {(3)}^{ - p} \times {(2)}^{ - p} \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: {(x \times y)}^{n} = {x}^{n} \times {y}^{n} \: }} \\ \\ [/tex]
[tex]\sf \: = \: {(5)}^{m + n + p - m} \times {(3)}^{m + n - p} \times {(2)}^{p - m - p} \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }} \\ \\ [/tex]
[tex]\sf \: = \: {(5)}^{n + p} \times {(3)}^{m + n - p} \times {(2)}^{- m} \\ \\ [/tex]
Hence,
[tex]\sf \: \dfrac{ {(15)}^{m + n} \times {(10)}^{n + p} \times {(6)}^{n} }{ {(10)}^{m + n} \times {(6)}^{n + p} } = {(5)}^{n + p}{(3)}^{m + n - p}{(2)}^{- m} \\ \\ [/tex]
Or
[tex]\boxed{ \sf{ \:\sf \: \dfrac{ {(15)}^{m + n} \times {(10)}^{n + p} \times {(6)}^{n} }{ {(10)}^{m + n} \times {(6)}^{n + p} } = \frac{{(5)}^{n + p}{(3)}^{m + n - p} }{ {2}^{m} } \: }} \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ {x}^{0} = 1}\\ \\ \bigstar \: \bf{ {x}^{m} \times {x}^{n} = {x}^{m + n} }\\ \\ \bigstar \: \bf{ {( {x}^{m})}^{n} = {x}^{mn} }\\ \\\bigstar \: \bf{ {x}^{m} \div {x}^{n} = {x}^{m - n} }\\ \\ \bigstar \: \bf{ {x}^{ - n} = \dfrac{1}{ {x}^{n} } }\\ \\\bigstar \: \bf{ {\bigg(\dfrac{a}{b} \bigg) }^{ - n} = {\bigg(\dfrac{b}{a} \bigg) }^{n} }\\ \\\bigstar \: \bf{ {x}^{m} = {x}^{n}\rm\implies \:m = n }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]
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Verified answer
Question :- Simplify the following :
[tex]\sf \: \dfrac{ {(15)}^{m + n} \times {(10)}^{n + p} \times {(6)}^{n} }{ {(10)}^{m + n} \times {(6)}^{n + p} } \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given expression is
[tex]\sf \: \dfrac{ {(15)}^{m + n} \times {(10)}^{n + p} \times {(6)}^{n} }{ {(10)}^{m + n} \times {(6)}^{n + p} } \\ \\ [/tex]
We know,
[tex]\sf \: \boxed{ \sf{ \: \frac{ {x}^{m} }{ {x}^{n} } = {x}^{m - n} \: }} \\ \\ [/tex]
So, using this law of exponents, we get
[tex]\sf \: = \: {(15)}^{m + n} \times {(10)}^{n + p - (m + n)} \times {(6)}^{n - (n + p)} \\ \\ [/tex]
[tex]\sf \: = \: {(15)}^{m + n} \times {(10)}^{n + p - m - n} \times {(6)}^{n - n - p} \\ \\ [/tex]
[tex]\sf \: = \: {(15)}^{m + n} \times {(10)}^{ p - m} \times {(6)}^{ - p} \\ \\ [/tex]
[tex]\sf \: = \: {(5 \times 3)}^{m + n} \times {(5 \times 2)}^{ p - m} \times {(3 \times 2)}^{ - p} \\ \\ [/tex]
[tex]\sf \: = \: {(5)}^{m + n} \times {(3)}^{m + n} \times {(5)}^{ p - m} \times {(2)}^{p - m} \times {(3)}^{ - p} \times {(2)}^{ - p} \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: {(x \times y)}^{n} = {x}^{n} \times {y}^{n} \: }} \\ \\ [/tex]
[tex]\sf \: = \: {(5)}^{m + n + p - m} \times {(3)}^{m + n - p} \times {(2)}^{p - m - p} \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }} \\ \\ [/tex]
[tex]\sf \: = \: {(5)}^{n + p} \times {(3)}^{m + n - p} \times {(2)}^{- m} \\ \\ [/tex]
Hence,
[tex]\sf \: \dfrac{ {(15)}^{m + n} \times {(10)}^{n + p} \times {(6)}^{n} }{ {(10)}^{m + n} \times {(6)}^{n + p} } = {(5)}^{n + p}{(3)}^{m + n - p}{(2)}^{- m} \\ \\ [/tex]
Or
[tex]\boxed{ \sf{ \:\sf \: \dfrac{ {(15)}^{m + n} \times {(10)}^{n + p} \times {(6)}^{n} }{ {(10)}^{m + n} \times {(6)}^{n + p} } = \frac{{(5)}^{n + p}{(3)}^{m + n - p} }{ {2}^{m} } \: }} \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ {x}^{0} = 1}\\ \\ \bigstar \: \bf{ {x}^{m} \times {x}^{n} = {x}^{m + n} }\\ \\ \bigstar \: \bf{ {( {x}^{m})}^{n} = {x}^{mn} }\\ \\\bigstar \: \bf{ {x}^{m} \div {x}^{n} = {x}^{m - n} }\\ \\ \bigstar \: \bf{ {x}^{ - n} = \dfrac{1}{ {x}^{n} } }\\ \\\bigstar \: \bf{ {\bigg(\dfrac{a}{b} \bigg) }^{ - n} = {\bigg(\dfrac{b}{a} \bigg) }^{n} }\\ \\\bigstar \: \bf{ {x}^{m} = {x}^{n}\rm\implies \:m = n }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]