Simplify and express the answer using positive exponents: (\frac{2x^{-1}}

Braxton Pugh 2021-09-17 Answered
Simplify and express the answer using positive exponents:
\(\displaystyle{\left({\frac{{{2}{x}^{{-{1}}}}}{{{3}{y}}}}\right)}^{{-{2}}}\)

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Expert Answer

yunitsiL
Answered 2021-09-18 Author has 22956 answers
Given expression:
\(\displaystyle{\left({\frac{{{2}{x}^{{-{1}}}}}{{{3}{y}}}}\right)}^{{-{2}}}\)
We know that \(\displaystyle{\left({a}^{{m}}\right)}^{{n}}={a}^{{{m}\times{n}}}\)
So, we have
\(\displaystyle{\left({\frac{{{2}{x}^{{-{1}}}}}{{{3}{y}}}}\right)}^{{-{2}}}={\frac{{{2}^{{-{2}}}\cdot{\left({x}\right)}^{{-{1}{x}-{2}}}}}{{{\left({3}{y}\right)}^{{-{2}}}}}}\)
\(\displaystyle{\left({\frac{{{2}{x}^{{-{1}}}}}{{{3}{y}}}}\right)}^{{-{2}}}={\frac{{{\left({x}\right)}^{{2}}}}{{{2}^{{2}}\cdot{3}^{{-{2}}}\cdot{y}^{{-{2}}}}}}\)
\(\displaystyle{\left({\frac{{{2}{x}^{{-{1}}}}}{{{3}{y}}}}\right)}^{{-{2}}}={\frac{{{\left({x}\right)}^{{2}}\cdot{y}^{{2}}\cdot{3}^{{2}}}}{{{2}^{{2}}}}}\)
\(\displaystyle{\left({\frac{{{2}{x}^{{-{1}}}}}{{{3}{y}}}}\right)}^{{-{2}}}={\frac{{{9}{x}^{{2}}{y}^{{2}}}}{{{4}}}}\)
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