How do you simplify the given term \[{{\left( {{x}^{2}}{{y}^{-1}} \right)}^{2}}\] and write it using only positive exponents?
Answer
622.5k+ views
Hint: We start solving the problem by equating the given term to a variable. We then make use of the law of exponents that ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$ to proceed through the problem. We then make use of the law of exponents that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ to proceed further through the problem. We then make use of the law of exponents that ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$ to get the required simplified form of the given term.
Complete step by step answer:
According to the problem, we are asked to simplify the given term \[{{\left( {{x}^{2}}{{y}^{-1}} \right)}^{2}}\] and write it using only positive exponents.
Let us assume $d={{\left( {{x}^{2}}{{y}^{-1}} \right)}^{2}}$ ---(1).
From the laws of exponents, we know that ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$. Let us use this result in equation (1).
$\Rightarrow d={{\left( {{x}^{2}} \right)}^{2}}{{\left( {{y}^{-1}} \right)}^{2}}$ ---(2).
From the laws of exponents, we know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$. Let us use this result in equation (2).
$\Rightarrow d=\left( {{x}^{2\times 2}} \right)\left( {{y}^{-1\times 2}} \right)$.
$\Rightarrow d=\left( {{x}^{4}} \right)\left( {{y}^{-2}} \right)$ ---(3).
From laws of exponents, we know that ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$. Let us use this result in equation (3).
$\Rightarrow d=\left( {{x}^{4}} \right)\left( \dfrac{1}{{{y}^{2}}} \right)$.
$\Rightarrow d=\dfrac{{{x}^{4}}}{{{y}^{2}}}$.
So, we have found the required simplified form of the given term \[{{\left( {{x}^{2}}{{y}^{-1}} \right)}^{2}}\] as $\dfrac{{{x}^{4}}}{{{y}^{2}}}$.
$\therefore $ The required simplified form of the given term \[{{\left( {{x}^{2}}{{y}^{-1}} \right)}^{2}}\]is $\dfrac{{{x}^{4}}}{{{y}^{2}}}$.
Note:
Whenever we get this type of problems, we try to make use of the laws of exponents to get the required answer. We should keep in mind that the exponents for variables x and y should be positive in the final result. Similarly, we can expect problems to find the simplified form of the given term ${{\left( \dfrac{{{x}^{4}}{{y}^{6}}}{{{x}^{2}}} \right)}^{\dfrac{1}{2}}}$.
Complete step by step answer:
According to the problem, we are asked to simplify the given term \[{{\left( {{x}^{2}}{{y}^{-1}} \right)}^{2}}\] and write it using only positive exponents.
Let us assume $d={{\left( {{x}^{2}}{{y}^{-1}} \right)}^{2}}$ ---(1).
From the laws of exponents, we know that ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$. Let us use this result in equation (1).
$\Rightarrow d={{\left( {{x}^{2}} \right)}^{2}}{{\left( {{y}^{-1}} \right)}^{2}}$ ---(2).
From the laws of exponents, we know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$. Let us use this result in equation (2).
$\Rightarrow d=\left( {{x}^{2\times 2}} \right)\left( {{y}^{-1\times 2}} \right)$.
$\Rightarrow d=\left( {{x}^{4}} \right)\left( {{y}^{-2}} \right)$ ---(3).
From laws of exponents, we know that ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$. Let us use this result in equation (3).
$\Rightarrow d=\left( {{x}^{4}} \right)\left( \dfrac{1}{{{y}^{2}}} \right)$.
$\Rightarrow d=\dfrac{{{x}^{4}}}{{{y}^{2}}}$.
So, we have found the required simplified form of the given term \[{{\left( {{x}^{2}}{{y}^{-1}} \right)}^{2}}\] as $\dfrac{{{x}^{4}}}{{{y}^{2}}}$.
$\therefore $ The required simplified form of the given term \[{{\left( {{x}^{2}}{{y}^{-1}} \right)}^{2}}\]is $\dfrac{{{x}^{4}}}{{{y}^{2}}}$.
Note:
Whenever we get this type of problems, we try to make use of the laws of exponents to get the required answer. We should keep in mind that the exponents for variables x and y should be positive in the final result. Similarly, we can expect problems to find the simplified form of the given term ${{\left( \dfrac{{{x}^{4}}{{y}^{6}}}{{{x}^{2}}} \right)}^{\dfrac{1}{2}}}$.
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