Rewrite \[\dfrac{1}{{{x}^{-2}}}\] with positive exponents?

Hint: In this question we have to rewrite the given expression with positive exponents. For that, As we know. A negative exponent means that the variable is “under the dividing bar” that is \[{{a}^{-1}}=\dfrac{1}{a}\]. So we will apply this rule to the given expression and after that we will be using one more rule that is “dividing by a ratio” it means that multiplying it by its inverse and simplifying it we will get a required answer.

Complete step by step solution:
In the question we have given expressions \[\dfrac{1}{{{x}^{-2}}}\]
We can be written as follows by using \[{{a}^{-1}}=\dfrac{1}{a}\].
Which implies \[\dfrac{1}{{{x}^{-2}}}=\dfrac{1}{\left(\dfrac{1}{x^2}\right)}\]
Now, apply the rule of dividing by a ratio is multiplying by its inverse and we get,
\[\dfrac{1}{{{x}^{-2}}}=1\times {{x}^{2}}\,\].
\[\Rightarrow \,\dfrac{1}{{{x}^{-2}}}\] with positive exponents can be written as \[{{x}^{2}}\].
Hence, \[\dfrac{1}{{{x}^{-2}}}={{x}^{2}}\] is the required answer.

Additional information
• Inverse of an exponent is called a logarithm.
• Logarithms are used in higher mathematics .
• As there are different laws of exponents there are different laws in logarithm to solve the expressions.

Note: There are some key points which we have to keep in mind while solving these types of questions.
Dividing by a ration is multiply by its inverse i.e.
• If we have $\dfrac{a}{\left(\dfrac{b}{c}\right)}$ then it is equal to $\dfrac{a}{b}\times c$
Negative exponent Rule:
• It says that negative exponents in the numerator get moved to the denominator and become positive exponents.
• Also, Negative exponents in the denominator get moved to the numerator and become positive exponents.
• Always keep in mind that we have to move only the negative exponents.