How do you simplify \[{\left( {2{x^2}} \right)^{ - 4}}\] and write it using only positive exponents \[?\]
Answer
623.4k+ views
Hint: In this type of question we need to know how to convert the negative exponent into a positive exponent. The final answer would be a simplified form of a given problem. This question describes the operation of addition/ subtraction/ multiplication/ division. Also, we need to know how to expand \[{a^n}\] the terms. We need to know how to combine the power of one term with the power of another term, for this we need to know the basic algebraic formulae
Complete step-by-step answer:
The given question is shown below,
\[{\left( {2{x^2}} \right)^{ - 4}} = ? \to \left( 1 \right)\]
We know that,
\[{a^{ - n}} = \dfrac{1}{{{a^n}}} \to \left( 2 \right)\]
By using the equation \[\left( 2 \right)\] , we can modify the equation \[\left( 1 \right)\] as follows,
\[{\left( {2{x^2}} \right)^{ - 4}} = \dfrac{1}{{{{\left( {2{x^2}} \right)}^4}}} \to \left( 3 \right)\]
We know that,
\[{\left( {ab} \right)^n} = {a^n}{b^n} \to \left( 4 \right)\]
By using the equation \[\left( 4 \right)\] , we can modify the equation \[\left( 3 \right)\] as follows,
\[{\left( {2{x^2}} \right)^{ - 4}} = \dfrac{1}{{{{\left( {2{x^2}} \right)}^4}}} = \dfrac{1}{{{2^4}{{\left( {{x^2}} \right)}^4}}}\]
We know that \[{2^4} = 2 \times 2 \times 2 \times 2 = 16\] . So, the above-mentioned equation becomes,
\[{\left( {2{x^2}} \right)^{ - 4}} = \dfrac{1}{{16{{\left( {{x^2}} \right)}^4}}} \to \left( 5 \right)\]
We know that,
\[{\left( {{a^m}} \right)^n} = {a^m}{a^n} \to \left( 6 \right)\]
By using the equation \[\left( 6 \right)\] we can modify the equation \[\left( 5 \right)\] as follows,
\[
{\left( {2{x^2}} \right)^{ - 4}} = \dfrac{1}{{16{{\left( {{x^2}} \right)}^4}}} = \dfrac{1}{{16{x^{2 \times 4}}}} \\
{\left( {2{x^2}} \right)^{ - 4}} = \dfrac{1}{{16{x^8}}} \\
\]
So, the final answer is,
\[{\left( {2{x^2}} \right)^{ - 4}} = \dfrac{1}{{16{x^8}}}\]
So, the correct answer is “$\dfrac{1}{{16{x^8}}}$”.
Note: In this type of question we won’t use negative exponents while solving the given problem. We would convert the negative exponents into positive exponents by moving the numerator term into the position of the denominator. We would compare the equations in the problem with the algebraic formulae to make the easy calculation. Note that, the final answer would be the most simplified form of a given problem. Also, these types of questions involve the process of addition/ subtraction/ multiplication/ division. Note that, when we have a negative power term in the denominator it converts into a positive power term when we move it into the position of the numerator.
Complete step-by-step answer:
The given question is shown below,
\[{\left( {2{x^2}} \right)^{ - 4}} = ? \to \left( 1 \right)\]
We know that,
\[{a^{ - n}} = \dfrac{1}{{{a^n}}} \to \left( 2 \right)\]
By using the equation \[\left( 2 \right)\] , we can modify the equation \[\left( 1 \right)\] as follows,
\[{\left( {2{x^2}} \right)^{ - 4}} = \dfrac{1}{{{{\left( {2{x^2}} \right)}^4}}} \to \left( 3 \right)\]
We know that,
\[{\left( {ab} \right)^n} = {a^n}{b^n} \to \left( 4 \right)\]
By using the equation \[\left( 4 \right)\] , we can modify the equation \[\left( 3 \right)\] as follows,
\[{\left( {2{x^2}} \right)^{ - 4}} = \dfrac{1}{{{{\left( {2{x^2}} \right)}^4}}} = \dfrac{1}{{{2^4}{{\left( {{x^2}} \right)}^4}}}\]
We know that \[{2^4} = 2 \times 2 \times 2 \times 2 = 16\] . So, the above-mentioned equation becomes,
\[{\left( {2{x^2}} \right)^{ - 4}} = \dfrac{1}{{16{{\left( {{x^2}} \right)}^4}}} \to \left( 5 \right)\]
We know that,
\[{\left( {{a^m}} \right)^n} = {a^m}{a^n} \to \left( 6 \right)\]
By using the equation \[\left( 6 \right)\] we can modify the equation \[\left( 5 \right)\] as follows,
\[
{\left( {2{x^2}} \right)^{ - 4}} = \dfrac{1}{{16{{\left( {{x^2}} \right)}^4}}} = \dfrac{1}{{16{x^{2 \times 4}}}} \\
{\left( {2{x^2}} \right)^{ - 4}} = \dfrac{1}{{16{x^8}}} \\
\]
So, the final answer is,
\[{\left( {2{x^2}} \right)^{ - 4}} = \dfrac{1}{{16{x^8}}}\]
So, the correct answer is “$\dfrac{1}{{16{x^8}}}$”.
Note: In this type of question we won’t use negative exponents while solving the given problem. We would convert the negative exponents into positive exponents by moving the numerator term into the position of the denominator. We would compare the equations in the problem with the algebraic formulae to make the easy calculation. Note that, the final answer would be the most simplified form of a given problem. Also, these types of questions involve the process of addition/ subtraction/ multiplication/ division. Note that, when we have a negative power term in the denominator it converts into a positive power term when we move it into the position of the numerator.
Recently Updated Pages
What is BLO What is the full form of BLO class 8 social science CBSE

Find the greatest six digit number that is exactly class 8 maths CBSE

What is the time difference between India and Cana class 8 social science CBSE

Compare LPG and wood as fuels class 8 chemistry CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

30 opposite words in English from a to z class 8 english CBSE

Trending doubts
Who commanded the Hector the first British trading class 8 social science CBSE

Write a letter to the Municipal Commissioner to inform class 8 english CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

Find the largest number of six digits which is a p-class-8-maths-CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE


