How do you simplify \[{\left( {16{x^{12}}} \right)^{\dfrac{3}{4}}}?\]
Answer
605.4k+ views
Hint: To simplify the given exponential, first simplify the base of the exponential with the help of law of indices for multiplication and brackets and then after simplifying the base use law of indices for brackets to open the parentheses and further simplify the exponential.
Law of indices for multiplication is given as:
${a^x}{a^y} = {a^{x + y}}$
And also law of indices for brackets is given as:
\[{\left( {{a^x}} \right)^b} = {a^{x \times b}}\]
Use these formulae to simplify the given exponential.
Complete step-by-step answer:
In order to simplify the given exponential \[{\left( {16{x^{12}}} \right)^{\dfrac{3}{4}}}\] , we will first simplify the base of the exponential that is \[16{x^{12}}\]
Now from the law of indices for multiplication we can write $16$ as follows:
$16 = 2 \times 2 \times 2 \times 2 = {2^4}$
And from the law of indices for brackets we can write \[{x^{12}}\] as
\[{x^{12}} = {x^{3 \times 4}} = {\left( {{x^3}} \right)^4}\]
Therefore the base of the exponential can be rewritten as follows
\[16{x^{12}} = {2^4}{\left( {{x^3}} \right)^4}\]
Since $2\;{\text{and}}\;{x^3}$ have same powers so we further write them as
\[{2^4}{\left( {{x^3}} \right)^4} = {\left( {2{x^3}} \right)^4}\]
Now coming to the exponential we can rewrite it with the new base as
\[{\left( {16{x^{12}}} \right)^{\dfrac{3}{4}}} = {\left( {{{\left( {2{x^3}} \right)}^4}} \right)^{\dfrac{3}{4}}}\]
Now with the help of law of indices for brackets we can write it as
\[{\left( {{{\left( {2{x^3}} \right)}^4}} \right)^{\dfrac{3}{4}}} = {\left( {2{x^3}} \right)^{4 \times \dfrac{3}{4}}} = {\left( {2{x^3}} \right)^3}\]
Using the commutative property of exponentials and the law of indices for brackets, we can further simplify it as
\[{\left( {2{x^3}} \right)^3} = {2^3}{x^{3 \times 3}} = 8{x^9}\]
Therefore \[8{x^9}\] is the simplified form of the given exponential term \[{\left( {16{x^{12}}} \right)^{\dfrac{3}{4}}}\]
So, the correct answer is “ \[8{x^9}\] ”.
Note: Law of indices is very helpful in simplifying this type of questions, there is law of indices for fractional power, for division, for negative powers and for power zero so learning about these laws will help students in simplifying and solving the problems easily.
Law of indices for multiplication is given as:
${a^x}{a^y} = {a^{x + y}}$
And also law of indices for brackets is given as:
\[{\left( {{a^x}} \right)^b} = {a^{x \times b}}\]
Use these formulae to simplify the given exponential.
Complete step-by-step answer:
In order to simplify the given exponential \[{\left( {16{x^{12}}} \right)^{\dfrac{3}{4}}}\] , we will first simplify the base of the exponential that is \[16{x^{12}}\]
Now from the law of indices for multiplication we can write $16$ as follows:
$16 = 2 \times 2 \times 2 \times 2 = {2^4}$
And from the law of indices for brackets we can write \[{x^{12}}\] as
\[{x^{12}} = {x^{3 \times 4}} = {\left( {{x^3}} \right)^4}\]
Therefore the base of the exponential can be rewritten as follows
\[16{x^{12}} = {2^4}{\left( {{x^3}} \right)^4}\]
Since $2\;{\text{and}}\;{x^3}$ have same powers so we further write them as
\[{2^4}{\left( {{x^3}} \right)^4} = {\left( {2{x^3}} \right)^4}\]
Now coming to the exponential we can rewrite it with the new base as
\[{\left( {16{x^{12}}} \right)^{\dfrac{3}{4}}} = {\left( {{{\left( {2{x^3}} \right)}^4}} \right)^{\dfrac{3}{4}}}\]
Now with the help of law of indices for brackets we can write it as
\[{\left( {{{\left( {2{x^3}} \right)}^4}} \right)^{\dfrac{3}{4}}} = {\left( {2{x^3}} \right)^{4 \times \dfrac{3}{4}}} = {\left( {2{x^3}} \right)^3}\]
Using the commutative property of exponentials and the law of indices for brackets, we can further simplify it as
\[{\left( {2{x^3}} \right)^3} = {2^3}{x^{3 \times 3}} = 8{x^9}\]
Therefore \[8{x^9}\] is the simplified form of the given exponential term \[{\left( {16{x^{12}}} \right)^{\dfrac{3}{4}}}\]
So, the correct answer is “ \[8{x^9}\] ”.
Note: Law of indices is very helpful in simplifying this type of questions, there is law of indices for fractional power, for division, for negative powers and for power zero so learning about these laws will help students in simplifying and solving the problems easily.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Give me the opposite gender of Duck class 8 english CBSE

What is the feminine gender of a stag class 8 english CBSE

What is the collective noun for soldiers class 8 english CBSE

What is the difference between rai and mustard see class 8 biology CBSE

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

Change He keeps me waiting to passive voice class 8 english CBSE


