Question

How do you simplify \[{16^{\dfrac{2}{3}}}?\]

Answer 1

Hint: This question describes the operation of addition/ subtraction/ multiplication/ division. Also, we need to know the basic formulae to solve the given problem. Also, we need to know how to solve cubic root terms. We need to know the formulae which are related to exponential components. Also, we need to know how to find the value of square and square root values of a particular number.

Complete step-by-step answer:
In this question, we would simplify the following term,
 \[{16^{\dfrac{2}{3}}} = ? \to \left( 1 \right)\]
We know that
 \[{\left( {{x^a}} \right)^b} = \left( {{x^{ab}}} \right)\]
The above equation can also be written as,
 \[\left( {{x^{ab}}} \right) = {\left( {{x^a}} \right)^b}\] \[ \to \left( 2 \right)\]
Let’s compare the equation \[\left( 1 \right)\] and \[\left( 2 \right)\] , we get
 \[\left( 1 \right) \to {16^{\dfrac{2}{3}}} = ?\]
 \[\left( 2 \right) \to \left( {{x^{ab}}} \right) = {\left( {{x^a}} \right)^b}\]
In the equation \[\left( 1 \right)\] , we have \[\dfrac{2}{3}\] the power of \[16\] .
 \[\dfrac{2}{3}\] can also be written as \[2 \times \dfrac{1}{3}\] .
So, the equation \[\left( 1 \right)\] becomes,
 \[{16^{2 \times \dfrac{1}{3}}} = ? \to \left( 3 \right)\]
By comparing the equation \[\left( 3 \right)\] and \[\left( 2 \right)\] , we get
 \[a = 2,b = \dfrac{1}{3}\] and \[x = 16\] .
Let’s substitute the above-mentioned values in the equation \[\left( 2 \right)\] , we get
 \[\left( 2 \right) \to \left( {{x^{ab}}} \right) = {\left( {{x^a}} \right)^b}\]
 \[{\left( {16} \right)^{2 \times \dfrac{1}{3}}} = {\left( {{{16}^2}} \right)^{\dfrac{1}{3}}} \to \left( 4 \right)\]
We know that \[\sqrt 2 \] can also be written as \[{\left( 2 \right)^{\dfrac{1}{2}}}\] , as same as \[\sqrt[3] {2}\] can be written as \[{\left( 2 \right)^{\dfrac{1}{3}}}\] . So, the equation \[\left( 4 \right)\] can also be written as,
 \[{\left( {{{16}^2}} \right)^{\dfrac{1}{3}}} = \sqrt[3] {{{{16}^2}}} \to \left( 5 \right)\]
We know that,
 \[{16^2} = 16 \times 16 = 4 \times 4 \times 4 \times 4\]
So the equation \[\left( 5 \right)\] becomes,
 \[
  \sqrt[3] {{{{16}^2}}} = \sqrt[3] {{4 \times 4 \times 4 \times 4}} \\
  \sqrt[3] {{{{16}^2}}} = 4\sqrt[3] {4} \;
 \]
So, the final answer is,
 \[{16^{\dfrac{2}{3}}} = 4\sqrt[3] {4}\]
So, the correct answer is “$4\sqrt[3] {4}$”.

Note: In this type of question we would involve the operation of addition/ subtraction/ multiplication/ division. We would try to compare the given problem with an algebraic formula. Note that if the \[{n^3}\] term is placed inside the cubic root, we can cancel the cube with a cubic root. Also, note that \[\sqrt n \] can be written as \[{\left( n \right)^{\dfrac{1}{2}}}\] and \[\sqrt[3] {n}\] can be written as \[{\left( n \right)^{\dfrac{1}{3}}}\] . If zero is placed inside the root the total value of the root will be zero.