Question

How do you simplify ${(\dfrac{{27}}{8})^{\dfrac{2}{3}}}$?

Answer 1

Hint: The given question is in the exponential form. Let us first know what an exponential form is. If $x$ is a positive integer and $y$ is a real number, then the exponential form can be expressed as ${x^y}$ where $x$ is multiplied by itself $y$ times. Here $x$ is known as the base and $y$ is known as the exponent or power or index.
The given question can be solved by using the Laws of Exponent. These laws simplify exponential expressions easily. Some of the important Laws of Exponent are given below as following:
1. ${a^m} \times {a^n} = {a^{m + n}}$
2. $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$
3. ${({a^m})^n} = {a^{mn}}$
4. ${(ab)^m} = {a^m}{b^m}$
5. ${(\dfrac{a}{b})^m} = \dfrac{{{a^m}}}{{{b^m}}}$
We will use either of these formulas to solve the given question.

Complete Step by Step Solution:
The given exponential form is ${(\dfrac{{27}}{8})^{\dfrac{2}{3}}}$. Here base $x = \dfrac{{27}}{8}$ and exponent $y = \dfrac{2}{3}$. Also considering Law 5 (as given in Hint), i.e. ${(\dfrac{a}{b})^m} = \dfrac{{{a^m}}}{{{b^m}}}$, we have $a = 27$; $b = 8$ and $m = \dfrac{2}{3}$.
On substituting the values of $a$, $b$ and $m$ and on using ${(\dfrac{a}{b})^m} = \dfrac{{{a^m}}}{{{b^m}}}$, we shall get
$ \Rightarrow {(\dfrac{a}{b})^m} = \dfrac{{{a^m}}}{{{b^m}}}$
$ \Rightarrow {(\dfrac{{27}}{8})^{\dfrac{2}{3}}} = \dfrac{{{{27}^{\dfrac{2}{3}}}}}{{{8^{\dfrac{2}{3}}}}}$
Now, we know that $27 = 3 \times 3 \times 3 = {3^3}$, i.e. $27 = {3^3}$.
And, $8 = 2 \times 2 \times 2 = {2^3}$, i.e. $8 = {2^3}$. On substituting these values of $27$and $8$, we shall get
$ \Rightarrow {(\dfrac{{27}}{8})^{\dfrac{2}{3}}} = \dfrac{{{{({3^3})}^{\dfrac{2}{3}}}}}{{{{({2^3})}^{\dfrac{2}{3}}}}}$
Now, on using Law 3, i.e. ${({a^m})^n} = {a^{mn}}$, we shall get
$ \Rightarrow {(\dfrac{{27}}{8})^{\dfrac{2}{3}}} = \dfrac{{{3^{3 \times \dfrac{2}{3}}}}}{{{2^{3 \times \dfrac{2}{3}}}}}$
On simplifying the fractions in the exponent, we shall get
$ \Rightarrow {(\dfrac{{27}}{8})^{\dfrac{2}{3}}} = \dfrac{{{3^2}}}{{{2^2}}}$
On further simplifying, we shall get
$ \Rightarrow {(\dfrac{{27}}{8})^{\dfrac{2}{3}}} = \dfrac{9}{4}$
Since there are no common factors of $9$ and $4$, therefore the fraction $\dfrac{9}{4}$ is in its simplest form.

Hence, when we simplify ${(\dfrac{{27}}{8})^{\dfrac{2}{3}}}$, we get $\dfrac{9}{4}$ as the answer.

Note:
Some more important Laws of Exponent are given below:
1) ${a^0} = 1$
2) ${a^{ - m}} = \dfrac{1}{{{a^m}}}$