Question

How do you simplify ${{\left( \dfrac{1}{8} \right)}^{\dfrac{4}{3}}}$?

Answer 1

Hint: To solve the above question we will use the concept of indices and exponents to simplify the above given term. At first, we will find the prime factor of 8 and since 2 is the prime factor of 8 so we can write it as ${{2}^{3}}$. Now, we know that ${{\left( {{x}^{a}} \right)}^{b}}={{x}^{a\times b}}$ . So, we will use it and write ${{\left( \dfrac{1}{8} \right)}^{\dfrac{4}{3}}}={{\left( \dfrac{1}{{{2}^{3}}} \right)}^{\dfrac{4}{3}}}={{\left( \dfrac{1}{2} \right)}^{\dfrac{4}{3}\times 3}}={{\left( \dfrac{1}{2} \right)}^{4}}$. And, we will also use the formula ${{\left( \dfrac{a}{b} \right)}^{c}}=\left( \dfrac{{{a}^{c}}}{{{b}^{c}}} \right)$ and write ${{\left( \dfrac{1}{2} \right)}^{4}}=\dfrac{{{1}^{4}}}{{{2}^{4}}}=\dfrac{1}{16}$ .

Complete step-by-step solution:
We will use the concepts of indices and exponent to simplify the above given term. Indices are the convenient ways of writing multiplication that have many repeated terms. For example: In ${{5}^{3}}$, 5 is the base and 3 is the index and here in ${{5}^{3}}$, 5 is repeated 3 times. That is it means ${{5}^{3}}=5\times 5\times 5$.
Also, note that index can be negative, zero or fractional.
Since, we have to simplify ${{\left( \dfrac{1}{8} \right)}^{\dfrac{4}{3}}}$ and we know that prime factor of 8 is ${{2}^{3}}$ so we can also write ${{\left( \dfrac{1}{8} \right)}^{\dfrac{4}{3}}}$ as ${{\left( \dfrac{1}{{{2}^{3}}} \right)}^{\dfrac{4}{3}}}$.
We know that ${{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}}$ , so we can write ${{\left( \dfrac{1}{{{2}^{3}}} \right)}^{\dfrac{4}{3}}}$ as:
$\Rightarrow {{\left( \dfrac{1}{{{2}^{3}}} \right)}^{\dfrac{4}{3}}}=\left( \dfrac{{{1}^{\dfrac{4}{3}}}}{{{\left( {{2}^{3}} \right)}^{\dfrac{4}{3}}}} \right)$
Since, we also know that ${{\left( 1 \right)}^{m}}=1$ and ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ so we can write the above expression as:
$\Rightarrow {{\left( \dfrac{1}{{{2}^{3}}} \right)}^{\dfrac{4}{3}}}=\left( \dfrac{1}{{{2}^{3\times \dfrac{4}{3}}}} \right)$
$\Rightarrow {{\left( \dfrac{1}{8} \right)}^{\dfrac{4}{3}}}=\left( \dfrac{1}{{{2}^{4}}} \right)$
Now, we know that ${{2}^{4}}$ means that multiply 2 with itself 4 times i.e. ${{2}^{4}}=2\times 2\times 2\times 2$
$\Rightarrow {{2}^{4}}=16$
So, we can write ${{\left( \dfrac{1}{8} \right)}^{\dfrac{4}{3}}}=\left( \dfrac{1}{{{2}^{4}}} \right)=\left( \dfrac{1}{16} \right)$
$\therefore {{\left( \dfrac{1}{8} \right)}^{\dfrac{4}{3}}}=\left( \dfrac{1}{16} \right)$
So, the simplified form of ${{\left( \dfrac{1}{8} \right)}^{\dfrac{4}{3}}}$ is $\dfrac{1}{16}$.
This is our required solution.

Note: Students are required to memorize all the index law and do not confuse between the laws as most of them are similar for example ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ but ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ . So, they should not make mistakes while applying laws otherwise they will get the wrong result.