Question

How do you simplify ${{64}^{\dfrac{3}{4}}}\times {{81}^{\dfrac{2}{3}}}$?

Answer 1

Hint: We try to form the given fraction in its simplest form. The given multiplication form is multiplication of two indices. We take help of indices and use identities like ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}},{{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}},{{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}}$. We complete the final power form and find the solution through multiplication.

Complete step by step solution:
We need to first simplify the given fraction $\dfrac{64}{27}$. We will try to convert it into its simplest form with real power or indices value.
We know the exponent form of the number $a$ with the exponent being $n$ can be expressed as ${{a}^{n}}$.
The simplified form of the expression ${{a}^{n}}$ can be written as the multiplied form of number $a$ of n-times.
In case the value of $n$ becomes negative, the value of the exponent takes its inverse value.
The formula to express the form is ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}},n\in {{\mathbb{R}}^{+}}$.
The multiplication of these exponents works as the addition of those indices.
For example, we take two exponential expressions where the exponents are $m$ and $n$.
Let the numbers be ${{a}^{m}}$ and ${{a}^{n}}$. We take multiplication of these numbers.
The indices get added. So, ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$.
The division works in an almost similar way. The indices get subtracted. So, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$.
Also, we have the identities where ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}},{{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}},{{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}}$.
We know $64={{2}^{6}},81={{3}^{4}}$.
Now applying \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\], we get ${{64}^{\dfrac{3}{4}}}={{\left( {{2}^{6}} \right)}^{\dfrac{3}{4}}}={{2}^{6\times \dfrac{3}{4}}}={{2}^{\dfrac{9}{2}}}$.
Similarly, ${{81}^{\dfrac{2}{3}}}={{\left( {{3}^{4}} \right)}^{\dfrac{2}{3}}}={{3}^{4\times \dfrac{2}{3}}}={{3}^{\dfrac{8}{3}}}$.
Using ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$, we get \[{{2}^{\dfrac{9}{2}}}={{2}^{4}}{{.2}^{\dfrac{1}{2}}}=16\sqrt{2}\] and \[{{3}^{\dfrac{8}{3}}}={{3}^{3}}{{.3}^{-\dfrac{1}{3}}}=\dfrac{27}{{{3}^{\dfrac{1}{3}}}}=\dfrac{27}{\sqrt[3]{3}}\]
Therefore, the expression ${{64}^{\dfrac{3}{4}}}\times {{81}^{\dfrac{2}{3}}}$ becomes ${{64}^{\dfrac{3}{4}}}\times {{81}^{\dfrac{2}{3}}}=\left( 16\sqrt{2} \right)\times \left( \dfrac{27}{\sqrt[3]{3}} \right)=\dfrac{432\sqrt{2}}{\sqrt[3]{3}}$.

The simplified form of ${{64}^{\dfrac{3}{4}}}\times {{81}^{\dfrac{2}{3}}}$ is $\dfrac{432\sqrt{2}}{\sqrt[3]{3}}$.

Note: If we are finding the square and cube root of any numbers, we don’t always need to find the all-possible roots. The problem gets more complicated in that case. Instead of simplifying the indices we first need to find the simplified form of the given fraction to find the actual indices value of the problem.