Question

How do you simplify ${\left( {64} \right)^{\dfrac{{ - 4}}{3}}}$ ?

Answer 1

Hint: In this question, we will solve by breaking the brackets and transform the number into its prime factors, and then we will get the expression in the exponents and now applying the exponent identity, ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$, and then simplify the expression to get the required result.

Complete step by step solution:
Exponents are defined as when an expression or a statement of specific natural numbers are represented as a repeated power by multiplication of its units then the resulting number is called as an exponent. The resulting set of numbers are the same as the original sequence.
Given expression ${\left( {64} \right)^{\dfrac{{ - 4}}{3}}}$,
Now firstly write $64$in prime factors, we get,
$ \Rightarrow 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2$,
Now here we can see that 64 can be written as 2 is multiplied 6 times, this can be written as,$ \Rightarrow 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = {2^6}$,
Therefore the given expression can be written as,
$ \Rightarrow {\left( {64} \right)^{\dfrac{{ - 4}}{3}}} = {\left( {{2^6}} \right)^{\dfrac{{ - 4}}{3}}}$,
Now using the exponent identity,${\left( {{a^m}} \right)^n} = {a^{m \times n}}$,
So here we have, $a = 2,m = - 6$and $n = \dfrac{{ - 4}}{3}$,
Now substituting then values in the identity we get,
$ \Rightarrow {\left( {64} \right)^{\dfrac{{ - 4}}{3}}} = \left( {{2^{6 \times \dfrac{{ - 4}}{3}}}} \right)$,
Now simplifying we get,
$ \Rightarrow {\left( {64} \right)^{\dfrac{{ - 4}}{3}}} = \left( {{2^{2 \times - 4}}} \right)$,
Now multiplying the powers we get,
$ \Rightarrow {\left( {64} \right)^{\dfrac{{ - 4}}{3}}} = \left( {{2^{ - 8}}} \right)$.
Now using the identity, $\dfrac{1}{{{a^m}}} = {a^{ - m}}$, we get,
Here$a = 2$ and $m = 8$, now substituting these values in the identity we get,
$ \Rightarrow {\left( {64} \right)^{\dfrac{{ - 4}}{3}}} = \left( {\dfrac{1}{{{2^8}}}} \right)$,
Now can be written as 2 multiplied 8 times, we get,
$ \Rightarrow {2^8} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$,
So the given expression is equal to 256.

$\therefore $The simplified form of ${\left( {64} \right)^{\dfrac{{ - 4}}{3}}}$will be equal to 256.

Note: There are various laws of exponents we should remember and practise in order to solve and understand the exponential concept. The following are some of the exponent laws:
${a^0} = 1$
${a^m} \times {a^n} = {a^{m + n}}$,
$\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$,
$\dfrac{1}{{{a^m}}} = {a^{ - m}}$,
${a^m} \times {b^m} = {\left( {ab} \right)^m}$,
$\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}$,