Question

Simplify ${\left( {\dfrac{1}{{64}}} \right)^{ - \dfrac{2}{3}}}$ ?

Answer 1

Hint: Use the property of the exponents.
First apply the property of negative exponents that is, \[{a^{ - n}} = \dfrac{1}{{{a^n}}}\], where $a \ne 0$.
Now, use ${a^{\dfrac{x}{y}}} = \sqrt[y]{{{a^x}}}$
We know that $\sqrt[3]{{4096}} = 16$therefore, $\dfrac{1}{{\sqrt[3]{{4096}}}} = \dfrac{1}{{16}}$

Complete step by step answer:
We have to simplify ${\left( {\dfrac{1}{{64}}} \right)^{ - \dfrac{2}{3}}}$.
First, apply the property of exponents;
For all nonzero element, \[{a^{ - x}} = \dfrac{1}{{{a^x}}}\].
Substitute $a = \dfrac{1}{{64}}$and $x = \dfrac{2}{3}$ into the formula,
\[{\left( {\dfrac{1}{{64}}} \right)^{ - \dfrac{2}{3}}} = \dfrac{1}{{{{\left( {\dfrac{1}{{64}}} \right)}^{\dfrac{2}{3}}}}} \ldots \ldots (1)\]
Now apply ${a^{\dfrac{x}{y}}} = \sqrt[y]{{{a^x}}}$to simplify \[{\left( {\dfrac{1}{{64}}} \right)^{\dfrac{2}{3}}}\].
Here $a = \dfrac{1}{{64}}$, $x = 2$ and $y = 3$.
${\left( {\dfrac{1}{{64}}} \right)^{\dfrac{2}{3}}} = \sqrt[3]{{{{\left( {\dfrac{1}{{64}}} \right)}^2}}}$
${\left( {\dfrac{1}{{64}}} \right)^{\dfrac{2}{3}}} = \sqrt[3]{{\dfrac{1}{{4096}}}}$
We know that $\sqrt[3]{{4096}} = 16$ therefore, $\dfrac{1}{{\sqrt[3]{{4096}}}} = \dfrac{1}{{16}}$.
$ \Rightarrow {\left( {\dfrac{1}{{64}}} \right)^{\dfrac{2}{3}}} = \dfrac{1}{{16}}$
Substitute ${\left( {\dfrac{1}{{64}}} \right)^{\dfrac{2}{3}}} = \dfrac{1}{{16}}$ into the equation $(1)$.
\[{\left( {\dfrac{1}{{64}}} \right)^{ - \dfrac{2}{3}}} = \dfrac{1}{{\dfrac{1}{{16}}}}\]
Simplify the expression further,
\[{\left( {\dfrac{1}{{64}}} \right)^{ - \dfrac{2}{3}}} = 16\]

Note: Here, is the list of the formulas of the exponents,
If \[n\;\] is a positive integer and \[x\] is any real number, then \[{x^n}\] corresponds to repeated multiplication
\[x \times x \times \cdots \times x = {x^n}\]
1.${x^a}{x^b} = {x^{a + b}}$
2.$\dfrac{{{x^a}}}{{{x^b}}} = {x^{a - b}}$
3.${({x^a})^b} = {x^{ab}}$
4.${x^0} = 1$
5.${x^{ - 1}} = \dfrac{1}{x}$