Question

How do you simplify \[{{\left( 64 \right)}^{\dfrac{-2}{3}}}\]?

Answer 1

Hint: We start solving the problem by equating \[{{\left( 64 \right)}^{\dfrac{-2}{3}}}\] to a variable and then make use of the result $64={{4}^{3}}$ to proceed through the problem. We then make use of the law of exponents ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ to proceed further through the problem. We then make use of the law of exponents ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$ to proceed further through the problem. We then make use of the result ${{4}^{2}}=16$ to get the required answer of the problem.

Complete step-by-step solution:
According to the problem, we are asked to find the value of \[{{\left( 64 \right)}^{\dfrac{-2}{3}}}\].
Let us assume \[a={{\left( 64 \right)}^{\dfrac{-2}{3}}}\] -------(1).
We know that $64={{4}^{3}}$. Let us use this result in equation (1).
$\Rightarrow a={{\left( {{4}^{3}} \right)}^{\dfrac{-2}{3}}}$ -------(2).
From the laws of exponents, we know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$. Let us use this result in equation (2).
$\Rightarrow a={{4}^{3\times \dfrac{-2}{3}}}$.
$\Rightarrow a={{4}^{-2}}$ --------(3).
From the law of exponents, we know that ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$. Let us use this result in equation (3).
$\Rightarrow a=\dfrac{1}{{{4}^{2}}}$ -------(4).
We know that ${{4}^{2}}=16$. Let us use this result in equation (4).
$\Rightarrow a=\dfrac{1}{16}$.
$\therefore $ We have found the value of \[{{\left( 64 \right)}^{\dfrac{-2}{3}}}\] as $\dfrac{1}{16}$.

Note: Whenever we get this type of problem, we should try to make use of the laws of exponents to proceed through the problem. We should not confuse ${{\left( {{a}^{m}} \right)}^{n}}$ with ${{a}^{{{m}^{n}}}}$ while solving this problem. We should not confuse the laws of exponents while solving this type of problem. Similarly, we can expect problems to find the value of ${{\left( 256 \right)}^{\dfrac{3}{4}}}\times {{\left( 16 \right)}^{\dfrac{3}{4}}}$ using the laws of exponents.