Question

How do you simplify $4{{\left( 8 \right)}^{-\dfrac{2}{3}}}$?

Answer 1

Hint: To simplify the given expression, we are going to write 8 as ${{2}^{3}}$. By doing that the 3 written in the denominator of the power of 8 will get cancelled out. After that we require the property of the exponent which says that ${{a}^{-1}}=\dfrac{1}{a}$. After applying this property, we are going to multiply this solution by 4 and in this way we will simplify the given expression.

Complete step-by-step solution:
The expression given in the above problem is as follows:
$4{{\left( 8 \right)}^{-\dfrac{2}{3}}}$
Now, in the above expression, we are going to write 8 as ${{2}^{3}}$ and then the above expression will look like:
$4{{\left( {{2}^{3}} \right)}^{-\dfrac{2}{3}}}$
There is a property of exponent which says that:
${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$
Using the above relation in $4{{\left( {{2}^{3}} \right)}^{-\dfrac{2}{3}}}$ we get,
$4{{\left( 2 \right)}^{3\times -\dfrac{2}{3}}}$
In the above expression, 3 will get cancelled out from the exponent of 2 and we get,
$4{{\left( 2 \right)}^{-2}}$
Putting the 2 in the power of 2 inside the bracket we get,
$\begin{align}
  & 4{{\left( {{2}^{2}} \right)}^{-1}} \\
 & =4{{\left( 4 \right)}^{-1}} \\
\end{align}$
We know the negative exponent property in the arithmetic which states that:
${{a}^{-1}}=\dfrac{1}{a}$
Substituting $a=4$ in the above equation we get,
${{4}^{-1}}=\dfrac{1}{4}$
Using the above relation in $4{{\left( 4 \right)}^{-1}}$ we get,
$4\left( \dfrac{1}{4} \right)$
In the above expression, 4 will get cancelled out from the numerator and the denominator and we get 1.
Hence, the solution of the above expression is 1.

Note: You might be thinking how we know that we should take 8 as ${{2}^{3}}$ because it will make 3 in the denominator of the exponent vanish. And this thing also clicks to you when you practice more problems like this so it would be better if you know the square and cube of 2 to 15.