Question

Find the value of $4{{\left( 512 \right)}^{-\dfrac{1}{9}}}$ .

Answer 1

Hint: To find the value of $4{{\left( 512 \right)}^{-\dfrac{1}{9}}}$ , we will write ${{\left( 512 \right)}^{-\dfrac{1}{9}}}$ in its fractional form. Then, we have to find the ninth root of 512 using ladder method. We will substitute this value in the given expression and simplify.

Complete step by step answer:
We have to find the value of $4{{\left( 512 \right)}^{-\dfrac{1}{9}}}$ . We know that ${{x}^{-1}}=\dfrac{1}{x}$ . Therefore, we can write the given expression as
$\Rightarrow 4{{\left( 512 \right)}^{-\dfrac{1}{9}}}=4\times \dfrac{1}{{{\left( 512 \right)}^{\dfrac{1}{9}}}}$
We know that ${{x}^{\dfrac{1}{n}}}=\sqrt[n]{x}$ . Therefore, we can write the above equation as
\[\Rightarrow 4{{\left( 512 \right)}^{-\dfrac{1}{9}}}=4\times \dfrac{1}{\sqrt[9]{512}}...\left( i \right)\]
Let us take the ninth root of 512. For this, we have to find the prime factors of 512 using ladder method.
$\begin{align}
  & 2\left| \!{\underline {\,
  512 \,}} \right. \\
 & 2\left| \!{\underline {\,
  256 \,}} \right. \\
 & 2\left| \!{\underline {\,
  128 \,}} \right. \\
 & 2\left| \!{\underline {\,
  64 \,}} \right. \\
 & 2\left| \!{\underline {\,
  32 \,}} \right. \\
 & 2\left| \!{\underline {\,
  16 \,}} \right. \\
 & 2\left| \!{\underline {\,
  8 \,}} \right. \\
 & 2\left| \!{\underline {\,
  4 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2 \,}} \right. \\
 & \text{ }\text{ }\text{ }1 \\
\end{align}$
We can write $512=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2={{2}^{9}}$ .
Therefore, \[\sqrt[9]{512}=2\] . We have to substitute this value in equation (i).
\[\Rightarrow 4{{\left( 512 \right)}^{-\dfrac{1}{9}}}=4\times \dfrac{1}{2}\]
Let us cancel the common factor 2.
\[\Rightarrow 4{{\left( 512 \right)}^{-\dfrac{1}{9}}}={{\require{cancel}\cancel{4}}^{2}}\times \dfrac{1}{\require{cancel}\cancel{2}}\]
We can write the result of the above simplification as
\[\Rightarrow 4{{\left( 512 \right)}^{-\dfrac{1}{9}}}=2\]
Hence, the value of $4{{\left( 512 \right)}^{-\dfrac{1}{9}}}$ is 2.

Note: Students must know to find the roots of numbers using ladder method. We can also find the value of $4{{\left( 512 \right)}^{-\dfrac{1}{9}}}$ using laws of exponents.
Firstly, we will write the given expression as follows.
$\Rightarrow 4{{\left( 512 \right)}^{-\dfrac{1}{9}}}=4\times \dfrac{1}{{{\left( 512 \right)}^{\dfrac{1}{9}}}}$
Now, we have to write the prime factors of 159. We can find the prime factors using ladder method.
Prime factors of $512=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2={{2}^{9}}$ . Let us substitute this value in the above expression.
$\Rightarrow 4{{\left( 512 \right)}^{-\dfrac{1}{9}}}=4\times \dfrac{1}{{{\left( {{2}^{9}} \right)}^{\dfrac{1}{9}}}}$
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ . Therefore, the above expression becomes
$\Rightarrow 4{{\left( 512 \right)}^{-\dfrac{1}{9}}}=4\times \dfrac{1}{{{2}^{9\times \dfrac{1}{9}}}}$
Let us simplify the powers of 2.
$\begin{align}
  & \Rightarrow 4{{\left( 512 \right)}^{-\dfrac{1}{9}}}=4\times \dfrac{1}{{{2}^{1}}} \\
 & \Rightarrow 4{{\left( 512 \right)}^{-\dfrac{1}{9}}}=4\times \dfrac{1}{2} \\
\end{align}$
Now, we have to simplify.
$\Rightarrow 4{{\left( 512 \right)}^{-\dfrac{1}{9}}}=2$
Students must deeply understand and learn the rules of exponents and where to apply these rules.