Question

How do you simplify ${\left( {\dfrac{1}{8}} \right)^{\dfrac{5}{2}}}$ ?

Answer 1

Hint: We have to simplify this we can write $\left( {\dfrac{5}{2}} \right)$ as the fifth power of the square-root of the given number in base. So, we first have to find out the square of the base number, $\left( {\dfrac{1}{8}} \right)$ and then the cube root of the square of $\left( {\dfrac{1}{8}} \right)$. The result thus obtained will be the answer to the given question.

Complete step-by-step answer:
So, the given question requires us to simplify $\left( {\dfrac{1}{8}} \right)$ to the power $\left( {\dfrac{1}{8}} \right)$. \[4\] to the power $\left( {\dfrac{5}{2}} \right)$ can be written as ${\left( {\dfrac{1}{8}} \right)^{\dfrac{5}{2}}}$ .
This is of the form ${a^{\left( {\dfrac{m}{n}} \right)}}$. But we can rewrite the expression using the laws of indices and powers ${a^{\left( {\dfrac{m}{n}} \right)}} = {\left( {{a^m}} \right)^{\dfrac{1}{n}}}$ .
Thus we will apply same on the question above, we get,
${\left( {\dfrac{1}{8}} \right)^{\dfrac{5}{2}}} = {\left[ {{{\left( {\dfrac{1}{8}} \right)}^5}} \right] ^{\dfrac{1}{2}}}$
Now we first solve the power inside the bracket and write $8$as ${2^3}$.
 \[ = {\left[ {\dfrac{1}{{{{\left( {{2^3}} \right)}^5}}}} \right] ^{\dfrac{1}{2}}}\]
Using the laws of indices and powers, we get, ${\left( {{a^m}} \right)^n} = \left( {{a^{mn}}} \right)$ .
 \[ = {\left[ {\dfrac{1}{{{2^3}^{ \times 5}}}} \right] ^{\dfrac{1}{2}}}\]
 \[ = {\left[ {\dfrac{1}{{{2^{15}}}}} \right] ^{\dfrac{1}{2}}}\]
Simplifying further, we get,
 \[ = \left( {\dfrac{1}{{{2^{\dfrac{{15}}{2}}}}}} \right)\]
Now, the power $\left( {\dfrac{{15}}{2}} \right)$ can be written as $\left( {7 + \dfrac{1}{2}} \right)$.
 \[ = \left( {\dfrac{1}{{{2^{7 + \dfrac{1}{2}}}}}} \right)\]
 \[ = \left( {\dfrac{1}{{{2^7}}}} \right)\left( {\dfrac{1}{{{2^{\dfrac{1}{2}}}}}} \right)\]
We know that \[{2^7} = 128\] and \[{2^{\dfrac{1}{2}}} = \sqrt 2 \]
 \[ = \dfrac{1}{{128\sqrt 2 }}\]
So, ${\left( {\dfrac{1}{8}} \right)^{\dfrac{5}{2}}}$ can be simplified as \[\dfrac{1}{{128\sqrt 2 }}\] .
So, the correct answer is “ \[\dfrac{1}{{128\sqrt 2 }}\] ”.

Note: These rules or laws of indices help us to minimize the problems and get the answer in very less time. These powers can be positive and negative but can be molded according to our convenience while solving the problem. Also note that cube-root, square-root is fractions with 1 as numerator and respective root in denominator.