Question

You are supposed to find the value of ${(16)^{\dfrac{1}{4}}}$

Answer 1

We have to simply solve the power which is placed over the number. After converting ‘16’ into index form an index of ‘4’ comes up and then we multiply it with the index placed over the bracket of the given expression i.e. $\dfrac{1}{4}$

Complete solution step by step:
Firstly, we write the given expression
${(16)^{\dfrac{1}{4}}}$

Simplifying the bracket part first using index conversion of a square number i.e. when we reduce a square number (16 in our question is a perfect square of 4), we replace it by putting index (power) of ‘2’ in this manner -

${(16)^{\dfrac{1}{4}}} = {({4^2})^{\dfrac{1}{4}}}$

Now to solve this expression we use the index properties i.e.
${({p^q})^r} = {({p^r})^q} = {p^{q \times r}}$

Using this property and taking first and third parts of it we solve our expression
${(16)^{\dfrac{1}{4}}} = {({4^2})^{\dfrac{1}{4}}} = {(4)^{2 \times \dfrac{1}{4}}} = {(4)^{\dfrac{1}{2}}}$

Converting the index into square root we have
${(4)^{\dfrac{1}{2}}} = \sqrt 4 = 2$

So our answer comes out to be ‘2’.

Additional information: We can solve the problem using basic index method also in which we solve the index part of the expression as explained below
${(16)^{\dfrac{1}{4}}}$

The expression has index of $\dfrac{1}{4}$ which means if we find the factors of 16 and check for a pair of four like numbers our 4 th root will come out be this factor of ‘16’ i.e.
$16 = 2 \times 2 \times 2 \times 2$

This is the prime factorization of ‘16’ which means it cannot be reduced further and we have to select a pair of four like factors. In this case only ‘2’ is the factor of ‘16’ and it is repeated four times which means using the definition of surd we get -
$

{(a)^{\dfrac{1}{n}}} = \sqrt[n]{a} \\
\Rightarrow {(16)^{\dfrac{1}{4}}} = \sqrt[4]{{16}} = \sqrt[4]{{\underbrace {2 \times 2 \times 2 \times
2}_{4;;times}}} = 2 \\
$

This means we have got ‘2’ as our answer which is the same as the previous method.

Note: In layman terms index is the power raised to a number and surd is the ‘nth’ root of a number i.e. $\sqrt[n]{a} = {(a)^{\dfrac{1}{n}}}$ and we can say both are opposite in nature because after solving index (power) of a number will be the raised value of the number whereas surd will become the reduced root value of the number.