How do you simplify \[{16^{\left( {\dfrac{3}{4}} \right)}}\]?

Hint:
To simplify exponent any number, one should solve it by method of prime factorisation. We can take the LCM of the number to simplify. Also, use the exponent rule, which says that when the bases are the same, the powers would add.

Complete step by step solution:
We know the LCM of 16 is $2\, \times \,2\, \times \,2\, \times 2$.
(2 is the smallest even prime number)
To simplify ${16^{\dfrac{3}{4}}}$, it can be written as, \[{16^{\dfrac{3}{4}}}\, = \,{(2 \times 2 \times 2 \times 2)^{\dfrac{3}{4}}}\]
Using the exponent rule, when the bases are same the powers get add. So we will get,
${16^{\dfrac{3}{4}}} = \,{\left( {{2^4}} \right)^{\dfrac{3}{4}}}$
Now using another exponent rule, ${\left( {{A^x}} \right)^y}$the total times A would get multiplied is $x \times y$ times.
Then, we can write
${16^{\dfrac{3}{4}}} = \,{2^{\left( {4 \times \dfrac{3}{4}} \right)}}$
The powers will cancel out and we will get
${16^{\dfrac{3}{4}}} = \,{2^3}$
Multiply the base number 2, three times
${16^{\dfrac{3}{4}}} = 8$
So, by solving ${16^{\dfrac{3}{4}}}$ , we will get 8.
Where, 8 is a positive integer.

So, the correct answer is 8.

Additional information: With ${A^x}.{A^{^y}}$, the number of times we multiply ‘A’ completely depends on the powers. This means multiply ‘A’ first "x" times, then by another "y" times, for a total of $x + y$ times. Also the law that ${\left( {{A^x}} \right)^y}$, means first you multiply "x" times. Then the number we will get will get multiplied "y" times, for a total of $x \times y$ times.

Note:
The exponent of a number says how many times to use the number in a multiplication and the number of times it gets multiplied becomes its power. To solve any problems we have several rules to deal with it. One should read the rules and try to use it in the given questions carefully.