Question

How do you simplify $ {27^{\dfrac{3}{4}}} $ ?

Answer 1

Hint: In this question, we are given a number that is raised to some power. The number has some power, the power is known as the exponent and the number is known as the base. Now for simplifying this kind of numbers we write it as a product of its prime factors, prime factors are those numbers that are divisible only by one and themselves, after prime factorization, we express the base in exponent form so that on using the laws of exponents, the expression can be simplified.

Complete step-by-step answer:
Prime factorization of 27 is done as –
 $
  27 = 3 \times 3 \times 3 \\
  27 = {3^3} \\
   \Rightarrow {27^{\dfrac{3}{4}}} = {({3^3})^{\dfrac{3}{4}}} \\
   \Rightarrow {27^{\dfrac{3}{4}}} = {3^{\dfrac{9}{4}}} \\
   \Rightarrow {27^{\dfrac{3}{4}}} = {(3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3)^{\dfrac{1}{4}}} \\
   \Rightarrow {27^{\dfrac{3}{4}}} = {(9 \times 9 \times 9 \times 9 \times 3)^{\dfrac{1}{4}}} \\
   \Rightarrow {27^{\dfrac{3}{4}}} = {({9^4}.3)^{\dfrac{1}{4}}} = {({9^4})^{\dfrac{1}{4}}}{(3)^{\dfrac{1}{4}}} \\
   \Rightarrow {27^{\dfrac{3}{4}}} = 9\sqrt[4]{3} \;
  $
Hence, the simplified form of $ {27^{\dfrac{3}{4}}} $ is $ 9\sqrt[4]{3} $ .
So, the correct answer is “ $ 9\sqrt[4]{3} $ ”.

Note: To solve the questions involving large powers or for other purposes, we can use one of the several laws of exponents, as they make the calculations a lot easier. When an exponential function is raised to another power, then keeping the base same we multiply the two powers with each other, that is, $ {({a^x})^y} = {a^{x \times y}} $ .In the question, we have to find the quad root of the cube of 27 so we express the cube of 27 as a product of its prime factors and write it as a power of four of a particular number. But the cube of 27 is not a perfect quad of any number so the expression obtained cannot be simplified further but it can be written in decimal form by putting the value of the quad root of 3.