Question

Find the value of:
1) ${9^{\dfrac{3}{2}}}$
2) ${32^{\dfrac{2}{5}}}$
3) ${16^{\dfrac{3}{4}}}$
4) ${125^{\dfrac{{ - 1}}{3}}}$

Answer 1

Hint:
A fractional exponent is an alternate way to express the powers and roots together. For example, the following are equivalent.
${81^{\dfrac{1}{2}}} = \sqrt[2]{{81}}$
The numerator denotes the power whereas the denominator denotes the index of the root. If there is no power being applied, write “1” in the numerator as a placeholder, which denotes the index of the root as shown below.
${81^{\dfrac{1}{2}\dfrac{{ \to power}}{{ \to \;\quad root}}}}$

Complete step by step solution:
${9^{\dfrac{3}{2}}}$, the expression states that we need to calculate the square root of 9 and then cube the resultant.
${9^{\dfrac{3}{2}}} = {\left( {\sqrt[2]{9}} \right)^3}$
Now we know that $9 = 3 \times 3$
So, which means 9 is a square of 3.
Hence, 3 is a square root of 9.
$ \Rightarrow {9^{\dfrac{3}{2}}} = {\left( {\sqrt[2]{9}} \right)^3} = {\left( {\sqrt[2]{{3 \times 3}}} \right)^3} = {3^3} = 3 \times 3 \times 3 = 27$
Therefore, $ \Rightarrow {9^{\dfrac{3}{2}}} = 27$
${32^{\dfrac{2}{5}}}$, the expression states that we need to calculate the fifth root of 32 and then square the resultant.
${32^{\dfrac{2}{5}}} = {\left( {\sqrt[5]{{32}}} \right)^2}$
Now we know that $32 = 2 \times 2 \times 2 \times 2 \times 2$
So, 32 is the fifth of 2.
Hence, 2 is a fifth root of 32.
$ \Rightarrow {32^{\dfrac{2}{5}}} = {\left( {\sqrt[5]{{32}}} \right)^2} = {\left( {\sqrt[5]{{2 \times 2 \times 2 \times 2 \times 2}}} \right)^2} = {2^2} = 2 \times 2 = 4$
Therefore, \[ \Rightarrow {32^{\dfrac{2}{5}}} = 4\]

${16^{\dfrac{3}{4}}}$, the expression states that we need to calculate the fourth root of 16 and then cube the resultant.
${16^{\dfrac{3}{4}}} = {\left( {\sqrt[4]{{16}}} \right)^3}$
Now we know that $16 = 2 \times 2 \times 2 \times 2$
So, which means 16 is fourth of 2.
Hence, 2 is a fourth root of 16.
$ \Rightarrow {16^{\dfrac{3}{4}}} = {\left( {\sqrt[4]{{16}}} \right)^3} = {\left( {\sqrt[4]{{2 \times 2 \times 2 \times 2}}} \right)^3} = {2^3} = 2 \times 2 \times 2 = 8$
Therefore, \[ \Rightarrow {16^{\dfrac{3}{4}}} = 8\]
${125^{\dfrac{{ - 1}}{3}}}$, the expression states that we need to calculate the cube root of reciprocal of 125.
${125^{\dfrac{{ - 1}}{3}}} = \sqrt[3]{{\dfrac{1}{{125}}}}$
Now we know that $125 = 5 \times 5 \times 5$
So, which means 125 is a cube of 5.
Hence, 5 is a cube root of 125.
$ \Rightarrow {125^{\dfrac{{ - 1}}{3}}} = \sqrt[3]{{\dfrac{1}{{125}}}} = \sqrt[3]{{\dfrac{1}{{5 \times 5 \times 5}}}} = \dfrac{1}{5}$

Therefore, $ \Rightarrow {125^{\dfrac{{ - 1}}{3}}} = \dfrac{1}{5}$

Note:
Please note when it comes to dealing with exponents students often make mistakes. Sometimes, they multiply the base and the exponent. For e.g. ${16^{\dfrac{1}{4}}}$is not equal to 4, it's 2. Then when using the multiplication rule don’t forget it only applies to expressions with the same base. For e.g.: - Four squared times two cubed is not the same as 8 raised to the power two plus three.${4^2} \times {2^3} \ne {8^{2 + 3}}$. The last one is the multiplication rule applies just to the product, not to the sum of two numbers.