Question

Find the value of (i) $ {\left( {32} \right)^{\dfrac{1}{5}}} $ (ii) $ {\left( {125} \right)^{\dfrac{1}{3}}} $ (iii) $ {\left( {32} \right)^{\dfrac{2}{5}}} $ (iv) $ {\left( {16} \right)^{\dfrac{1}{4}}} $

Answer 1

Hint: In this problem, first we will find the prime factorization of given number. Then, we will use the law of exponent which is given by $ {\left( {{a^m}} \right)^n} = {a^{m\; \times \;n}} $ .

Complete step-by-step answer:
To solve the given problem, first we will write the prime factorization of given number.
(i) Let us find the prime factorization of number $ 32 $ . Note that $ 32 $ is an even number. So, we can start the prime factorization of $ 32 $ with number $ 2 $ .

$ 2 $	$ 32 $
$ 2 $	$ 16 $
$ 2 $	$ 8 $
$ 2 $	$ 4 $
$ 2 $	$ 2 $
	$ 1 $

Therefore, we can write $ 32 = 2 \times 2 \times 2 \times 2 \times 2 = {2^5} $ . Now we need to find the value of $ {\left( {32} \right)^{\dfrac{1}{5}}} $ but we have $ 32 = {2^5} $ . So, now we can write $ {\left( {32} \right)^{\dfrac{1}{5}}} = {\left( {{2^5}} \right)^{\dfrac{1}{5}}} \cdots \cdots \left( 1 \right) $ . Let us compare $ {\left( {{2^5}} \right)^{\dfrac{1}{5}}} $ with $ {\left( {{a^m}} \right)^n} $ . So, we can say that $ a = 2,m = 5 $ and $ n = \dfrac{1}{5} $ . Now we are going to use the law of exponent which is given by $ {\left( {{a^m}} \right)^n} = {a^{m\; \times \;n}} $ . Therefore, from $ \left( 1 \right) $ we get $ {\left( {32} \right)^{\dfrac{1}{5}}} = {\left( {{2^5}} \right)^{\dfrac{1}{5}}} = {2^{5\; \times \;\dfrac{1}{5}}} = {2^1} = 2 $ . Hence, we can say that the value of $ {\left( {32} \right)^{\dfrac{1}{5}}} $ is equal to $ 2 $ .
(ii) Let us find the prime factorization of number $ 125 $ .

$ 5 $	$ 125 $
$ 5 $	$ 25 $
$ 5 $	$ 5 $
	$ 1 $

Therefore, we can write $ 125 = 5 \times 5 \times 5 = {5^3} $ . Now we need to find the value of $ {\left( {125} \right)^{\dfrac{1}{3}}} $ but we have $ 125 = {5^3} $ . So, now we can write $ {\left( {125} \right)^{\dfrac{1}{3}}} = {\left( {{5^3}} \right)^{\dfrac{1}{3}}} $ . Now we are going to use the law of exponent which is given by $ {\left( {{a^m}} \right)^n} = {a^{m\; \times \;n}} $ . Therefore, we get $ {\left( {125} \right)^{\dfrac{1}{3}}} = {\left( {{5^3}} \right)^{\dfrac{1}{3}}} = {5^{3\; \times \;\dfrac{1}{3}}} = {5^1} = 5 $ . Hence, we can say that the value of $ {\left( {125} \right)^{\dfrac{1}{3}}} $ is equal to $ 5 $ .
(iii) Let us find the prime factorization of number $ 32 $ .

$ 2 $	$ 32 $
$ 2 $	$ 16 $
$ 2 $	$ 8 $
$ 2 $	$ 4 $
$ 2 $	$ 2 $
	$ 1 $

Therefore, we can write $ 32 = 2 \times 2 \times 2 \times 2 \times 2 = {2^5} $ . Now we need to find the value of $ {\left( {32} \right)^{\dfrac{2}{5}}} $ but we have $ 32 = {2^5} $ . So, now we can write $ {\left( {32} \right)^{\dfrac{2}{5}}} = {\left( {{2^5}} \right)^{\dfrac{2}{5}}} $ . Now we are going to use the law of exponent which is given by $ {\left( {{a^m}} \right)^n} = {a^{m\; \times \;n}} $ . Therefore, we get $ {\left( {32} \right)^{\dfrac{2}{5}}} = {\left( {{2^5}} \right)^{\dfrac{2}{5}}} = {2^{5\; \times \;\dfrac{2}{5}}} = {2^2} = 4 $ . Hence, we can say that the value of $ {\left( {32} \right)^{\dfrac{2}{5}}} $ is equal to $ 4 $ .

(iv) Let us find the prime factorization of number $ 16 $ .

$ 2 $	$ 16 $
$ 2 $	$ 8 $
$ 2 $	$ 4 $
$ 2 $	$ 2 $
	$ 1 $

Therefore, we can write $ 16 = 2 \times 2 \times 2 \times 2 = {2^4} $ . Now we need to find the value of $ {\left( {16} \right)^{\dfrac{1}{4}}} $ but we have $ 16 = {2^4} $ . So, now we can write $ {\left( {16} \right)^{\dfrac{1}{4}}} = {\left( {{2^4}} \right)^{\dfrac{1}{4}}} $ . Now we are going to use the law of exponent which is given by $ {\left( {{a^m}} \right)^n} = {a^{m\; \times \;n}} $ . Therefore, we get $ {\left( {16} \right)^{\dfrac{1}{4}}} = {\left( {{2^4}} \right)^{\dfrac{1}{4}}} = {2^{4\; \times \;\dfrac{1}{4}}} = {2^1} = 2 $ . Hence, we can say that the value of $ {\left( {16} \right)^{\dfrac{1}{4}}} $ is equal to $ 2 $ .

Note: In this problem, the numbers $ 16 $ and $ 32 $ are even numbers. So, we can start the prime factorization of these numbers with number $ 2 $ . The number $ 125 $ is not an even number and it is not divisible by $ 3 $ because the sum of digits of the number $ 125 $ is not divisible by $ 3 $ . If the unit digit of the number is $ 0 $ or $ 5 $ then that number is divisible by $ 5 $ . So, we can start the prime factorization of $ 125 $ with the number $ 5 $ .