Question

Find the cube root of $ \left(4^{\dfrac{17}{24}}\right) $ :
 $ \left(a\right)4^{\dfrac{17}{48}} $
 $ \left(b\right)4^{\dfrac{17}{36}} $
 $ \left(c\right)2^{\dfrac{17}{48}} $
 $ \left(d\right)2^{\dfrac{17}{36}} $

Answer 1

Hint: We are required to find the cube root of a number already given in powers. We will use the definition of cube root in this case. We will use the fact that the cube root of any number is the number raised to the power $ \dfrac{1}{3} $ . After that we will make some more modifications, so that the answer becomes equal to one of the options given.

Complete step by step answer:
Suppose we have a number $ a $ . We multiply this by itself three times. The number thus obtained will be $ a^3 $ . Suppose this number is equal to $ y $ .
 $ \implies y=a^3 $
Then, $ y $ is said to be the cube of the number $ a $ . Now, if we take the power $ \dfrac{1}{3} $ on both sides then we obtain:
 $ y^{\dfrac{1}{3}}=a $
So, $ a $ is said to be the cube root of $ y $ . So, if we want to find the cube root of a number, we simply raise it to the power $ \dfrac{1}{3} $ .
We have $ \left(4^{\dfrac{17}{24}}\right) $ . Raising it to the power of $ \dfrac{1}{3} $ , we get:
 $ \left(4^{\dfrac{17}{24}}\right)^{\dfrac{1}{3}} $
We use the following formula that holds true for any three reals $ a $ , $ b $ and $ c $ :
 $ \left(a^b\right)^c=a^{bc} $
We get:
 $ \left(4^{\dfrac{17}{24}}\right)^{\dfrac{1}{3}}=4^{\dfrac{17}{24}\times\dfrac{1}{3}} $
Now, we know that $ 4=2^2 $ . Applying this here we get:
 $ \left(4^{\dfrac{17}{24}}\right)^{\dfrac{1}{3}}=2^{2\times\dfrac{17}{24}\times\dfrac{1}{3}} $
 $ =2^{\dfrac{17}{36}} $

So, the correct answer is “Option d”.

Note: Note that while you raise one power to another power you are supposed to multiply those powers. It is a common mistake to add those powers instead of multiplying while you raise one power to another. Also while multiplying, be aware because a calculation mistake while multiplying can lead to a wrong answer.