Question

What is the fourth root of $80$?

Answer 1

Hint: In this problem we need to calculate the fourth root of the given number. For this we need to have the given number in the exponential form. We know that the exponential form of a number is simply obtained by prime factorization of the number. After prime factorization of the number, we can use the exponential formula $a\times a\times a\times a.....\text{n times}={{a}^{n}}$ to get the exponential form of the number. Once we have the exponential form, we can find the fourth root of the number by calculating the value of ${{\left( \dfrac{1}{4} \right)}^{th}}$ power of the number. Here we may use some of the exponential formulas like ${{\left( {{a}^{m}}\times {{b}^{n}} \right)}^{p}}={{a}^{mp}}\times {{b}^{np}}$, ${{a}^{m+n}}={{a}^{m}}\times {{a}^{n}}$ etc to simplify the value.

Complete step by step answer:
Given number is $80$.
We have the prime factorization of the number $80$ as
$80=2\times 2\times 2\times 2\times 5$
Applying the exponential formula $a\times a\times a\times a.....\text{n times}={{a}^{n}}$ in the above equation, then we will get
$80={{2}^{4}}\times 5$
We are going to calculate the ${{\left( \dfrac{1}{4} \right)}^{th}}$ power of the number to get the fourth root of the number, then we will have
$\sqrt[4]{80}={{\left( {{2}^{4}}\times 5 \right)}^{\dfrac{1}{4}}}$
Applying the exponential formula ${{\left( {{a}^{m}}\times {{b}^{n}} \right)}^{p}}={{a}^{mp}}\times {{b}^{np}}$ in the above equation, then we will get
$\sqrt[4]{80}={{2}^{4\times \dfrac{1}{4}}}\times {{\left( 5 \right)}^{\dfrac{1}{4}}}$
Simplifying the above equation, then we will have
$\sqrt[4]{80}=2\sqrt[4]{5}$

Hence the fourth root of the number $80$ is $2\sqrt[4]{5}$.

Note: In the above solution we have directly used the prime factorization of the number $80$ as $80=2\times 2\times 2\times 2\times 5$. To get this mean prime factorization we need to check whether the given number is divisible by the prime numbers $2$, $3$, $5$, $7$, $11...$. If yes then we can write the number as a product of the prime number as well as a reminder. We will continue this process until we will get a reminder as one. In this process we can get the prime factorization of any number.