Question

How do you write $\dfrac{8}{\sqrt[7]{{{x}^{15}}}}$ in exponential form?

Answer 1

Hint: We are given a term as $\dfrac{8}{\sqrt[7]{{{x}^{15}}}}$ , we have to change it into exponential, we will start our solution by learning about $\dfrac{8}{\sqrt[7]{{{x}^{15}}}}$ , what type of term it is then we will learn about exponential form, we will use that nth root of x is written as $\sqrt[n]{x}$ in radical form and ${{x}^{\dfrac{1}{n}}}$ in exponential form. We will use that $\dfrac{1}{{{x}^{a}}}={{x}^{-a}}$ , we will have one that ${{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}}$ .

Complete step-by-step solution:
We are given that $\dfrac{8}{\sqrt[7]{{{x}^{15}}}}$ , we have seen that one expression $\dfrac{8}{\sqrt[7]{{{x}^{15}}}}$ contain a $\left( \sqrt{\left( {} \right)} \right)$ radial symbol. So, clearly we are given a radical expression, we are asked to change it into the exponential.
Exponential form of a number is given as ${{a}^{b}}$ where ‘a’ denotes base and ‘b’ denotes power.
We have to change $\dfrac{8}{\sqrt[7]{{{x}^{15}}}}$ into this form.
We will be using the following exponential identity $\dfrac{1}{{{x}^{a}}}={{x}^{-a}}$ .
We will also need radical identity which says that nth root in radical can also express in exponential.
We have $\sqrt[n]{x}={{x}^{\dfrac{1}{n}}}$ and lastly we will be using ${{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}}$ .
So we will start –
We have $\dfrac{8}{\sqrt[7]{{{x}^{15}}}}$
First we change this radical $\sqrt[7]{{{x}^{15}}}$ into the exponential.
As we know the $\sqrt[n]{x}={{x}^{\dfrac{1}{n}}}$ , so –
$\sqrt[7]{{{x}^{15}}}$ will be ${{\left( {{x}^{15}} \right)}^{\dfrac{1}{7}}}$
$\Rightarrow \sqrt[7]{{{x}^{15}}}={{\left( {{x}^{15}} \right)}^{\dfrac{1}{7}}}$
Now as we know ${{\left( {{x}^{a}} \right)}^{b}}={{x}^{a\times b}}$ .
So we will use this on ${{\left( {{x}^{15}} \right)}^{\dfrac{1}{7}}}$
So, we get –
$\Rightarrow {{\left( {{x}^{15}} \right)}^{\dfrac{1}{7}}}={{x}^{15\times \dfrac{1}{7}}}$
As $15\times \dfrac{1}{7}=\dfrac{15}{7}$ so –
$\Rightarrow {{\left( {{x}^{15}} \right)}^{\dfrac{1}{7}}}={{x}^{\dfrac{15}{7}}}$
So our equation $\dfrac{8}{\sqrt[7]{{{x}^{15}}}}$ will become $\dfrac{8}{{{x}^{{}^{15}/{}_{7}}}}$ .
Now as we know that $\dfrac{1}{{{x}^{a}}}={{x}^{-a}}$ so –
$\Rightarrow \dfrac{1}{{{x}^{{}^{15}/{}_{7}}}}={{x}^{-{}^{15}/{}_{7}}}$ .
Hence, $\dfrac{8}{{{x}^{{}^{15}/{}_{7}}}}=8.{{x}^{-{}^{15}/{}_{7}}}$ .
Now we can further use that $8={{2}^{3}}$ , so –
$\Rightarrow \dfrac{8}{{{x}^{{}^{15}/{}_{7}}}}={{2}^{3}}{{x}^{-{}^{15}/{}_{7}}}$ ,
Hence, $\Rightarrow \dfrac{8}{\sqrt[7]{{{x}^{15}}}}={{2}^{3}}{{x}^{-{}^{15}/{}_{7}}}$.

Note: It is necessary to split the term into denominator and numerator for easy working.
Exponential factor as changed given as ${{x}^{a}}$ .
There are properties of exponential –
1, ${{x}^{a}}\times {{x}^{b}}={{x}^{a+b}}$
2, ${{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}}$
3, $\dfrac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}$
They also work when we use them to find the radical. We can simplify our term using these before changing them to the radical form.