Question

How do you write ${{7}^{\dfrac{3}{4}}}$ in radical form?

Answer 1

Hint: To write ${{7}^{\dfrac{3}{4}}}$ in the radical form. For that, we should see the exponent of the number 7. The number written in the numerator of the rational exponent is the power and the denominator of the rational exponent is the root over of the number. So, the radical form of ${{7}^{\dfrac{3}{4}}}$ is writing this number in the form of power and root over.

Complete step by step answer:
The number which we are asked to write in the radical form:
${{7}^{\dfrac{3}{4}}}$
We know that in the rational exponent of a number, the numerator is the power of the number and the denominator of this rational exponent is the root.
The root expression is:
$\sqrt{{}}$
And in the above, we have given the 4th root of a number so writing the fourth root in terms of the above root expression we get,
$\Rightarrow \sqrt[4]{{}}$
Now, writing 3 in the power of 7 and then putting the above fourth root over expression we get,
$\Rightarrow \sqrt[4]{{{7}^{3}}}$
Hence, we have written the given number ${{7}^{\dfrac{3}{4}}}$ into radical form as $\sqrt[4]{{{7}^{3}}}$.

Note: There is also another way in which we can write ${{7}^{\dfrac{3}{4}}}$ in the radical form. Instead of writing the power of 7 inside the root over expression, we can write the power of 3 outside the root over expression in the following way:
${{\left( \sqrt[4]{7} \right)}^{3}}$
You might think how come we can write the same number into two radical forms. The answer lies in the multiplication of root and the power in the exponent. When you write the above two radical forms which we have shown below:
$\sqrt[4]{{{7}^{3}}}$
${{\left( \sqrt[4]{7} \right)}^{3}}$
Conversion of the above two radical forms into the rational exponent we get,
$\begin{align}
  & \Rightarrow \sqrt[4]{{{7}^{3}}} \\
 & ={{\left( {{7}^{3}} \right)}^{\dfrac{1}{4}}} \\
 & \Rightarrow {{\left( \sqrt[4]{7} \right)}^{3}} \\
 & ={{\left( {{7}^{\dfrac{1}{4}}} \right)}^{3}} \\
\end{align}$
And we know the exponent multiplication in a number is as follows:
$\begin{align}
  & {{\left( {{n}^{a}} \right)}^{\dfrac{1}{b}}} \\
 & ={{n}^{a\times \dfrac{1}{b}}} \\
 & ={{n}^{\dfrac{a}{b}}} \\
\end{align}$
So, using this exponent multiplication in simplifying the above radical forms:
$\begin{align}
  & \Rightarrow \sqrt[4]{{{7}^{3}}}={{\left( {{7}^{3}} \right)}^{\dfrac{1}{4}}} \\
 & ={{7}^{\dfrac{3}{4}}} \\
 & \Rightarrow {{\left( \sqrt[4]{7} \right)}^{3}}={{\left( {{7}^{\dfrac{1}{4}}} \right)}^{3}} \\
 & ={{7}^{\dfrac{3}{4}}} \\
\end{align}$
Hence, we are getting the same radical conversions in both the cases.