So we’ve worked with exponents. We know the basic exponent properties. However, the lingering question is; What do we do with a rational exponent? We will derive a Rational Exponents/Radicals Formula that will help us easily convert between radicals and exponents.
What happens when our exponent is not a whole number? What shall we do if our exponent is a rational number in the form of a fraction?
First, let’s take a look at an example. What happens when we decompose a whole number exponent into the sum of two or more fractions.
Before we get started let’s review briefly;
\Large{x^{a+b}=x^a+x^b }\Large{(x^a)^b=x^{ab} }\Large{ 1=\frac{n}{n} \\ such that}\Large{ \begin{align} \sqrt[n]{x^n} &= x \\
\text{for }x\ge&0, n\isin \reals \end{align}} \Large \sqrt{x}=\sqrt[2]{x}The Plan
To the right, we start with 4^1 and decompose the exponent into the sum, \frac{1}{2}+\frac{1}{2}. Using the product property of exponents we can turn a power with a sum for an exponent into a square product. And finally, taking the square root, we are left with;
\Large\bold{\sqrt[2]{4^1}=4^{\frac{1}{2}}}
\Large{ \begin{aligned} 4 &= 4^{1} \\
4&= 4^{ \frac{2}{2} } \\
4&=4^{\frac{1}{2}+\frac{1}{2}} \\
4&=4^{\frac{1}{2}} \cdot 4^{\frac{1}{2}} \\
\sqrt[2]{4} &= \sqrt[2]{4^{\frac{1}{2}} \cdot 4^{\frac{1}{2}}} \\
\sqrt[2]{4} &=4^{\frac{1}{2}} \\
\sqrt[2]{4^1}&=4^{\frac{1}{2}}
\end{aligned} }
Generalizing
We need to generalize this process so that, independent of values, we have a identity relating radicals and exponents. We will start by applying this concept, of decomposing the exponent into a sum, to powers whose base is a variable.
Square Roots
\Large{ \begin{aligned} x &= x^{1}\\
x&=x^{ \frac{2}{2} }\\
x&=x^{(\frac{1}{2}+\frac{1}{2})}\\
x&=x^{\frac{1}{2}} \cdot x^{\frac{1}{2}} \\
\sqrt[2]{x} &= \sqrt[2]{x^{\frac{1}{2}} \cdot x^{\frac{1}{2}}} \\
\sqrt[2]{x} &=x^{\frac{1}{2}} \\
\sqrt[2]{x} &=x^{\frac{1}{2}} \\
\end{aligned} }Cube Roots
\Large{ \begin{aligned} x &= x^{1}\\
x&=x^{ \frac{3}{3} } \\
x&=x^{(\frac{1}{3}+\frac{1}{3}+\frac{1}{3})}\\
x&=x^{\frac{1}{3}} \cdot x^{\frac{1}{3}} \cdot x^{\frac{1}{3}}\\
\sqrt[3]{x} &= \sqrt[3]{x^{\frac{1}{3}} \cdot x^{\frac{1}{3}} \cdot x^{\frac{1}{3}}} \\
\sqrt[3]{x} &=x^{\frac{1}{3}} \\
\end{aligned} }bth roots
\Large{ \begin{aligned} x &= x^{1}\\
x&=x^{ \frac{b}{b} } \\
x&=x^{(\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+...)}\\
x&=x^{\frac{1}{b}} \cdot x^{\frac{1}{b}} \cdot x^{\frac{1}{b}} \cdot ...\\
\sqrt[b]{x} &= \sqrt[b]{x^{\frac{1}{b}} \cdot x^{\frac{1}{b}} \cdot x^{\frac{1}{b}} \cdot ...} \\
\sqrt[b]{x} &=x^{\frac{1}{b}} \end{aligned} }\boxed{ \LARGE{ x^{\frac{1}{\color{blue}b}}= \sqrt[\color{blue}b]{x} }}
One More Step
Now, let’s take this and generalize it further for a fraction of the form \Large{ \frac{a}{b} }.
\Large{ \begin{aligned}
x^{\frac{a}{b}} &= x^{(a \cdot \frac{1}{b})} \text{ ... or ...}&&=x^{(\frac{1}{b}\cdot a)}\\
&= (x^{a})^{\frac{1}{b}} &&=(x^{\frac{1}{b}})^{a}\\
&=\sqrt[b]{x^{a}} &&=(\sqrt[b]{x})^{a}\\
\end{aligned} }
The Fractional Exponent/Radical Formula
With these two conclusions, we are left with the final rule that we use for converting from exponential form to radical form and vice versa:
\boxed{ \huge{ x^{\frac{\color{red}a}{\color{blue}b}}= \sqrt[\color{blue}b]{x^{\color{red}a}}=(\sqrt[\color{blue}b]{x})^{\color{red}a} }}
