# How to Merging of Exponents and Roots

Merging Exponents and Roots

In this section, we shall use the properties of integral exponents to motivate definitions for the use of rational numbers as exponents. These definitions will tie together the concepts of exponent and root.

Let’s consider the following comparisons.

From our study of radicals we know that $(\sqrt{5})^2=5$

If $(b^n)^m=b^{mn}$ is to hold when n equals a rational number of the form $\frac{1}{p}$, where p is a positive integer greater than one, then $(5^{\frac{1}{2}})^2=5^{2(\frac{1}{2}}=5^1=5$

It would seem reasonable to make the following definition.

Definition 1 :

If b is a real number, n a positive integer greater than one, and $\sqrt[n]{b}$ exists, then $b^{\frac{1}{n}}=\sqrt[n]{b}$.

Definition 1 states that : $b^{\frac{1}{n}}$ means the nth root of b. We shall assume that b and n are chosen so that $\sqrt[n]{b}$ exists.

For example, $(-25)^{\frac{1}{2}}$ is not meaningful at this time because $\sqrt{-25}$ is not a real number.

Consider the following examples that illustrate the use of definition 1. $25^{\frac{1}{2}}=\sqrt{25}=5$ $8^{\frac{1}{3}}=\sqrt{8}=2$ $(-27)^{\frac{1}{3}}=\sqrt{-27}=-3$ $16^{\frac{1}{4}}=\sqrt{16}=2$ $(\frac{36}{49})^{\frac{1}{2}}=\sqrt{\frac{36}{49}}=\frac{6}{7}$

The following definition provides the basic for the use of all rational numbers as exponents.

Definition 2 :

If $\frac{m}{n}$ is a rational number, where n is a positive integer greater than one, and b is a real number such that $\sqrt[n]{b}$ exists, then $b^{\frac{m}{n}}=\sqrt[n]{b^m}=(\sqrt[n]{b})^m$

In definition 2, notice that denominator of the exponent is the index of the radical and the numerator of the exponent is either the exponent of the radicand or the exponent of the root.

Whether we use the form $\sqrt[n]{b^m}\text{ or }(\sqrt[n]{b})^m$ for computational purposes depends somewhat on the magnitude of the problem. Let’s use both forms on two problems to illustrate this point. $8^{\frac{2}{3}}=\sqrt{8^2}=\sqrt{64}=4$

Or $8^{\frac{2}{3}}=(\sqrt{8})^2=2^2=4$ $27^{\frac{2}{3}}=\sqrt{27^2}=\sqrt{729}=9$

Or $27^{\frac{2}{3}}=(\sqrt{27})^2=3^2=9$

To compute $8^{\frac{2}{3}}$, either form seems to work about as well as the other one. However, to compute $27^{\frac{2}{3}}$, it should be obvious that $(\sqrt{27})^2$ is much easier to handle than $\sqrt{27^2}$.