Apply the Rules of Exponents and Properties of Radicals to simplify expressions with fractional and arbitrary real exponents, and to solve equations involving roots.
In this section, we will review the concepts of roots and radicals. Please note that in this book the focus is on real numbers — complex numbers are not considered. In particular, when we talk about roots, we mean real roots.
We say that a number \(y\) is a root of order \(2\) (a square root) of a number \(a\) if \(y^2 = a\text{.}\) Similarly, we say that a number \(y\) is a root of order \(3\) (a cube root) of \(a\) if \(y^3 = a\text{.}\) The definition of a root of order \(n\) is given below.
Let \(a\) be a given number and let \(n\) be a positive integer. We say that \(y\) is a root of order \(n\) of \(a\text{,}\) or an \(n\) th root of \(a\text{,}\) if \(y^n = a\text{.}\)
For example, let \(a = 4\) and \(n = 2\text{.}\) The number \(4\) has two roots of order \(2\text{:}\)\(y = 2\) and \(y = -2\text{,}\) since \(2^2 = (-2)^2 = 4\text{.}\) However, if \(a = -4\) and \(n = 2\text{,}\) the situation would be different. The number \(-4\) has no roots of order \(2\text{,}\) since there are no real numbers whose square is \(-4\text{.}\) We say that the square root of \(-4\)does not exist.
Now, take \(a = -64\) and \(n = 3\text{.}\) The number \(-64\) has exactly one root of order 3, which is \(y = -4\text{.}\) Indeed, \((-4)^3 = -64\text{.}\) Observe that \(4^3 = 64\) so 4 is not a cube root of \(-64\text{.}\)
If \(a \gt 0\text{,}\) then \(a\) has two \(n\) th roots, one positive and one of the same magnitude but negative. We denote the positive\(n\) th root as:
\begin{equation*}
\sqrt[n]{a}
\end{equation*}
(We use the “radical” symbol \(\sqrt{\,\,}\) to denote a root.) The two \(n\) th roots can then be written as
We are looking for all numbers \(y\) such that \(y^4 = 81\text{.}\) The order, \(4\text{,}\) is even. Hence, we have two roots, one positive and one negative. The positive root is denoted by \(\sqrt[4]{81}\) and the two roots are:
The number \(7\) has two roots of order \(2\text{:}\)\(\sqrt{7}\) and \(-\sqrt{7}\text{.}\) We cannot easily guess them as they are not integers. Using a calculator, though, \(\sqrt{7} \approx 2.65\text{.}\) The two roots are then approximately \(2.65\) and \(-2.65\text{.}\)
The order, \(3\text{,}\) is odd. Hence, there is only one root of order \(3\) of \(-27\) denoted as \(\sqrt[3]{-27}\text{.}\) As \((-3)^3 = -27\text{,}\) we have
By definition, \(x\) is a root of order \(n\) of \(a\) if \(x\) is a solution to the equation:
\begin{equation*}
x^n = a
\end{equation*}
Hence, roots and radicals appear naturally when solving equations containing powers of the unknown. You may recall seeing many radicals in Chapter 3 in the context of quadratic equations.
Let \(a\text{,}\)\(b\) be given numbers. Let \(n\text{,}\)\(m\) be positive integers. Then the following equalities hold provided that the roots involved exist, and both sides are defined:
To extend the definition to fractional exponents \(a^{\frac{m}{n}}\text{,}\) we will use roots. In the first step, for every positive integer \(n\text{,}\) we define \(a^{\frac{1}{n}}\) as:
Does this make sense? Recall that \(\left(\sqrt[n]{a}\right)^n = a\text{.}\) Therefore,
\begin{equation*}
(a^{\frac{1}{n}})^n = a
\end{equation*}
which is what Rules of Exponents would dictate. In the next step, we define \(a^{\frac{m}{n}} = (a^{\frac{1}{n}})^m = (\sqrt[n]{a})^m\text{.}\) Here is a precise definition of a power with a fractional exponent.
Note that \(\sqrt[n]{a}\) exists unless \(n\) is even and \(a\) is negative. The combination of negative radicands (numbers under radicals) with even roots causes a number of problems for the behavior of fractional exponents, and the Rules of Exponents do not always hold.
Therefore, when we talk about fractional powers, we will most often assume that bases are positive except for some simple cases where no issues arise, such as
By definition \(4^{\frac{1}{2}} = \sqrt[2]{4} = \sqrt{4} = 2\text{.}\) Note that \(4^{\frac{1}{2}}\) is the positive of the two square roots of \(4\) as is \(\sqrt{4}\text{.}\)
We have defined powers \(a^p\) for integer and fractional exponents \(p\text{.}\) As you may know, not all real numbers can be expressed as fractions (irrational numbers, for example). Can we define powers \(a^p\) for all real numbers \(p\text{?}\) The answer is yes, provided the base \(a\) is positive. The construction falls outside the scope of this course. It suffices to know that \(a^p\) can be defined for all exponents and Rules of Exponents are preserved. Therefore, we have the following result.
Use Rules of Exponents to simplify the following. Write your answers in terms of powers and not radicals. Assume \(a\text{,}\)\(b\text{,}\)\(x\text{,}\) and \(y\) are all positive.
First, observe that \(14 = 2 \cdot 7\) and \(21 = 3 \cdot 7\text{.}\) This allows us to simplify the fraction \(14/21\) as \(2/3\text{.}\) Since \(x^{-2} = \dfrac{1}{x^2}\) and \(\sqrt{x^7} = x^{7/2}\text{,}\) then
Solve the equation for \(x\) or for \(b\text{.}\) Be sure to list all solutions. Give exact and approximate values rounded off to three decimal places. If there are no solutions say so.
