Page 1.3: Rules of Exponents and Radicals
A big part of learning calculus involves being able to understand the notation that is used. More generally, anytime a new symbol is introduced, we should question it and see several examples and non-examples. Often, the purpose of a symbol is to provide a means of simplifying the notation and this is also the case when it comes to exponents and radicals. For instance, it would be rather cumbersome to write \( 5 \times 5 \times 5 \times 5 \) and even more cumbersome if we wished to multiply \(5\) by itself twenty times. Thus, we introduce the notion of an exponent . Instead of writing \( 5 \times 5 \times 5 \times 5 \), we can simply denote this as \( 5^4\) which is read as "\(5\) raised to the fourth power" and in the case of multiplying \(5\) by itself twenty times, rather than writing \( 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \), we may simply denote this by \(5^{20}\). We call \(5\) the base and we call \(20\) the exponent .
More generally, we make the following definition.
Definition: If \(a\) is any real number and \(n\) is any positive integer, then the \(n^{th} \) power of \(a\) is denoted by \( a^n \) where
\[ a^n = \underbrace{a \times a \times \ldots \times a}_{n ~ \text{times} } \]
The number \(a\) is called the base and \(n\) is called the exponent .
Whenever a definition is provided, we should reflect on the assumptions and requirements of the definition. Doing so will lead to us a deeper understanding of the material.
Here, we notice that the definition requires \(a\) to be a real number and pretty much all the numbers we will see in calculus will be real numbers (the main exception to this will be noted later on). Since most of the numbers we see will be real numbers, we will not place too much attention on \(a\). Rather, our focus is on \(n\). We see our definition requires \(n\) to be a positive integer . This immediately raises two questions:
- Is it possible for \(n\) to be negative? For instance, how do we define \(5^{-20} \)?
- Is it possible for \(n\) to not be an integer? For instance, what does \(5^{20.23} \) mean?
We will consider the answer to these two questions in just a moment. For now, allow us to consider some examples.
Compute the following:
1) \( \bigg(\dfrac{1}{2} \bigg)^5 \)
2) \(2^4 \)
3) \( (-2)^{4} \)
4) \( - 2^4 \)
5) \( (-2)^3 \)
6) \( - 2^3 \)
- Answer
-
1) \( (\frac{1}{2} )^5 = (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{32} \)
2) \(2^4 = (2)(2)(2)(2) = 16 \)
3) \( (-2)^4 = (-2)(-2)(-2)(-2) = 16 \)
4) \( - 2^4 = - (2)(2)(2)(2) = - 16 \)
5) \( (-2)^3 = (-2)(-2)(-2) = -8 \)
6) \( - 2^3 = - (2)(2)(2) = -8 \)
We now turn our attention to the first question posed earlier: is it possible for \(n\) to be negative? The answer is yes:
Definition: If \(a \neq 0\) is a real number and \(n\) is a positive integer, then
\[ a^{-n} = \dfrac{1}{a^n} ~~ \text{and}~~ a^0 = 1 \nonumber\ \]
As mentioned above, whenever a definition is provided, we should reflect on the assumptions and requirements of the definition. Doing so will lead to us a deeper understanding of the material. Here, we see that \(a\) is not allowed to be equal to zero and we may wonder why. That is, what does \( 0^0 \) equal to? The answer is that for us, \(0^0 \) is not defined! We will touch a little bit more upon this in Day 8 but for those who are interested, a good exercise is to graph the functions \( f(x) = x^0 \) and \(g(x) = 0^x \). What happens to the values of \(f\) and \(g\) as \(x\) gets closer and closer to zero from th right hand side? It is okay if we cannot answer this now since the solution will be apparent by Day 8. In the meantime, allow us to consider the following example.
Compute the following:
1) \( \bigg(\dfrac{1}{2} \bigg)^{-5} \)
2) \(2^{-4} \)
3) \( (-2)^{-4} \)
4) \( -2^{-4} \)
5) \( (-2)^{-3} \)
6) \( -2^{-3} \)
7) \( x^{-1} \) you may assume that \(x \neq 0 \).
8) \( x^{-2} \) you may assume that \(x \neq 0 \).
9) \( x^{-3} \) you may assume that \(x \neq 0 \).
10) \( x^{-2023} \) you may assume that \(x \neq 0 \).
11) \(2023^0\)
- Answer
-
1) \( \bigg(\dfrac{1}{2} \bigg)^{-5} = \dfrac{1}{(\frac{1}{2})^5} = \dfrac{1}{\frac{1}{32}} = 32 \)
2) \(2^{-4} = \dfrac{1}{2^4} = \dfrac{1}{16} \)
3) \( (-2)^{-4} = \dfrac{1}{(-2)^4} = \dfrac{1}{16} \)
4) \( -2^{-4} = - \dfrac{1}{2^{4}} = - \dfrac{1}{16} \)
5) \( (-2)^{-3} = \dfrac{1}{(-2)^3} = \dfrac{1}{-8} = - \dfrac{1}{8} \)
6) \( -2^{-3}= - (2^{-3}) = - \dfrac{1}{(2)^3} = - \dfrac{1}{8} \)
7) \( x^{-1} = \dfrac{1}{x} \)
8) \( x^{-2} = \dfrac{1}{x^2} \)
9) \( x^{-3} = \dfrac{1}{x^3} \)
10) \( x^{-2023} = \dfrac{1}{x^{2023}} \)
11) \(2023^0 = 1 \)
It is important that we understand the rules when it comes to working with exponents. These rules are summarized below.
Theorem: Let \(a\) and \(b\) be real numbers and \(m\) and \(n\) be integers. Then we have the following:
1) \( a^m a^n = a^{m+n} \)
2) \( \frac{a^m}{a^n} = a^{m-n} \)
3) \( \bigg( a^m \bigg)^n = a^{mn} \)
4) \( (ab)^n = a^n b^n \)
5) \( \displaystyle \bigg( \dfrac{a}{b} \bigg)^n = a^n b^n \)
Whenever a theorem is presented to you, a good idea is to take a few moments and translate the symbols into words. The above theorem is saying the following:
1) In order to multiply two powers with the same base, we add the exponents.
2) In order to divide two powers with the same base, we subtract the exponents.
3) Whenever we raise a power to a new power, we multiply the exponents.
4) To raise a product to a power, raise each factor to the power.
5) To raise a quotient to a power, raise both numerator and denominator to the power.
With these ideas in mind, allow us to try a few examples.
Perform the given operation using the multiplication properties of exponents and write your answer in simplest form:
- \(b^{2} \cdot b^{3}=b^{2+3}=b^{5}\)
- \(x^{8} x^{7}=x^{8+7}=x^{15}\)
- \(a^{8} a^{9} a^{14}=a^{8+9+14}=a^{31}\)
- \(4 x^{4} \cdot 7 x^{6}=(4 \cdot 7)\left(x^{4} \cdot x^{6}\right)=28 x^{10}\)
- \(5 x y^{3} \cdot 6 y=(5 \cdot 6)(x)\left(y^{3} \cdot y\right)=30 x y^{4}\)
- \(\left(5 x^{4} y^{2}\right)\left(2 x^{7} y^{3}\right)=10 x^{4+7} y^{2+3}=10 x^{11} y^{5}\)
- \((2 x)^{3}=2^{3} x^{3}=8 x^{3}\)
- \(\left(-4 a^{2} b^{5}\right)^{3}=(-4)^{3} a^{2 \cdot 3} b^{5 \cdot 3}=-64 a^{6} b^{15}\)
- \(\begin{align*}\left(-5 r^{3} s\right)^{2} \cdot\left(2 r^{4} s^{3}\right)^{3} \cdot\left(-r^{2} s^{2}\right) &=(-5)^{2} r^{3 \cdot 2} s^{2} \cdot 2^{3} r^{4 \cdot 3} s^{3 \cdot 3} \cdot(-1) r^{2} s^{2} \\&=25 r^{6} s^{2} \cdot 8 r^{12} s^{9} \cdot(-1) r^{2} s^{2} \\&=-200 r^{20} s^{13}\end{align*}\)
- \(\dfrac{x^{5}}{x^{3}}=\dfrac{x \cdot x \cdot x \cdot x \cdot x}{x \cdot x \cdot x}=\dfrac{x \cdot x}{1}=\dfrac{x^{2}}{1}=x^{2}\)
- \(\dfrac{x^{3}}{x^{5}}=\dfrac{x \cdot x \cdot x}{x \cdot x \cdot x \cdot x \cdot x}=\dfrac{1}{x \cdot x}=\dfrac{1}{x^{2}}\)
- \(\left(\dfrac{x}{y}\right)^{5}=\dfrac{x}{y} \cdot \dfrac{x}{y} \cdot \dfrac{x}{y} \cdot \dfrac{x}{y} \cdot \dfrac{x}{y}=\dfrac{x \cdot x \cdot x \cdot x \cdot x}{y \cdot y \cdot y \cdot y \cdot y}=\dfrac{x^{5}}{y^{5}}\)
Perform the given operation using the division properties of exponents and state your answer in simplest form:
- \(\dfrac{b^{8}}{b^{7}}=b^{8-7}=b^{1}=b\)
- \(\dfrac{x^{12} y^{2}}{x^{8} y}=x^{12-8} y^{2-1}=x^{4} y\)
- \(\dfrac{8 x^{6} y^{2}}{6 x^{5} y^{7}}=\dfrac{4 x^{6-5}}{3 y^{7-2}}=\dfrac{4 x^{1}}{3 y^{5}}=\dfrac{4 x}{3 y^{5}}\)
- \(\dfrac{\left(m^{2}\right)^{3}\left(n^{4}\right)^{5}}{\left(m^{3}\right)^{3}}=\dfrac{m^{6} n^{20}}{m^{9}}=\dfrac{n^{20}}{m^{3}}\)
- \(\left(\dfrac{3 a^{2} b^{4}}{9 c^{3}}\right)^{2}=\left(\dfrac{a^{2} b^{4}}{3 c^{3}}\right)^{2}=\dfrac{a^{2 \cdot 2} b^{4 \cdot 2}}{3^{2} c^{3 \cdot 2}}=\dfrac{a^{4} b^{8}}{9 c^{6}}\)Another way to simplify this correctly is this: \(\left(\dfrac{3 a^{2} b^{4}}{9 c^{3}}\right)^{2}=\dfrac{3^{2} a^{2 \cdot 2} b^{4 \cdot 2}}{9^{2} c^{3 \cdot 2}}=\dfrac{9 a^{4} b^{8}}{81 c^{6}}=\dfrac{a^{4} b^{8}}{9 c^{6}}\)
Perform the given operation and write your answer using positive exponents only
- \(\dfrac{x^{4} y^{2}}{x^{3} y^{-3}}=x^{4-3} y^{2-(-3)}=x y^{5}\)
- \(\left(x^{2} y\right) \cdot\left(x y^{-4}\right)=x^{2+1} y^{1+(-4)}=\dfrac{x^{3}}{y^{3}}\)
We will now turn our attention to the second question asked earlier. Namely, is it possible for \(n\) to not be an integer? For instance, what does \(5^{20.23} \) mean? In order to answer this, we must introduce the notion of a radical or root.
Definition: Let \(a\) denote a non-negative real number (meaning \(a\) is either positive or zero). The square root of \(a\), denoted by \( \sqrt{a} \) or \( \sqrt[2]{a} \) is said to equal \(b\) provided that \(b \geq 0\) and \(b^2 = a\). That is,
\[ \sqrt[2]{a} = b ~~ \text{means} ~~ b^2 = a ~~\text{and} ~~ b \geq 0\nonumber\ \]
Again, allow us to reflect on this definition. We may wonder, why does \(a\) have to be positive or zero? Notice that \(a = b^2 \) and for all real numbers \(b\), \(b^2 \geq 0 \). Hence \(a\) must also be greater than or equal to 0 and so \(a\) must either be positive or zero. Or quite simply, if \( b^2 = a \) and since and real number raised to the power of 2 is positive or zero, then \(a\) must be positive or zero. This will be important later on!
Additionally, we may wonder why is it that we require \(b \geq 0 \). The answer to this will be discussed on Day 2 since the explanation requires us to understand what is a function . Allow us to now consider an example.
\[ \sqrt{16} = 4 ~~ \text{because} ~~ 4^2 = 16 ~~\text{and} ~~ 4 \geq 0 \nonumber\ \]
Notice that
\[ \sqrt{16} \neq -4 \nonumber\ \]
Although it is true that \( (-4)^2 = 16 \), it is not true that \( -4 \geq 0\) and so \( \sqrt{16} \neq -4 \nonumber\ \). In summary, the square root symbol, \( \sqrt{~~} \) means "the positive square root of".
With regards to square roots, we will now pose an interesting question.
Suppose \(a\) is some real number. Is it always true that \( \sqrt{a^2} = a\)? Discuss.
- Answer
-
To answer questions like these, allow us to substitute in a few values for \(a\) in order to investigate the statement. If \(a\) is any real number, then \(a\) can be positive, negative or zero. Let us first choose a positive value for \(a\), say 3. Is it true that \( \sqrt{3^2} = 3\)? Yes!
Let us now choose a negative value for \(a\), say \( -3 \). Is it true that \( \sqrt{(-3)^2} = - 3\)? No!
And so the equation \( \sqrt{a^2} = a\) is not always true. In fact, the equation is only true when \(a \geq 0 \).
More generally, we should note the following:
\[ \text{For all real numbers} ~ a, \sqrt{a^2} = |a| \nonumber\ \]
However, there is nothing special about the square root. That is, we can extend this idea. If \[ \sqrt[2]{a} = b ~~ \text{means} ~~ b^2 = a ~~\text{and} ~~ b \geq 0\nonumber\ \] then it makes sense to say \[ \sqrt[4]{a} = b ~~ \text{means} ~~ b^4 = a \nonumber\ \] and \[ \sqrt[6]{a} = b ~~ \text{means} ~~ b^6 = a ~~\text{and} ~~ b \geq 0\nonumber\ \] and so on. More generally, we present the following definition:
Definition: Let \(a\) denote a non-negative real number (meaning \(a\) is either positive or zero) and let \(n\) denote any positive even integer. Then the \(n^{th} \) root of \(a\) is defined as follows:
\[ \sqrt[n]{a} = b ~~ \text{means} ~~ b^n = a ~~\text{and} ~~ b \geq 0\nonumber\ \]
The above definition is saying if \(n\) is even, then to find \( \sqrt[n]{a} \), we require both \(a\) and \(b\) to be positive.
Evaluate the following:
1) \( \sqrt{25} \)
2) \( \sqrt[4]{81} \)
3) \( \sqrt[6]{64} \)
4) \( \sqrt[6]{-64} \)
- Answer
-
1) To find \( \sqrt{25} \) we must ask ourselves can we find a real number \(b\) such that \(b^2 = 25 \) and \(b \geq 0\)? The answer is yes - if we take \(b = 5 \) then \(b^2 = 25 \) and \(b \geq 0\).
2) To find \( \sqrt[4]{81} \) we must ask ourselves can we find a real number \(b\) such that \(b^4 = 81 \) and \(b \geq 0\)? The answer is yes - if we take \(b = 3 \) then \(b^4 = 81 \) and \(b \geq 0\).
3) To find \( \sqrt[6]{64} \) we must ask ourselves can we find a real number \(b\) such that \(b^6 = 64 \) and \(b \geq 0\)? The answer is yes - if we take \(b = 2 \) then \(b^6 = 64 \) and \(b \geq 0\).
4) \( \sqrt[6]{-64} \) we must ask ourselves can we find a real number \(b\) such that \(b^6 = -64 \) and \(b \geq 0\)? The answer here is no. Notice for any real number \(b\), \(b^6 \) will be positive or zero and so \(b^6 \) can never be equal to \( - 64 \). Hence the answer is that \( \sqrt[6]{-64} \) is not defined. Similarly, \( \sqrt[2]{-64} \) and \( \sqrt[4]{-64} \) are not defined.
Notice that the last part of the above example suggests that for even roots, the number inside the root cannot be negative. Allow us to now discuss odd roots.
Definition: Let \(a\) denote a real number (meaning \(a\) is either positive, negative or zero) and let \(n\) denote any positive odd integer. Then the \(n^{th} \) root of \(a\) is defined as follows:
\[ \sqrt[n]{a} = b ~~ \text{means} ~~ b^n = a \nonumber\ \]
Notice in the above definition for odd roots, we now allow the number inside the root to be negative and we dropped the condition that \(b \geq 0 \).
Evaluate the following:
1) \( \sqrt[3]{8} \)
2) \( \sqrt[3]{-8} \)
3) \( \sqrt[3]{10^3} \)
4) \( \sqrt[3]{(-10)^3} \)
- Answer
-
1) To find\( \sqrt[3]{8} \) we must ask ourselves can we find a real number \(b\) such that \(b^3 = 8 \)? The answer is yes - if we take \(b = 2 \) then \(b^3 = 8 \).
2) To find \( \sqrt[3]{-8} \) we must ask ourselves can we find a real number \(b\) such that \(b^3 = -8 \)? The answer is yes - if we take \(b = -2 \) then \(b^3 = -8 \).
-
3) To find \( \sqrt[3]{10^3} \) we must ask ourselves can we find a real number \(b\) such that \(b^3 = 10^3 \)? The answer is yes - if we take \(b = 10 \) then \(b^3 = 10^3 \).
4) To find \( \sqrt[3]{(-10)^3} \) we must ask ourselves can we find a real number \(b\) such that \(b^3 = (-10)^3 \)? The answer is yes - if we take \(b =- 10 \) then \(b^3= (-10)^3 \).
The above examples can be briefly summarized as the following:
1) \( \sqrt[n]{a^n} = a \nonumber\ \) if \(n\) is a positive odd integer.
2) \( \sqrt[n]{a^n} = |a| \nonumber\ \) if \(n\) is a positive even integer.
2) We cannot take the \(n^{th}\) root of a negative number when \(n\) is a positive even integer. However, we can take the \(n^{th}\) root of a negative number when \(n\) is a positive odd integer.
When it comes to \(n^{th} \) roots, we should be familiar with their properties:
Theorem:
1) \( \sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b} \)
2) \( \sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} \)
3) \( \sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a} \)
Take a moment and try to translate the above theorem into words.
- Answer
-
1) The \(n^{th} \) root of a product is the product of the individual \(n^{th} \) roots.
2) The \(n^{th} \) root of a quotient is the quotient of the individual \(n^{th} \) roots.
3) The \(m^{th} \) root of the \(n^{th} \) root is the same as the \(mn^{th} \) root.
Simplifythe following expressions as much as possible:
1) \( \sqrt[3]{x^4} \)
2) \( \sqrt[4]{81x^8 y^4 } \)
- Answer
-
1) \begin{align*} \sqrt[3]{x^4} &= \sqrt[3]{x^3 x } \\ &= \sqrt[3]{x^3 } \sqrt[3]{x } \\ &= x \sqrt[3]{x } \end{align*}
2) \begin{align*} \sqrt[4]{81x^8 y^4 } &= \sqrt[4]{81 } \sqrt[4]{x^8 } \sqrt[4]{y^4 } \\ &= 3 \sqrt[4]{ (x^2)^4 } |y| \\ &= 3 |x^2| |y| \\ &= 3x^2 |y| \end{align*}
We now make one final extension via the following definition.
Definition: For any rational exponent \( \frac{m}{n} \) in lowest terms and \(n > 0 \), we define
\[ a^{\dfrac{m}{n}} = (\sqrt[n]{a})^m \nonumber\ \] or equivalently \[ a^{\dfrac{m}{n}} = (\sqrt[n]{a^m}) \nonumber\ \]
If \(n\) is even, then we require \(a \geq 0\).
Express the following roots in terms of a rational exponent.
1) \( \sqrt{x} \)
2) \( \sqrt[3]{x} \)
3) \( \sqrt[3]{x^4} \)
4) \( 8^{\frac{2}{3}} \)
5) \( 125^{-\frac{1}{3}} \)
- Answer
-
1) \( \sqrt{x} = x^{\frac{1}{2}} \)
2) \( \sqrt[3]{x} = x^{\frac{1}{3}} \)
3) \( \sqrt[3]{x^4} = x^{\frac{4}{3}} \)
4) \( 8^{\frac{2}{3}} = \sqrt[3]{8^2} = \sqrt[3]{64} = 4 \) or \( 8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 2^2 = 4 \)
5) \( 125^{-\frac{1}{3}} = \frac{1}{125^{\frac{1}{3}}} = \frac{1}{\sqrt[3]{125}} = \frac{1}{5} \)
Note that The Law of Exponents also applies to when the powers are rational numbers and not just integers.
Theorem: Let \(a\) and \(b\) be real numbers and \(m\) and \(n\) be rational numbers. Then we have the following:
1) \( a^m a^n = a^{m+n} \)
2) \( \frac{a^m}{a^n} = a^{m-n} \)
3) \( \bigg( a^m \bigg)^n = a^{mn} \)
4) \( (ab)^n = a^n b^n \)
5) \( \displaystyle \bigg( \dfrac{a}{b} \bigg)^n = a^n b^n \)
Allow us to now work through a few examples.
Simplify the following:
1) \( x^{\frac{1}{3}} x^{\frac{7}{3}} \)
2) \( \frac{10x^{\frac{5}{6}} }{2x^{\frac{7}{6}}} \)
3) \( ( 4a^5 b^7)^{\frac{1}{2}} \)
- Answer
-
1) \[ x^{\frac{1}{3}} x^{\frac{7}{3}} = x^{ \frac{1}{3} + \frac{7}{3} } = x^{ \frac{8}{3}} \nonumber \]
2) \[\frac{10x^{\frac{5}{6}} }{2x^{\frac{7}{6}}} = \frac{10}{2} \frac{x^{\frac{5}{6}}}{x^{\frac{7}{6}}} = 5 x^{\frac{5}{6} - \frac{7}{6} } = 5x^{- \frac{2}{6}} = 5x^{- \frac{1}{3}} \nonumber \]
3) \[ ( 4a^5 b^7)^{\frac{1}{2}} = 4^{\frac{1}{2}} (a^5)^{\frac{1}{2}} (b^7)^{\frac{1}{2}} = 2 a^{\frac{5}{2}} b^{\frac{7}{2}} \nonumber \]