An exponent is a shorthand notation for a number or variable that is multiplied by itself. The number that is being multiplied is known as the exponent's base, while the number indicating how many times you multiply is known as the superscript (used interchangeably with "exponent"). Using the definition of an exponent, many expressions can be much easier to write than with expanded notation. Those shortcuts, and ways to use exponents, will be discussed and demonstrated in this article.
One of the basic examples of converting between exponents and standard notation is where the exponent is a 2 and the base is multiplied by itself twice. For instance:
Example 1: By the definition of the exponent:
\(2^2 = 2 \times 2\)
Of course, this means \(2^2 = 4\).
We can see that, in general,
$$a^2 = a \times a$$
Here is one more example:
Example 2: Since \(6^2 = 6 \times 6\), we know that the shorthand \(6^2\) is still equivalent to \(36\).
The exponent can be anything, really (even a function with variables in it). But for now, we'll stick with positive integer exponents. An example where the exponent is not \(2\) is shown below.
Example 3: We can write the expression \(4^5\) as follows:
$$4 \times 4 \times 4 \times 4 \times 4$$
The cases of \(0\) and \(1\) are special and deserve special attention. When dealing with either of these numbers, there are special rules to consider:
When the base is \(0\), the expression equals zero, unless the exponent is also \(0\).
When the exponent is \(0\), the expression equals \(1\), unless the base is also \(0\).
When the base is \(1\), the expression equals \(1\), regardless of the exponent.
When the exponent is \(1\), the expression is equal to the base, regardless of what the base is.
We can break down and explain each of these rules:
Rule 1: This is the same as multiplying zero by itself any number of times, which will always be zero (excluding the special case).
Rule 2: This is like multiplying a number by itself zero times, which, not including \(0^0\), will always give an answer of \(1\).
Rule 3: Similar to Rule 1, this is like multiplying \(1\) by itself a number of times, which will always give you an answer of \(1\).
Rule 4: This rule resembles being given a number and leaving it alone. You are not multiplying it by itself, but you are not using an exponent of \(0\), so we get the base (these three quantities have a special relationship which will be explained later).
The aforementioned exception is \(0^0\). This expression is undefined, and there is a reason. The first rule suggests that \(0^0 = 0\), while the second rule suggests that \(0^0 = 1\), but since \(0 \neq 1\), neither of these can be true, and so \(0^0\) is undefined.
I will now present a string of examples that illustrate these rules:
Example 4: \(0^6 = 0\)
Example 5: \(15^0 = 1\)
Example 6: \(24^1 = 24\)
Example 7: \(1^{x} = 1\), regardless of the real value of \(x\).
We can also multiply exponents together, which is most easily done when the exponents have the same base. To see how this is done, consider the following example:
Example 8: We will work with the expression \(3^4 \cdot 3^5\). We can expand \(3^4\) and \(3^5\) separately and multiply the results:
\(3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3\)
This is the same as \(3^9\), as it is a \(3\) multiplied by itself eight more times.
This example helps to illustrate the following law of exponents:
If \(b\), \(x\), and \(y\) are numbers, then:
$$b^{x} \cdot b^{y} = b^{x + y}$$
The exception is where \(b = 0\) and \(x = -y\) because this gives us the dreaded \(0^0\).
Here is another example of using this law.
Example 9: Simplify \(11^{15} \cdot 11^{4}\).
Solution: The bases are the same, so we can add the exponents:
$$11^{15} \cdot 11^{4} = 11^{15 + 4} = 11^{19}$$
Sometimes factoring the base of an exponent will allow you to simplify it further. But first, we must first show an example of how base factoring works.
Example 10: Since \(9 = 3 \cdot 3\), we can rewrite \(9^2\) as
$$(3 \cdot 3)^2$$
We can distribute the exponent just like we would do to a factor:
$$3^2 \cdot 3^2$$
We also see that this can be rewritten as
$$(3^2)^2$$
As a final note, notice that \(9^2 = 9 \cdot 9 = 81 = 3^4\). This is an example of a special pattern that will be discussed later.
Example 10 illustrates the following rule:
If \(a\), \(b\), and \(x\) are numbers,
$$(ab)^x = a^x \cdot b^x$$
The exception is where \(x = 0\) and either \(a = 0\) or \(b = 0\), because that gives \(0^0\). Every exponent rule will have an exception where the relevant expression becomes \(0^0\) or \(\frac{0}{0}\), because both of these expressions are undefined.
Here is another example, keeping this exponent law in mind.
Example 11: Simplify the expression \(\frac{15^6}{5^8}\).
Solution: We realize that \(15\) factors as \(3 \cdot 5\), so we can rewrite the expression as
$$\frac{(5 \cdot 3)^6}{5^8} = \frac{5^6 \cdot 3^6}{5^8}$$
Both the numerator and denominator have an exponent with a base of \(5\), so write all of the \(5\)s out:
$$\frac{5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 3^6}{5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5}$$
It is now clear that the numerator and denominator have six common factors of \(5\), or a common factor of \(5^6\), so cancel out the corresponding factors:
$$\frac{3^6}{5 \cdot 5} = \frac{3^6}{5^2}$$
Since the bases do not share any common factors besides \(1\), we cannot simplify this further.
Here is one more example involving variables. It is important to include examples with variables since many problems will be in more general forms.
Example 12: Simplify the expression \((2a)^{x + 7} \cdot a^{x + 7}\).
Solution: \(a\) is a factor of \(2a\), so we can factor the first base:
$$(2 \cdot a)^{x + 7} \cdot a^{x + 7}$$
Now, distribute the exponent to each of the two factors:
$$2^{x + 7} \cdot a^{x + 7} \cdot a^{x + 7}$$
The factor \(a^{x + 7}\) appears twice, so just like with any duplicate factor, we can condense it as follows:
$$a^{x + 7} \cdot a^{x + 7} = (a^{x + 7})^2$$
Thus the expression becomes
$$2^{x + 7} \cdot (a^{x + 7})^2$$
There is something else we can do. We can use the rule \(b^x \cdot b^y = b^{x + y}\) on the first factor, but use it backwards, giving the following:
$$2^x \cdot 2^7 \cdot (a^{x + 7})^2$$
The factor without variables can be written as an integer (which is considered simplifying if it is relatively easy to do):
$$128 \cdot 2^x \cdot (a^{x + 7})^2$$
This is as far as we can simplify the expression for now. We will learn another rule to help us simplify it further later on.
You can also divide exponents, provided they have the same base. Otherwise, nothing can be done to simplify the expression unless you convert between bases (which is rather hard in most cases). Since division and multiplication are inverse (opposite) operations, the laws of multiplication and division of exponents are similar.
If \(b\), \(x\), and \(y\) are numbers, then
$$\frac{b^x}{b^y} = b^{x - y}$$
This is another rule that has common uses.
Example 13: Simplify the expression \(\frac{8^6}{8^4}\).
Solution: We subtract the exponents, since the bases are the same:
$$\frac{8^6}{8^4} = 8^{6 - 4} = 8^2 = 8 \cdot 8 = 64$$
Here is an example that combines two of the laws we have learned up to this point. We are going to do Example 11 over again, but in a slightly different and more efficient way.
Example 14: Simplify \(\frac{15^6}{5^8}\).
Solution: The first step remains the same: we factor the base of the numerator:
$$\frac{(5 \cdot 3)^6}{5^8} = \frac{5^6 \cdot 3^6}{5^8}$$
Now we subtract the superscripts of the exponents with the same base, giving
$$5^{-2} \cdot 3^6$$
Example 14 had a negative exponent. While these probably look intimidating to readers that are unfamiliar with them, they are really simple. Just follow this one rule, established by the definition of [negative] exponents:
$$a^{-x} = \frac{1}{a^x}$$
Of course this fails if \(a = 0\), because then the denominator is \(0\).
As a basic example of this process, we return to where we ended Example 14:
Example 15: When trying to rewrite \(5^{-2} \cdot 3^6\) without the use of any negative exponents, we ignore the \(3^6\) since the superscript is positive. Then we get
$$5^{-2} \cdot 3^6 = \frac{1}{5^2} \cdot 3^6 = \frac{3^6}{5^2} = \frac{729}{25}$$
Here is another example that asks for something different:
Example 16: Simplify and write in terms of negative superscripts: \(\frac{108^7}{6^{11}}\).
Solution: This is a multi-step problem. We can factor the base of the numerator as
$$108 = 36 \cdot 3 = 6 \cdot 6 \cdot 3$$
Now implement this factorization:
$$\frac{3^7 \cdot 6^7 \cdot 6^7}{6^{11}}$$
Now we just consider part of the fraction:
$$\frac{6^7}{6^{11}}$$
We subtract the exponents since the bases are the same:
$$6^{-4} \cdot 3^7 \cdot 6^7$$
Now we can combine the two factors with the same base:
$$6^{7 - 4} \cdot 3^7 = 6^3 \cdot 3^7$$
Lastly, write the exponents with negative superscripts by taking their reciprocals and changing the sign of the superscripts:
$$\frac{1}{6^{-3}} \cdot \frac{1}{3^{-7}} = \frac{1}{6^{-3} \cdot 3^{-7}}$$
Fractional exponents are even more intimidating than negative exponents at first, but follow all the same rules that integer superscripts do. Fractional superscripts can be converted as follows:
If \(b\), \(x\), and \(y\) are real numbers:
$$b^{\frac{x}{y}} = \sqrt[y]{b^x}$$
The exception is where \(y = 0\) or both \(b\) and \(x\) are equal to \(0\).
Problems involving fractional exponents are not much different from problems that have integer superscripts.
Example 17: Simplify \(8^{\frac{3}{8}} \cdot 8^{\frac{1}{2}}\).
Solution: The bases are the same, so just combine the superscripts:
$$8^{\frac{3}{8} + \frac{1}{2}} = 8^{\frac{7}{8}}$$
Example 18: Simplify the expression \(\frac{12^{\frac{22}{3}}}{6^{\frac{7}{2}}}\).
Solution: Factor the base:
$$\frac{2^{\frac{22}{3}} \cdot 6^{\frac{22}{3}}}{6^{\frac{7}{2}}}$$
Subtract the superscripts of the exponents with the same base:
$$2^{\frac{22}{3}} \cdot 6^{\frac{22}{3} - \frac{7}{2}} = 2^{\frac{22}{3}} \cdot 6^{\frac{23}{6}}$$
The next example involves negative fractional exponents.
Example 19: Simplify \(3^{\frac{5}{3}} \cdot 9^{\frac{-4}{3}}\) and write it with positive superscripts.
Solution: \(9\) factors as \(3 \cdot 3\), so factor the base. Now all the bases will be the same:
$$3^{\frac{5}{3}} \cdot (3 \cdot 3)^{\frac{-4}{3}} = 3^{\frac{5}{3}} \cdot 3^{\frac{-4}{3}} \cdot 3^{\frac{-4}{3}}$$
All the bases are the same, so we can combine all the exponents at once:
$$3^{-1} = \frac{1}{3}$$
Some "rules" in exponent algebra appear to make common sense to people who are new to exponents based on the mechanics of more simple arithmetic. Some of those fake rules will be presented here. Each will be presented with a counterexample, or an example that proves a general relationship false.
Fake Rule 1: \(a^x + a^y = a^{x + y}\)
We already know that \(a^x \cdot a^y = a^{x + y}\), so rarely will both be true.
Counterexample:
$$3^6 + 3^9 \neq 3^{15}$$
Fake Rule 2: \(a^x \cdot a^y = a^{xy}\)
This suggests that \(a^{x + y} = a^{xy}\), according to a real rule that was already established.
Counterexample:
$$7^5 \cdot 7^3 \neq 7^{15}$$
In reality, the left-hand side is equivalent to \(7^8\).
Fake Rule 3: \(a^{-x} = -a^x\)
All exponential functions with a positive base are always nonnegative over the real numbers (just look at graphs), so this one doesn't make much sense. Of course it works for \(a = 0\) and some cases of \(a = -1\) but it is not always true.
Counterexample:
$$2^8 \neq -2^{-8}$$
The right hand side is the same as \(-\frac{1}{2^8}\), and so the two sides are obviously not equivalent.
If you study the rules of exponents carefully and do not fall for the pitfalls demonstrated here, you should do well in your studies of exponents. The material used in algebra courses is used again repeatedly in precalculus and calculus courses, so it is worthwhile to keep this information in your head for a long time. The basics will get you very far.