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20 Exponential Expressions

Learning Objectives

Rules of Exponents

  • Use the rules to simplify exponential expressions involving products, quotients and exponents
  • Demonstrate an understanding of fractional exponents
  • Define and use the Negative and Zero exponent rules

Introduction to exponential terms

We use exponential notation to write repeated multiplication. For example [latex]10\cdot10\cdot10[/latex] can be written more succinctly as [latex]10^{3}[/latex]. The 10 in [latex]10^{3}[/latex] is called the base. The 3 in [latex]10^{3}[/latex] is called the exponent. The expression [latex]10^{3}[/latex] is called the exponential expression. Knowing the names for the parts of an exponential expression or term will help you learn how to perform mathematical operations on them.

[latex]\text{base}\rightarrow10^{3\leftarrow\text{exponent}}[/latex]

[latex]10^{3}[/latex] is read as “10 to the third power” or “10 cubed.” It means [latex]10\cdot10\cdot10[/latex], or 1,000.

[latex]8^{2}[/latex] is read as “8 to the second power” or “8 squared.” It means [latex]8\cdot8[/latex], or 64.

[latex]5^{4}[/latex] is read as “5 to the fourth power.” It means [latex]5\cdot5\cdot5\cdot5[/latex], or 625.

[latex]b^{5}[/latex] is read as “b to the fifth power.” It means [latex]{b}\cdot{b}\cdot{b}\cdot{b}\cdot{b}[/latex]. Its value will depend on the value of b.

The exponent applies only to the number that it is next to. Therefore, in the expression [latex]xy^{4}[/latex], only the y is affected by the 4. [latex]xy^{4}[/latex] means [latex]{x}\cdot{y}\cdot{y}\cdot{y}\cdot{y}[/latex]. The x in this term is a coefficient of y.

If the exponential expression is negative, such as [latex]−3^{4}[/latex], it means [latex]–\left(3\cdot3\cdot3\cdot3\right)[/latex] or [latex]−81[/latex].

If [latex]−3[/latex] is to be the base, it must be written as [latex]\left(−3\right)^{4}[/latex], which means [latex]−3\cdot−3\cdot−3\cdot−3[/latex], or 81.

Likewise, [latex]\left(−x\right)^{4}=\left(−x\right)\cdot\left(−x\right)\cdot\left(−x\right)\cdot\left(−x\right)=x^{4}[/latex], while [latex]−x^{4}=–\left(x\cdot x\cdot x\cdot x\right)[/latex].

You can see that there is quite a difference, so you have to be very careful! The following examples show how to identify the base and the exponent, as well as how to identify the expanded and exponential format of writing repeated multiplication.

Sample Exercises 20.1

Identify the exponent and the base in the following terms, then simplify:

  1. [latex]7^{2}[/latex]
  2. [latex]{\left(\frac{1}{2}\right)}^{3}[/latex]
  3. [latex]2x^{3}[/latex]
  4. [latex]\left(-5\right)^{2}[/latex]
Solution

1) [latex]7^{2}[/latex]

The exponent in this term is 2 and the base is 7. To simplify, expand the term: [latex]7^{2}=7\cdot{7}=49[/latex]

2) [latex]{\left(\frac{1}{2}\right)}^{3}[/latex]

The exponent on this term is 3, and the base is [latex]\frac{1}{2}[/latex]. To simplify, expand the multiplication and remember how to multiply fractions: [latex]{\left(\frac{1}{2}\right)}^{3}=\frac{1}{2}\cdot{\frac{1}{2}}\cdot{\frac{1}{2}}=\frac{1}{8}[/latex]

3)  [latex]2x^{3}[/latex]

The exponent on this term is 3, and the base is x, the 2 is not getting the exponent because there are no parentheses that tell us it is.  This term is in its most simplified form.

4) [latex]\left(-5\right)^{2}[/latex]

The exponent on this terms is 2 and the base is [latex]-5[/latex]. To simplify, expand the multiplication: [latex]\left(-5\right)^{2}=-5\cdot{-5}=25[/latex]

Caution!

Whether to include a negative sign as part of a base or not often leads to confusion. To clarify whether a negative sign is applied before or after the exponent, here is an example.

What is the difference in the way you would evaluate these two terms?

  1. [latex]-{3}^{2}[/latex]
  2. [latex]{\left(-3\right)}^{2}[/latex]

To evaluate 1), you would apply the exponent to the three first, then apply the negative sign last, like this:

[latex]\begin{array}{c}-\left({3}^{2}\right)\\=-\left(9\right) = -9\end{array}[/latex]

To evaluate 2), you would apply the exponent to the 3 and the negative sign:

[latex]\begin{array}{c}{\left(-3\right)}^{2}\\=\left(-3\right)\cdot\left(-3\right)\\={ 9}\end{array}[/latex]

The key to remembering this is to follow the order of operations. The first expression does not include parentheses so you would apply the exponent to the integer 3 first, then apply the negative sign. The second expression includes parentheses, so hopefully you will remember that the negative sign also gets squared.

Simplifying Exponential expressions

Exponential notation was developed to write repeated multiplication more efficiently. There are times when it is easier or faster to leave the expressions in exponential notation when multiplying or dividing. Let’s look at rules that will allow you to do this.

For example, the notation [latex]5^{4}[/latex] can be expanded and written as [latex]5\cdot5\cdot5\cdot5[/latex], or 625. And don’t forget, the exponent only applies to the number immediately to its left, unless there are parentheses.

What happens if you multiply two numbers in exponential form with the same base? Consider the expression [latex]{2}^{3}{2}^{4}[/latex]. Expanding each exponent, this can be rewritten as [latex]\left(2\cdot2\cdot2\right)\left(2\cdot2\cdot2\cdot2\right)[/latex] or [latex]2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2[/latex]. In exponential form, you would write the product as [latex]2^{7}[/latex]. Notice that 7 is the sum of the original two exponents, 3 and 4.

What about [latex]{x}^{2}{x}^{6}[/latex]? This can be written as [latex]\left(x\cdot{x}\right)\left(x\cdot{x}\cdot{x}\cdot{x}\cdot{x}\cdot{x}\right)=x\cdot{x}\cdot{x}\cdot{x}\cdot{x}\cdot{x}\cdot{x}\cdot{x}[/latex] or [latex]x^{8}[/latex]. And, once again, 8 is the sum of the original two exponents. This concept can be generalized in the following way:

The Product Rule for Exponents

For any number x and any integers a and b, [latex]\left(x^{a}\right)\left(x^{b}\right) = x^{a+b}[/latex].

To multiply exponential terms with the same base, add the exponents.

Caution!

When you are reading mathematical rules, it is important to pay attention to the conditions on the rule.  For example, when using the product rule, you may only apply it when the terms being multiplied have the same base and the exponents are integers. Conditions on mathematical rules are often given before the rule is stated, as in this example it says “For any number x, and any integers a and b.”

Sample Exercise 20.2

Simplify.

[latex](a^{3})(a^{7})[/latex]

Solution

The base of both exponents is a, so the product rule applies.

[latex]\left(a^{3}\right)\left(a^{7}\right)[/latex]

Add the exponents with a common base.

[latex]a^{3+7}[/latex]

Answer

[latex]\left(a^{3}\right)\left(a^{7}\right) = a^{10}[/latex]

When multiplying more complicated terms, multiply the coefficients and then multiply the variables.

Sample Exercise 20.3

Simplify.

[latex]5a^{4}\cdot7a^{6}[/latex]

Solution

Multiply the coefficients.

[latex]35\cdot{a}^{4}\cdot{a}^{6}[/latex]

The base of both exponents is a, so the product rule applies. Add the exponents.

[latex]35\cdot{a}^{4+6}[/latex]

Add the exponents with a common base.

[latex]35\cdot{a}^{10}[/latex]

Answer

[latex]5a^{4}\cdot7a^{6}=35a^{10}[/latex]

Caution!

Do not try to apply this rule to sums.

Think about the expression [latex]\left(2+3\right)^{2}[/latex]

Does [latex]\left(2+3\right)^{2}[/latex] equal [latex]2^{2}+3^{2}[/latex]?

No, it does not because of the order of operations!

[latex]\left(2+3\right)^{2}=5^{2}=25[/latex]

and

[latex]2^{2}+3^{2}=4+9=13[/latex]

Therefore, you can only use this rule when the numbers inside the parentheses are being multiplied (or divided, as we will see next).

The Quotient Rule for Exponents

Let’s look at dividing terms containing exponential expressions. What happens if you divide two numbers in exponential form with the same base? Consider the following expression.

[latex]{\displaystyle 1+ \frac{4^5}{4^2} }[/latex]

You can rewrite the expression as: [latex]\frac{4\cdot 4\cdot 4\cdot 4\cdot 4}{4\cdot 4}[/latex]. Then you can cancel the common factors of 4 in the numerator and denominator: [latex]\displaystyle[/latex]

Finally, this expression can be rewritten as [latex]4^{3}[/latex] using exponential notation. Notice that the exponent, 3, is the difference between the two exponents in the original expression, 5 and 2.

So, [latex]{\displaystyle \frac{4^5}{4^2}} = 4^{5-2} = 4^{3}.[/latex]

Be careful that you subtract the exponent in the denominator from the exponent in the numerator.

So, to divide two exponential terms with the same base, subtract the exponents.

For any non-zero number x and any integers a and b:

[latex]{\displaystyle\frac{x^a}{x^b}} = x^{a-b}[/latex]

Sample Exercise 20.4

Evaluate. [latex]\frac{{{4}^{9}}}{{{4}^{4}}}[/latex]

Solution

These two exponents have the same base, 4. According to the Quotient Rule, you can subtract the power in the denominator from the power in the numerator.

[latex]\displaystyle {{4}^{9-4}}[/latex]

Answer

[latex]\displaystyle \frac{{{4}^{9}}}{{{4}^{4}}}=4^{5}[/latex]

[/hidden-answer]

When dividing terms that also contain coefficients, divide the coefficients and then divide variable powers with the same base by subtracting the exponents.

Sample Exercise 20.5

Simplify. [latex]\displaystyle \frac{12{{x}^{4}}}{2x}[/latex]

Solution

Separate into numerical and variable factors.

[latex]\displaystyle \left( \frac{12}{2} \right)\left( \frac{{{x}^{4}}}{x} \right)[/latex]

Since the bases of the exponents are the same, you can apply the Quotient Rule. Divide the coefficients and subtract the exponents of matching variables.

[latex]\displaystyle 6\left( {{x}^{4-1}} \right)[/latex]

Answer

[latex]\displaystyle \frac{12{{x}^{4}}}{2x}[/latex]=[latex]\displaystyle 6{{x}^{3}}[/latex]

Raise Powers to Powers (or the Exponent of an Exponent)

Another word for exponent is power.  You have likely seen or heard an example such as [latex]3^5[/latex] can be described as 3 raised to the 5th power. In this section we will further expand our capabilities with exponents. We will learn what to do when a term with a power is raised to another power, and what to do when two numbers or variables are multiplied and both are raised to an exponent.  We will also learn what to do when numbers or variables that are divided are raised to a power.  We will begin by raising powers to powers.

Let’s simplify [latex]\left(5^{2}\right)^{4}[/latex]. In this case, the base is [latex]5^2[/latex] and the exponent is 4, so you multiply [latex]5^{2}[/latex] four times: [latex]\left(5^{2}\right)^{4}=5^{2}\cdot5^{2}\cdot5^{2}\cdot5^{2}=5^{8}[/latex] (using the Product Rule—add the exponents).

[latex]\left(5^{2}\right)^{4}[/latex] is a power of a power. It is the fourth power of 5 to the second power. And we saw above that the answer is [latex]5^{8}[/latex]. Notice that the new exponent is the same as the product of the original exponents: [latex]2\cdot4=8[/latex].

So, [latex]\left(5^{2}\right)^{4}=5^{2\cdot4}=5^{8}[/latex] (which equals 390,625, if you do the multiplication).

Likewise, [latex]\left(x^{4}\right)^{3}=x^{4\cdot3}=x^{12}[/latex]

This leads to another rule for exponents—the Power Rule for Exponents. To simplify a power of a power, you multiply the exponents, keeping the base the same. For example, [latex]\left(2^{3}\right)^{5}=2^{15}[/latex].

For any positive number x and integers a and b: [latex]\left(x^{a}\right)^{b}=x^{a\cdot{b}}[/latex].

Take a moment to contrast how this is different from the product rule for exponents found on the previous page.

Sample Exercise 20.6

Simplify [latex]6\left(c^{4}\right)^{2}[/latex].

Solution

Since you are raising a power to a power, apply the Power Rule and multiply exponents to simplify. The coefficient remains unchanged because it is outside of the parentheses.

[latex]6\left(c^{4}\right)^{2}[/latex]

Answer

[latex]6\left(c^{4}\right)^{2}=6c^{8}[/latex]

Raise a product to a power

Simplify this expression.

[latex]\left(2a\right)^{4}=\left(2a\right)\left(2a\right)\left(2a\right)\left(2a\right)=\left(2\cdot2\cdot2\cdot2\right)\left(a\cdot{a}\cdot{a}\cdot{a}\cdot{a}\right)=\left(2^{4}\right)\left(a^{4}\right)=16a^{4}[/latex]

Notice that the exponent is applied to each factor of 2a. So, we can eliminate the middle steps.

[latex]\begin{array}{l}\left(2a\right)^{4} = \left(2^{4}\right)\left(a^{4}\right)\text{, applying the }4\text{ to each factor, }2\text{ and }a\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,=16a^{4}\end{array}[/latex]

The product of two or more numbers raised to a power is equal to the product of each number raised to the same power.

A Product Raised to a Power

For any nonzero numbers a and b and any integer x, [latex]\left(ab\right)^{x}=a^{x}\cdot{b^{x}}[/latex].

How is this rule different from the power raised to a power rule? How is it different from the product rule for exponents on the previous page?

Sample Exercise 20.7

Simplify. [latex]\left(2yz\right)^{6}[/latex]

Solution

Apply the exponent to each number in the product.

[latex]2^{6}y^{6}z^{6}[/latex]

Answer

[latex]\left(2yz\right)^{6}=64y^{6}z^{6}[/latex]

If the variable has an exponent with it, use the Power Rule: multiply the exponents.

Sample Exercise 20.8

Simplify. [latex]\left(−7a^{4}b\right)^{2}[/latex]

Solution

Apply the exponent 2 to each factor within the parentheses.

[latex]\left(−7\right)^{2}\left(a^{4}\right)^{2}\left(b\right)^{2}[/latex]

Square the coefficient and use the Power Rule to square [latex]\left(a^{4}\right)^{2}[/latex].

[latex]49a^{4\cdot2}b^{2}[/latex]

Simplify.

[latex]49a^{8}b^{2}[/latex]

Answer

[latex]\left(-7a^{4}b\right)^{2}=49a^{8}b^{2}[/latex]

Raise a quotient to a power

Now let’s look at what happens if you raise a quotient to a power. Remember that quotient means divide. Suppose you have [latex]\displaystyle \frac{3}{4}[/latex] and raise it to the 3rd power.

[latex]\displaystyle {{\left( \frac{3}{4} \right)}^{3}}=\left( \frac{3}{4} \right)\left( \frac{3}{4} \right)\left( \frac{3}{4} \right)=\frac{3\cdot 3\cdot 3}{4\cdot 4\cdot 4}=\frac{{{3}^{3}}}{{{4}^{3}}}[/latex]

You can see that raising the quotient to the power of 3 can also be written as the numerator (3) to the power of 3, and the denominator (4) to the power of 3.

Similarly, if you are using variables, the quotient raised to a power is equal to the numerator raised to the power over the denominator raised to power.

[latex]\displaystyle {{\left( \frac{a}{b} \right)}^{4}}=\left( \frac{a}{b} \right)\left( \frac{a}{b} \right)\left( \frac{a}{b} \right)\left( \frac{a}{b} \right)=\frac{a\cdot a\cdot a\cdot a}{b\cdot b\cdot b\cdot b}=\frac{{{a}^{4}}}{{{b}^{4}}}[/latex]

When a quotient is raised to a power, you can apply the power to the numerator and denominator individually, as shown below.

[latex]\displaystyle {{\left( \frac{a}{b} \right)}^{4}}=\frac{{{a}^{4}}}{{{b}^{4}}}[/latex]

For any number a, any non-zero number b, and any integer x, [latex]\displaystyle {\left(\frac{a}{b}\right)}^{x}=\frac{a^{x}}{b^{x}}[/latex].

Sample Exercise 20.9

Simplify. [latex]\displaystyle {{\left( \frac{2{x}^{2}y}{x} \right)}^{3}}[/latex]

Solution

Apply the power to each factor individually.

[latex]\displaystyle \frac{{{2}^{3}{\left({x}^{2}\right)}^{3}{y}^{3}}}{{{x}^{3}}}[/latex]

Separate into numerical and variable factors.

[latex]\displaystyle {{2}^{3}}\cdot \frac{{{x}^{3\cdot2}}}{{{x}^{3}}}\cdot \frac{{{y}^{3}}}{1}[/latex]

Simplify by taking 2 to the third power and applying the Power and Quotient Rules for exponents—multiply and subtract the exponents of matching variables.

[latex]\displaystyle 8\cdot {{x}^{(6-3)}}\cdot {{y}^{3}}[/latex]

Simplify.

[latex]\displaystyle 8{{x}^{3}}{{y}^{3}}[/latex]

Answer

[latex]\displaystyle {{\left( \frac{2{x}^{2}y}{x} \right)}^{3}}=8{{x}^{3}}{{y}^{3}}[/latex]

fractional exponents

A fractional exponent is a power that is written as a fraction. It is also an alternative way to write roots of numbers.

For example: [latex]25^{½} = \sqrt{25}[/latex]

When fractional exponents are used, it is important to note that the numerator is the representation of a traditional exponent and the denominator represents the root.

Therefore,

[latex]8^{⅔} = \sqrt[3]{8^2} = (\sqrt[3]{8})^2[/latex]

Regardless of which method was used, in each scenario [latex]8^{⅔} = 4[/latex]

Where [latex]8^2 = 64[/latex] and [latex]\sqrt[3]{64} = 4[/latex]

or [latex]\sqrt[3]{8} = 2[/latex] and [latex]2^2 = 4[/latex]

When the numerator of fractional exponent is one, the expression is similar to a root. Raising a number to the power of ½ is the same as finding the square root.
Raising a number to the power of is the same as finding the cubic root.
Raising a number to the power of ¼ is the same as finding the fourth root, and so on.

Define and use the zero exponent rule

When we defined the quotient rule, we only worked with expressions like the following: [latex]\frac{{{4}^{9}}}{{{4}^{4}}}[/latex], where the exponent in the numerator (up) was greater than the one in the denominator (down), so the final exponent after simplifying was always a positive number, and greater than zero. In this section, we will explore what happens when we apply the quotient rule for exponents and get a negative or zero exponent.

What if the exponent is zero?

To see how this is defined, let us begin with an example. We will use the idea that dividing any number by itself gives a result of 1.

[latex]\frac{t^{8}}{t^{8}}=\frac{\cancel{t^{8}}}{\cancel{t^{8}}}=1[/latex]

If we were to simplify the original expression using the quotient rule, we would have

[latex]\frac{{t}^{8}}{{t}^{8}}={t}^{8 - 8}={t}^{0}[/latex]

If we equate the two answers, the result is [latex]{t}^{0}=1[/latex]. This is true for any nonzero real number, or any variable representing a real number.

[latex]{a}^{0}=1[/latex]

The sole exception is the expression [latex]{0}^{0}[/latex]. This appears later in more advanced courses, but for now, we will consider the value to be undefined, or DNE (Does Not Exist).

Exponents of 0 or 1

Any number or variable raised to a power of 1 is the number itself.

[latex]n^{1}=n[/latex]

Any non-zero number or variable raised to a power of 0 is equal to 1

[latex]n^{0}=1[/latex]

The quantity [latex]0^{0}[/latex] is undefined.

As done previously, to evaluate expressions containing exponents of 0 or 1, substitute the value of the variable into the expression and simplify.

Sample Exercise 20.10

Evaluate [latex]2x^{0}[/latex] if [latex]x=9[/latex]

Solution

Substitute 9 for the variable x.

[latex]2\cdot9^{0}[/latex]

Evaluate [latex]9^{0}[/latex]. Multiply.

[latex]2\cdot1=2[/latex]

Answer

[latex]2x^{0}=2[/latex], if [latex]x=9[/latex]

Sample Exercise 20.11

Simplify [latex]\frac{{c}^{3}}{{c}^{3}}[/latex].

Solution

Use the quotient and zero exponent rules to simplify the expression.

[latex]\begin{array}\text{ }\frac{c^{3}}{c^{3}} \,\,\,= \,\,\,c^{3-3} \\ \,\,\,\,\,\,\,\,\,\,\,=\,\,\,c^{0} \\\,\,\,\,\,\,\,\,\,\,\,=\,\,\,1\end{array}[/latex]

Answer

1

Define and use the negative exponent rule

We proposed another question at the beginning of this section.  Given a quotient like [latex]\displaystyle \frac{{{2}^{m}}}{{{2}^{n}}}[/latex] what happens when n is larger than m? We will need to use the negative rule of exponents to simplify the expression so that it is easier to understand.

Let’s look at an example to clarify this idea. Given the expression:

[latex]\frac{{h}^{3}}{{h}^{5}}[/latex]

Expand the numerator and denominator, all the terms in the numerator will cancel to 1, leaving two hs multiplied in the denominator, and a numerator of 1.

[latex]\begin{array}{l} \frac{{h}^{3}}{{h}^{5}}\,\,\,=\,\,\,\frac{h\cdot{h}\cdot{h}}{h\cdot{h}\cdot{h}\cdot{h}\cdot{h}} \\ \,\,\,\,\,\,\,\,\,\,\,=\,\,\,\frac{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}}{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}\cdot {h}\cdot {h}}\\\,\,\,\,\,\,\,\,\,\,\,=\,\,\,\frac{1}{h\cdot{h}}\\\,\,\,\,\,\,\,\,\,\,\,=\,\,\,\frac{1}{{h}^{2}} \end{array}[/latex]

We could have also applied the quotient rule from the last section, to obtain the following result:

[latex]\begin{array}{r}\frac{h^{3}}{h^{5}}\,\,\,=\,\,\,h^{3-5}\\\\=\,\,\,h^{-2}\,\,\end{array}[/latex]

Putting the answers together, we have [latex]{h}^{-2}=\frac{1}{{h}^{2}}[/latex]. This is true when h, or any variable, is a real number and is not zero.

The Negative Rule of Exponents

For any nonzero real number [latex]a[/latex] and natural number [latex]n[/latex], the negative rule of exponents states that
[latex]{a}^{-n}=\frac{1}{{a}^{n}}[/latex]
Let’s looks at some examples of how this rule applies under different circumstances.

Sample Exercise 20.12

Evaluate the expression [latex]{4}^{-3}[/latex].

Solution

First, write the expression with positive exponents by putting the term with the negative exponent in the denominator.

[latex]{4}^{-3} = \frac{1}{{4}^{3}} = \frac{1}{4\cdot4\cdot4}[/latex]

Now that we have an expression that looks somewhat familiar.

[latex]\frac{1}{4\cdot4\cdot4} = \frac{1}{64}[/latex]

Answer

[latex]\frac{1}{64}[/latex]

Sample Exercise 20.13

Write [latex]\frac{{\left({t}^{3}\right)}}{{\left({t}^{8}\right)}}[/latex] with positive exponents.

Solution

Use the quotient rule to subtract the exponents of terms with like bases.

[latex]\begin{array}{r}\frac{{\left({t}^{3}\right)}}{{\left({t}^{8}\right)}}={t}^{3-8}\\={t}^{-5}\,\,\end{array}[/latex]

Write the expression with positive exponents by putting the term with the negative exponent in the denominator.

[latex]=\frac{1}{{t}^{5}}[/latex]

Answer

[latex]\frac{1}{{t}^{5}}[/latex]

Sample Exercise 20.14

Simplify [latex]{\left(\frac{1}{3}\right)}^{-2}[/latex].

Solution

Apply the power property of exponents.

[latex]\frac{{1}^{-2}}{{3}^{-2}}[/latex]

Write each term with a positive exponent, the numerator will go to the denominator and the denominator will go to the numerator.

[latex]\frac{{3}^{2}}{{1}^{2}}{ = }\frac{{3}\cdot{3}}{{1}\cdot{1}}[/latex]

Simplify.

[latex]\frac{{3}\cdot{3}}{{1}\cdot{1}}{ = }\frac{9}{1}{ = }{9}[/latex]

Answer

[latex]9[/latex]

Sample Exercise20.15

Simplify.[latex]\frac{1}{4^{-2}}[/latex] Write your answer using positive exponents.

Solution

Write each term with a positive exponent, the denominator will go to the numerator.

[latex]\frac{1}{4^{-2}}=1\cdot\frac{4^{2}}{1}=\frac{16}{1}=16[/latex]

Answer

16

Simplify expressions using a combination of exponent rules

Once the rules of exponents are understood, you can begin simplifying more complicated expressions. There are many applications and formulas that make use of exponents, and sometimes expressions can get pretty cluttered. Simplifying an expression before evaluating can often make the computation easier, as you will see in the following example which makes use of the quotient rule to simplify before substituting 4 for x.

Sample Exercise 20.16

Evaluate [latex]\displaystyle \frac{24{{x}^{8}}}{2{{x}^{5}}}[/latex] when [latex]x=4[/latex].

Solution

Separate into numerical and variable factors.

[latex]\displaystyle \left( \frac{24}{2} \right)\left( \frac{{{x}^{8}}}{{{x}^{5}}} \right)[/latex]

Divide coefficients, and subtract the exponents of the variables.

[latex]\displaystyle 12\left( {{x}^{8-5}} \right)[/latex]

Simplify.

[latex]\displaystyle 12{{x}^{3}}[/latex]

Substitute the value 4 for the variable x.

[latex]\displaystyle (12)({{4}^{3}})=12\cdot 64[/latex]

Answer

[latex]\displaystyle \frac{24{{x}^{8}}}{2{{x}^{5}}}[/latex] = 768

Sample Exercise 20.17

Evaluate [latex]\displaystyle \frac{24{{x}^{8}}{{y}^{2}}}{{{(2{{x}^{3}}y)}^{2}}}[/latex] when [latex]x=4[/latex] and [latex]y=-2[/latex].

Solution

In the denominator, notice that a product is being raised to a power.

Use the rules of exponents to simplify the denominator.

[latex]\displaystyle{\left(2{x}^{3}y\right)}^{2}={2}^{2}{x}^{3\cdot 2}{y}^{2}={2}^{2}{x}^{6}{y}^{2}={4x^{6}y^{2}}[/latex]

Here is the fraction with a simplified denominator:

[latex]\frac{24x^{8}y^{2}}{4x^{6}y^{2}}[/latex]

Separate into numerical and variable factors to simplify further.

[latex]\displaystyle \left( \frac{24}{4} \right)\left( \frac{{{x}^{8}}}{{{x}^{6}}} \right)\left( \frac{{{y}^{2}}}{{{y}^{2}}}\right)[/latex]

Divide coefficients, use the Quotient Rule to divide the variables—subtract the exponents.

[latex]\displaystyle 6\left( {{x}^{8-6}} \right)\left( {{y}^{2-2}} \right)[/latex]

Simplify. Remember that [latex]y^{0}[/latex] is 1.

[latex]\displaystyle 6{{x}^{2}}{{y}^{0}}=6{{x}^{2}}[/latex]

Substitute the value 4 for the variable x.

[latex]\displaystyle (6)({{4}^{2}})=6\cdot 16[/latex]

Answer

[latex]\displaystyle \frac{24{{x}^{8}}{{y}^{2}}}{{{(2{{x}^{3}}y)}^{2}}}=96[/latex] when [latex]x=4[/latex] and [latex]y=-2[/latex]

Notice that you could have worked this problem by substituting 4 for x and 2 for y in the original expression. You would still get the answer of 96, but the computation would be much more complex. Notice that you didn’t even need to use the value of y to evaluate the above expression.

Usually, it is easier to simplify the expression before substituting any values for your variables, but you will get the same answer either way. In the next examples, you will see how to simplify expressions using different combinations of the rules for exponents.

Sample Exercise 20.18

Simplify. [latex]a^{2}\left(a^{5}\right)^{3}[/latex]

Solution

Raise [latex]a^{5}[/latex] to the power of 3 by multiplying the exponents together (the Power Rule).

[latex]\displaystyle {{a}^{2}}{{a}^{5\cdot 3}}[/latex]

Since the exponents share the same base, a, they can be combined (the Product Rule).

[latex]\displaystyle {{a}^{2}}{{a}^{15}}\\{{a}^{2+15}}[/latex]

Answer

[latex]\displaystyle {{a}^{2}}{{({{a}^{5}})}^{3}}={{a}^{17}}[/latex]

The following examples require the use of all the exponent rules we have learned so far. Remember that the product, power, and quotient rules apply when your terms have the same base.

Sample Exercise 20.19

Simplify. [latex]\displaystyle \frac{{{a}^{2}}{{({{a}^{5}})}^{3}}}{8{{a}^{8}}}[/latex]

Solution

Use the order of operations.

Parentheses, Exponents, Multiply/ Divide, Add/ Subtract

There is nothing inside parentheses or brackets that we can simplify further, so we will evaluate exponents first.

Use the Power Rule to simplify [latex]\left(a^{5}\right)^{3}[/latex].

[latex]\left(a^{5}\right)^{3}=a^{5\cdot{3}}=a^{15}[/latex]

The expression now looks like this:

[latex]\displaystyle\frac{{{a}^{2}}{{a}^{15}}}{8{{a}^{8}}}[/latex]

Now we can multiply, using the Product Rule to simplify the numerator because the bases are the same.

[latex]\displaystyle{{a}^{2}}{{a}^{15}}=a^{17}[/latex], and the expression looks like this:

[latex]\displaystyle\frac{{{{a}^{17}}}}{{8{{a}^{8}}}}[/latex]

Now we can divide using the Quotient Rule.

[latex]\displaystyle \frac{{{a}^{17-8}}}{8}[/latex]

Answer

[latex]\displaystyle \frac{{{a}^{2}}{{({{a}^{5}})}^{3}}}{8{{a}^{8}}}=\frac{{{a}^{9}}}{8}[/latex]

Simplify Expressions With Negative Exponents

Now we will add the last layer to our exponent simplifying skills and practice simplifying compound expressions that have negative exponents in them. It is standard convention to write exponents as positive because it is easier for the user to understand the value associated with positive exponents, rather than negative exponents.

Use the following summary of negative exponents to help you simplify expressions with negative exponents.

Rules for Negative Exponents

With a, b, m, and n not equal to zero, and and n as integers, the following rules apply:

[latex]a^{-m}=\frac{1}{a^{m}}[/latex]

[latex]\frac{1}{a^{-m}}=a^{m}[/latex]

[latex]\frac{a^{-n}}{b^{-m}}=\frac{b^m}{a^n}[/latex]

When you are simplifying expressions that have many layers of exponents, it is often hard to know where to start. It is common to start in one of two ways:

  • Rewrite negative exponents as positive exponents
  • Apply the product rule to eliminate any “outer” layer exponents such as in the following term: [latex]\left(5y^3\right)^2[/latex]

We will explore this idea with the following example:

Simplify. [latex]\displaystyle {{\left( 4{{x}^{3}} \right)}^{5}}\cdot \,\,{{\left( 2{{x}^{2}} \right)}^{-4}}[/latex]

Write your answer with positive exponents. The table below shows how to simplify the same expression in two different ways, rewriting negative exponents as positive first, and applying the product rule for exponents first. You will see that there is a column for each method that describes the exponent rule or other steps taken to simplify the expression.

Rewrite with positive Exponents First Description of Steps Taken Apply the Product Rule for Exponents First Description of Steps Taken
[latex]\frac{\left(4x^{3}\right)^{5}}{\left(2x^{2}\right)^{4}}[/latex] move the term [latex]{{\left( 2{{x}^{2}} \right)}^{-4}}[/latex] to the denominator with a positive exponent [latex]\left(4^5x^{15}\right)\left(2^{-4}x^{-8}\right)[/latex]  Apply the exponent of 5 to each term in expression on the left, and the exponent of -4 to each term in the expression on the right.
 [latex]\frac{\left(4^5x^{15}\right)}{\left(2^4x^{8}\right)}[/latex] Use the product rule to apply the outer exponents to the terms inside each set of parentheses. [latex]\left(4^5\right)\left(2^{-4}\right)\\\left(x^{15}\right)\left({x^{-8}}\right)[/latex] Regroup the numerical terms and the variables to make combining like terms easier
 [latex]\left(\frac{4^5}{2^4}\right)\left(\frac{x^{15}}{x^{8}}\right)[/latex] Regroup the numerical terms and the variables to make combining like terms easier [latex]\left(4^5\right)\left(2^{-4}\right)\left(x^{15-8}\right)[/latex]  Use the rule for multiplying terms with exponents to simplify the x terms
 [latex]\left(\frac{4^5}{2^4}\right)\left(x^{15-8}\right)[/latex] Use the quotient rule to simplify the x terms [latex]\left(\frac{4^5}{2^4}\right)\left(x^{7}\right)[/latex]  Rewrite all the negative exponents with positive exponents
 [latex]\left(\frac{1,024}{16}\right)\left(x^{7}\right)[/latex] Expand the numerical terms [latex]\left(\frac{1,024}{16}\right)\left(x^{7}\right)[/latex]  Expand the numerical terms
  [latex]64x^{7}[/latex] Divide the numerical terms  [latex]64x^{7}[/latex]  Divide the numerical terms

If you compare the two columns that describe the steps that were taken to simplify the expression, you will see that they are all nearly the same, except the order is changed slightly. Neither way is better or more correct than the other, it truly is a matter of preference.

Sample Exercise 20.20

Simplify [latex]\frac{\left(t^{3}\right)^2}{\left(t^2\right)^{-8}}[/latex]

Write your answer with positive exponents.

Solution

We can either rewrite this expression with positive exponents first or use the Product Raised to a Power Rule first.

Let’s start by simplifying the numerator and denominator using the Product Raised to a Power Rule.

Numerator: [latex]\left(t^{3}\right)^2=t^{3\cdot{2}}=t^6[/latex]

Denominator: [latex]\left(t^2\right)^{-8}=t^{2\cdot{-8}}=t^{-16}[/latex]

Now the expression looks like this:

[latex]\frac{t^6}{t^{-16}}[/latex]

We can use the quotient rule because we have the same base.

Quotient Rule: [latex]\frac{t^6}{t^{-16}}=t^{6-\left(-16\right)}=t^{6+16}=t^{22}[/latex]

Answer

[latex]\frac{\left(t^{3}\right)^2}{\left(t^2\right)^{-8}}=t^{22}[/latex]

Sample Exercise 20.21

Simplify [latex]\frac{\left(5x\right)^{-2}y}{x^3y^{-1}}[/latex]

Write your answer with positive exponents.

Solution

This time, let’s start by rewriting the terms in the expression so they have positive exponents. The terms with negative exponents in the top will go to the bottom of the fraction, and the terms with negative exponents in the bottom will go to the top.

[latex]\begin{array}{c}\frac{\left(5x\right)^{-2}y}{x^3y^{-1}}\\\text{ }\\=\frac{\left({y^{1}}\right)y}{x^3\left(5x\right)^{2}}\end{array}[/latex]

Note how we left the single y term in the top because it did not have a negative exponent on it, and we left the [latex]x^3[/latex] term in the bottom because it did not have a negative exponent on it.

Now we can apply the Product Raised to a Power Rule:

[latex]\frac{yy^{1}}{5^{2}x^3x^{2}}[/latex]

Use the product rule to simplify further:

[latex]\frac{yy^{1}}{5^{2}x^3x^{2}}=\frac{y^2}{25x^{3+2}}=\frac{y^2}{25x^5}[/latex]

We can’t simplify any further, so our answer is

Answer

[latex]\frac{\left(5x\right)^{-2}y}{x^3y^{-1}}=\frac{y^2}{25x^5}[/latex]

In the next section, you will learn how to write very large and very small numbers using exponents. This practice is widely used in science and engineering.

Summary

  • Evaluating expressions containing exponents is the same as evaluating any expression. You substitute the value of the variable into the expression and simplify.
  • The product rule for exponents: For any number x and any integers a and b, [latex]\left(x^{a}\right)\left(x^{b}\right) = x^{a+b}[/latex].
  • The quotient rule for exponents: For any non-zero number x and any integers a and b: [latex]\displaystyle \frac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}[/latex]
  • Fractional exponents are another way of writing the roots of numbers
  • The power rule for exponents:
    1. For any nonzero numbers a and b and any integer x, [latex]\left(ab\right)^{x}=a^{x}\cdot{b^{x}}[/latex].
    2. For any number a, any non-zero number b, and any integer x, [latex]\displaystyle {\left(\frac{a}{b}\right)}^{x}=\frac{a^{x}}{b^{x}}[/latex]

Case Studies

Scenario A: Product Rule for Exponents

A medical office administrator is tasked with calculating the total medication dose based on two prescriptions for the same drug. The first prescription is for 2 doses per day for 3 days, and the second prescription is for 2 doses per day for 4 days.

  • Write the total number of doses taken as an exponential expression and simplify it.
Scenario A Solution

The total number of doses is the product of the two prescriptions:

  • First prescription: [latex]2^3[/latex] (2 doses per day for 3 days)
  • Second prescription: [latex]2^4[/latex] (2 doses per day for 4 days)

Using the Product Rule for exponents:

[latex]2^3 \times 2^4 = 2^{3+4} = 2^7[/latex]

So, the total number of doses is [latex]2^7 = 128.[/latex]

Scenario B:Quotient Rule for Exponents

A nurse is reviewing a patient’s medical records and calculating the difference in the number of doses prescribed over two weeks. The patient was prescribed 4 doses of medication per day for 5 days and 4 doses per day for 2 days the following week.

  • Calculate the difference in the total doses prescribed using exponential notation.
Scenario B Solution
  • Total doses in the first week: [latex]4^5[/latex]
  • Total doses in the second week: latex]4^2[/latex]

Using the Quotient Rule for exponents:

[latex]{4^5}{4^2}= 4^{5−2} = 4^3[/latex]

So, the difference in doses prescribed is [latex]4^3 = 64.[/latex]

Scenario C: Zero Rule for Exponents

A pharmaceutical technician is reviewing a formula where the amount of an active ingredient is represented by 70.

  • Simplify the expression and explain the meaning of the result in the context of the formula.
Scenario C Solution

Using the Zero Exponent Rule:

[latex]7^0 = 1[/latex]This means that the active ingredient is considered neutral in this formula, contributing a value of 1, regardless of the actual number of doses or units involved.

Chapter Credit

Chapter (including Case Studies) consists of original content.

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