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8.2: Exponential Terms Raised to an Exponent

Difficulty Level: At Grade Created by: CK-12
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What if you had an exponential expression that was raised to a secondary power, like \begin{align*}(2^3)^2\end{align*}? How could you simplify it? After completing this Concept, you'll be able to use the power of a product property to simplify exponential expressions like this one.

Watch This

Foundation: 0802S Power of a Product

The following video from YourTeacher.com may make it clearer how the power rule works for a variety of exponential expressions:

YourTeacher: Power Rule- Multiplying Exponents

Guidance

What happens when we raise a whole expression to a power? Let’s take \begin{align*}x\end{align*} to the power of 4 and cube it. Again we’ll use the full factored form for each expression:

\begin{align*}(x^4)^3 & = x^4 \times x^4 \times x^4 \qquad 3 \ factors \ of \ \{x \ to \ the \ power \ 4\} \\ (x \cdot x \cdot x \cdot x) \cdot (x \cdot x \cdot x \cdot x) \cdot (x \cdot x \cdot x \cdot x) & = x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x = x^{12}\end{align*}

So \begin{align*}(x^4)^3 = x^{12}\end{align*}. You can see that when we raise a power of \begin{align*}x\end{align*} to a new power, the powers multiply.

Power Rule for Exponents: \begin{align*}(x^n)^m = x^{(n\cdot m)}\end{align*}

If we have a product of more than one term inside the parentheses, then we have to distribute the exponent over all the factors, like distributing multiplication over addition. For example:

\begin{align*}(x^2y)^4 = (x^2)^4 \cdot (y)^4 = x^8y^4.\end{align*}

Or, writing it out the long way:

\begin{align*}(x^2 y)^4 = (x^2 y) (x^2 y) (x^2 y) (x^2 y) & = (x \cdot x \cdot y) (x \cdot x \cdot y) (x \cdot x \cdot y) (x \cdot x \cdot y)\\ & = x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y = x^8 y^4\end{align*}

Note that this does NOT work if you have a sum or difference inside the parentheses! For example, \begin{align*}(x+y)^2 \neq x^2+y^2\end{align*}. This is an easy mistake to make, but you can avoid it if you remember what an exponent means: if you multiply out \begin{align*}(x+y)^2\end{align*} it becomes \begin{align*}(x+y)(x+y)\end{align*}, and that’s not the same as \begin{align*}x^2+y^2\end{align*}. We’ll learn how we can simplify this expression in a later chapter.

Example A

Simplify the following expressions.

a) \begin{align*}3^5 \cdot 3^7\end{align*}

b) \begin{align*}2^6 \cdot 2\end{align*}

c) \begin{align*}(4^2)^3\end{align*}

Solution

When we’re just working with numbers instead of variables, we can use the product rule and the power rule, or we can just do the multiplication and then simplify.

a) We can use the product rule first and then evaluate the result: \begin{align*}3^5 \cdot 3^7 = 3^{12}=531441\end{align*}.

OR we can evaluate each part separately and then multiply them: \begin{align*}3^5 \cdot 3^7 = 243 \cdot 2187 = 531441\end{align*}.

b) We can use the product rule first and then evaluate the result: \begin{align*}2^6 \cdot 2 = 2^7 = 128\end{align*}.

OR we can evaluate each part separately and then multiply them: \begin{align*}2^6 \cdot 2 = 64 \cdot 2 = 128\end{align*}.

c) We can use the power rule first and then evaluate the result: \begin{align*}(4^2)^3 = 4^6 = 4096\end{align*}.

OR we can evaluate the expression inside the parentheses first, and then apply the exponent outside the parentheses: \begin{align*}(4^2)^3 = (16)^3 = 4096\end{align*}.

Example B

Simplify the following expressions.

a) \begin{align*}x^2 \cdot x^7\end{align*}

b) \begin{align*}(y^3)^5\end{align*}

Solution

When we’re just working with variables, all we can do is simplify as much as possible using the product and power rules.

a) \begin{align*}x^2 \cdot x^7 = x^{2+7} = x^9\end{align*}

b) \begin{align*}(y^3)^5 = y^{3 \times 5}= y^{15}\end{align*}

Example C

Simplify the following expressions.

a) \begin{align*}(3x^2 y^3) \cdot (4xy^2)\end{align*}

b) \begin{align*}(4xyz) \cdot (x^2 y^3) \cdot (2y z^4)\end{align*}

c) \begin{align*}(2a^3b^3)^2\end{align*}

Solution

When we have a mix of numbers and variables, we apply the rules to each number and variable separately.

a) First we group like terms together: \begin{align*}(3x^2y^3) \cdot (4xy^2) = (3 \cdot 4) \cdot (x^2 \cdot x) \cdot (y^3 \cdot y^2)\end{align*}

Then we multiply the numbers or apply the product rule on each grouping: \begin{align*}=12 x^3y^5\end{align*}

b) Group like terms together: \begin{align*}(4xy z) \cdot (x^2 y^3) \cdot (2y z^4) = (4 \cdot 2) \cdot (x \cdot x^2) \cdot (y \cdot y^3 \cdot y) \cdot (z \cdot z^4)\end{align*}

Multiply the numbers or apply the product rule on each grouping: \begin{align*}= 8x^3 y^5 z^5\end{align*}

c) Apply the power rule for each separate term in the parentheses: \begin{align*}(2a^3b^3)^2 = 2^2 \cdot (a^3)^2 \cdot (b^3)^2\end{align*}

Multiply the numbers or apply the power rule for each term \begin{align*}=4a^6 b^6\end{align*}

Watch this video for help with the Examples above.

CK-12 Foundation: Power of a Product

Vocabulary

  • Exponent: An exponent is a power of a number that shows how many times that number is multiplied by itself. An example would be \begin{align*}2^3\end{align*}. You would multiply 2 by itself 3 times: \begin{align*}2 \times 2 \times 2.\end{align*} The number 2 is the base and the number 3 is the exponent. The value \begin{align*}2^3\end{align*} is called the power.
  • Product of Powers Property: For all real numbers \begin{align*}\chi\end{align*},

\begin{align*}\chi^n \cdot \chi^m = \chi^{n+m}\end{align*}.

  • Power of a Product Property: For all real numbers \begin{align*}\chi\end{align*},

\begin{align*}(\chi^n)^m = \chi^{n \cdot m}\end{align*}.

Guided Practice

Simplify the following expressions.

a) \begin{align*}(x^2)^2 \cdot x^3\end{align*}

b) \begin{align*}(2x^2y) \cdot (3xy^2)^3\end{align*}

c) \begin{align*}(4a^2 b^3)^2 \cdot (2ab^4)^3\end{align*}

Solution

In problems where we need to apply the product and power rules together, we must keep in mind the order of operations. Exponent operations take precedence over multiplication.

a) We apply the power rule first: \begin{align*}(x^2)^2 \cdot x^3 = x^4 \cdot x^3\end{align*}

Then apply the product rule to combine the two terms: \begin{align*}x^4 \cdot x^3 = x^7\end{align*}

b) Apply the power rule first: \begin{align*}(2x^2 y) \cdot (3xy^2)^3 = (2x^2y) \cdot (27x^3y^6)\end{align*}

Then apply the product rule to combine the two terms: \begin{align*}(2x^2 y) \cdot (27 x^3 y^6) = 54x^5y^7\end{align*}

c) Apply the power rule on each of the terms separately: \begin{align*}(4a^2 b^3)^2 \cdot (2ab^4)^3 = (16a^4 b^6) \cdot (8a^3 b^{12})\end{align*}

Then apply the product rule to combine the two terms: \begin{align*}(16a^4 b^6) \cdot (8a^3 b^{12}) = 128a^7 b^{18}\end{align*}

Practice

Simplify:

  1. \begin{align*}(a^3)^4\end{align*}
  2. \begin{align*}(xy)^2\end{align*}
  3. \begin{align*}(-5y)^3\end{align*}
  4. \begin{align*}(3a^2b^3)^4\end{align*}
  5. \begin{align*}(-2xy^4z^2)^5\end{align*}
  6. \begin{align*}(-8x)^3(5x)^2\end{align*}
  7. \begin{align*}(-x)^2(xy)^3\end{align*}
  8. \begin{align*}(4a^2)(-2a^3)^4\end{align*}
  9. \begin{align*}(12xy)(12xy)^2\end{align*}
  10. \begin{align*}(2xy^2)(-x^2y)^2(3x^2y^2)\end{align*}

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Date Created:
Aug 13, 2012
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Apr 11, 2016
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MAT.ALG.932.4.L.1