Exponent of a Multiplication

πŸ†Practice power of a multiplication

When finding an expression with multiplication or an exercise that has only multiplication operations inside a parenthesis and the wholes expression is raised to a certain exponent, we can take the exponent and apply it to each of the terms of the expression or exercise.
We must not forget to keep the multiplication signs between the terms.
Property formula:
(aΓ—b)n=anΓ—bn (a\times b)^n=a^n\times b^n
This property also pertains to algebraic expressions.

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Test yourself on power of a multiplication!

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\( (2\times8\times7)^2= \)

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Example of the power of a multiplication

First example

(5Γ—2)3= (5\times2)^3=
Notice that we have a power of a product of whole numbers. We can see that the exponent 3 3 is applied to the entire expression enclosed within the parentheses, therefore, we can raise each of the terms while maintaining the multiplication sign between them.
We will obtain:
5323=Γ— 5^32^3=\times
125Γ—8=1000 125\times8=1000


If we have a multiplication exercise with an exponent, we can apply the exponent to each of the terms separately.

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Let's look at some examples

(2X)3 (2X)^3

We will notice that the exponent applies to the entire expression enclosed in parentheses and that between the 2 2 and the X X there is, in fact, a multiplication operation.

We can raise each of the terms of the expression to the exponent and we will obtain:

23β‹…X3= 2^3\cdot X^3=

8β‹…X3= 8\cdot X^3=

We will multiply the X X by its coefficient and we will get:

8X3 8X^3


Now let's move on to a slightly more complicated exercise.

(22β‹…X3)2 (2^2\cdot X^3)^2

We see that the exponent 2 2 is outside the parentheses, therefore, it applies to the entire expression. The terms of the expression are multiplied, therefore, it is the power of a multiplication.

Well, now we can apply the exponent to each of the terms separately and we will not forget to maintain the multiplication between them.

We obtain:

(22)2β‹…(x3)2 \left(2^2\right)^2\cdot\left(x^3\right)^2

We have obtained what is called: Power of a power

To be able to continue solving the exercise we will remember the following property:

Power of power

If we have a power of a power we must perform a multiplication of exponents, that is, multiply powers.

In symbols:

(an)k=anβ‹…k (a^n)^k=a^{n\cdot k}

First we will multiply the exponent located outside the parentheses by the exponent of the base 2 2 and then, we will do it by the exponent of the base X X .

We will get:

24β‹…X6 2^4\cdot X^6

16β‹…X6 16\cdot X^6

Now we will multiply the X X by its coefficient and we will have:

16X6 16X^6


Do you know what the answer is?

Let's move on to see another example

which will demonstrate to us that no matter how many terms the exercise includes, as long as there is multiplication between them and as long as they are raised to a certain power located outside of the parentheses, in such a case we can apply the power to each of the terms separately and maintain the multiplication operation between them.

(22Xβ‹…X5β‹…X2β‹…2β‹…5)3 (2^2X\cdot X^5\cdot X^2\cdot2\cdot5)^3

Recommendation:

Before applying the power located outside of the parentheses to each of the terms separately, carefully observe the exercise.

If you look closely and are thoroughly familiar with the properties of powers or laws of exponents, you will immediately see that, you can first act inside the parentheses using the property of multiplication of powers with the same base to simplify the expression.

Let's remember that, when we have a multiplication of powers with the same base, we can add the exponents and obtain a single base with a single exponent.

If there is no exponent it means that the exponent is one and it is important that we remember to add it.

Upon observing the exercise we will discover that there are some equal bases: 2 2 and X X .

Since the operation between all the terms is multiplication we can add the relevant exponents and in this way we will arrive at:

(23β‹…X8β‹…5)3= (2^3\cdot X^8\cdot5)^3=


Now, with an already abbreviated expression between the parentheses, it will be easier and faster for us to apply the power located outside of the parentheses to each of the terms.

We will do it and obtain:

(23)3β‹…(x8)3β‹…(5)3= \left(2^3\right)^3\cdot\left(x^8\right)^3\cdot\left(5\right)^3=

29β‹…X24β‹…53= 2^9\cdot X^{24}\cdot5^3=

We can continue solving and we will arrive at:

512β‹…X24β‹…125= 512\cdot X^{24}\cdot125=

64,000β‹…X24= 64,000\cdot X^{24}=

Let's multiply the X X by its coefficient and it will give us:

64,000X24= 64,000X^{24}=


Exercises with the Power of a Multiplication

(2Γ—5)3= (2\times5)^3=

(1Γ—2)3= (1\times2)^3=

(3Γ—3)3= (3\times3)^3=

(2Γ—2)2= (2\times2)^2=

(7Γ—4)2= (7\times4)^2=


(32XΓ—X5Γ—X3)3= \left(3^2X\times X^5\times X^3\right)^3=

(52XΓ—X2Γ—X2)2= \left(5^2X\times X^2\times X^2\right)^2=

(82XΓ—X2Γ—X25Γ—3X)3= \left(8^2X\times X^2\times X^25\times3X\right)^3=

(24Γ—5X2Γ—X37Γ—3X)3= \left(2^4\times5X^2\times X37\times3X\right)^3=

(24Γ—5X2Γ—XΓ—8Γ—X2)3= \left(2^4\times5X^2\times X\times8\times X^2\right)^3=


Check your understanding

Review Questions

How to solve a power of a multiplication?

To solve a power of a multiplication we must raise each of the factors to the indicated power and maintain the multiplication sign between the terms.


How to solve multiplication of powers with different bases?

The law of exponents indicates that when we have multiplication of powers with the same base, we must add the exponents, however, if we have different bases we cannot apply this law.


How to multiply powers with different bases and the same exponent?

If the base is different we cannot add the exponents, although in some exercises it will be possible to manipulate the expressions to equalize the bases and apply the law of exponents.


How to solve addition of powers with different bases?

There is no law for the addition of powers. It does not matter if they have the same base or not.


How are powers multiplied with the same exponent and different bases?

If the base is different, we cannot directly apply the law of exponents. Sometimes we may manipulate the bases to try to equalize them and apply the law.


Do you think you will be able to solve it?

examples with solutions for power of a multiplication

Exercise #1

(2Γ—8Γ—7)2= (2\times8\times7)^2=

Video Solution

Step-by-Step Solution

We use the power property for the product inside parentheses:

(zβ‹…t)n=znβ‹…tn (z\cdot t)^n=z^n\cdot t^n That is, the power applied to a product inside parentheses is applied to each term of it when the parentheses are opened,

We apply the property to the problem:

(2β‹…8β‹…7)2=22β‹…82β‹…72 (2\cdot8\cdot7)^2=2^2\cdot8^2\cdot7^2 Therefore, the correct answer is option d.

Note:

From the formula of the power property inside parentheses mentioned above, it might seem that it refers only to two terms of the product inside parentheses, but in reality, it is also valid for the power over a multiplication of many terms inside parentheses, as was done in this problem and in other problems.

A good exercise is to demonstrate that if the previous property is valid for a power over a product of two terms inside parentheses (as formulated above), then it is also valid for a power over several terms of the product inside parentheses (for example - three terms, etc.).

Answer

12,544 12,544

Exercise #2

(9Γ—2Γ—5)3= (9\times2\times5)^3=

Video Solution

Step-by-Step Solution

We use the power property for a power that is applied to parentheses in which terms are multiplied:

(xβ‹…y)n=xnβ‹…yn (x\cdot y)^n=x^n\cdot y^n We apply the property in the problem:

(9β‹…2β‹…5)3=93β‹…23β‹…53 (9\cdot2\cdot5)^3=9^3\cdot2^3\cdot5^3 When we apply the power within parentheses to the product of the terms to each term of the product separately and maintain the multiplication,

Therefore, the correct answer is option B.

Answer

93Γ—23Γ—53 9^3\times2^3\times5^3

Exercise #3

(3Γ—4Γ—5)4= (3\times4\times5)^4=

Video Solution

Step-by-Step Solution

We use the power law for multiplication within parentheses:

(xβ‹…y)n=xnβ‹…yn (x\cdot y)^n=x^n\cdot y^n We apply it to the problem:

(3β‹…4β‹…5)4=34β‹…44β‹…54 (3\cdot4\cdot5)^4=3^4\cdot4^4\cdot5^4 Therefore, the correct answer is option b.

Note:

From the formula of the power property mentioned above, we understand that it refers not only to two terms of the multiplication within parentheses, but also for multiple terms within parentheses.

Answer

344454 3^44^45^4

Exercise #4

(4Γ—7Γ—3)2= (4\times7\times3)^2=

Video Solution

Step-by-Step Solution

We use the power law for multiplication within parentheses:

(xβ‹…y)n=xnβ‹…yn (x\cdot y)^n=x^n\cdot y^n We apply it to the problem:

(4β‹…7β‹…3)2=42β‹…72β‹…32 (4\cdot7\cdot3)^2=4^2\cdot7^2\cdot3^2 Therefore, the correct answer is option a.

Note:

From the formula of the power property mentioned above, we understand that we can apply it not only to the multiplication of two terms within parentheses, but is also for multiple terms within parentheses.

Answer

42Γ—72Γ—32 4^2\times7^2\times3^2

Exercise #5

(3Γ—2Γ—4Γ—6)βˆ’4= (3\times2\times4\times6)^{-4}=

Video Solution

Step-by-Step Solution

We use the power property for the product inside parentheses:

(zβ‹…t)n=znβ‹…tn (z\cdot t)^n=z^n\cdot t^n That is, the power applied to a product inside parentheses is applied to each term of it when the parentheses are opened,

We apply the property to the problem:

(3β‹…2β‹…4β‹…6)βˆ’4=3βˆ’4β‹…2βˆ’4β‹…4βˆ’4β‹…6βˆ’4 (3\cdot2\cdot4\cdot6)^{-4}=3^{-4}\cdot2^{-4}\cdot4^{-4}\cdot6^{-4} Therefore, the correct answer is option d.

Note:

From the formula of the power property inside parentheses mentioned above, it can be understood that it refers only to two terms of the product inside parentheses, but in reality, it is also valid for the power over a multiplication of many terms inside parentheses, as was done in this problem and in other problems.

A good exercise is to demonstrate that if the previous property is valid for a power over a product of two terms inside parentheses (as formulated above), then it is also valid for a power over several terms of the product inside parentheses (for example - three terms, etc.).

Answer

3βˆ’4Γ—2βˆ’4Γ—4βˆ’4Γ—6βˆ’4 3^{-4}\times2^{-4}\times4^{-4}\times6^{-4}

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