When finding an expression with multiplication or an exercise that has only multiplication operationsinside 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: (aΓb)n=anΓbn This property also pertains to algebraic expressions.
(5Γ2)3= Notice that we have a power of a product of whole numbers. We can see that the exponent3 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 125Γ8=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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We will notice that the exponent applies to the entire expression enclosed in parentheses and that between the 2 and the 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=
8β X3=
We will multiply theX by its coefficient and we will get:
8X3
Now let's move on to a slightly more complicated exercise.
(22β X3)2
We see that the exponent 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
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
First we will multiply the exponent located outside the parentheses by the exponent of the base2 and then, we will do it by the exponent of the baseX.
We will get:
24β X6
16β X6
Now we will multiply theX by its coefficient and we will have:
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
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 and 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=
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=
29β X24β 53=
We can continue solving and we will arrive at:
512β X24β 125=
64,000β X24=
Let's multiply theX by its coefficient and it will give us:
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.
If you are interested in this article, you might also be interested in the following articles:
examples with solutions for power of a multiplication
Exercise #1
(2Γ8Γ7)2=
Video Solution
Step-by-Step Solution
We use the power property for the product inside parentheses:
(zβ t)n=znβ tnThat 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 72Therefore, 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
Exercise #2
(9Γ2Γ5)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 ynWe apply the property in the problem:
(9β 2β 5)3=93β 23β 53When 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
93Γ23Γ53
Exercise #3
(3Γ4Γ5)4=
Video Solution
Step-by-Step Solution
We use the power law for multiplication within parentheses:
(xβ y)n=xnβ ynWe apply it to the problem:
(3β 4β 5)4=34β 44β 54Therefore, 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
Exercise #4
(4Γ7Γ3)2=
Video Solution
Step-by-Step Solution
We use the power law for multiplication within parentheses:
(xβ y)n=xnβ ynWe apply it to the problem:
(4β 7β 3)2=42β 72β 32Therefore, 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
42Γ72Γ32
Exercise #5
(3Γ2Γ4Γ6)β4=
Video Solution
Step-by-Step Solution
We use the power property for the product inside parentheses:
(zβ t)n=znβ tnThat 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)β4=3β4β 2β4β 4β4β 6β4Therefore, 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.).