Power of a Product - Examples, Exercises and Solutions

We begin by using the power rule for parentheses:

$(z\cdot t)^n=z^n\cdot t^n$That is, the power applied to a product inside parentheses, is applied to each of the terms within, when the parentheses are opened.

We then apply the above rule to the problem:

$(2\cdot8\cdot7)^2=2^2\cdot8^2\cdot7^2$Therefore, the correct answer is option d.

Note:

From the formula of the power property inside parentheses mentioned above, it might seem as though it refers to only two terms of the product inside of the parentheses, but in reality, it is also valid for the power over a multiplication of many terms inside parentheses, as was seen above.

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.).

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

$(x\cdot y)^n=x^n\cdot y^n$We apply the rule to the problem:

$(9\cdot2\cdot5)^3=9^3\cdot2^3\cdot5^3$When we apply the power within parentheses to the product of the terms we do so separately and maintain the multiplication,

Therefore, the correct answer is option B.

We use the power law for multiplication within parentheses:

$(x\cdot y)^n=x^n\cdot y^n$We apply it to the problem:

$(3\cdot4\cdot5)^4=3^4\cdot4^4\cdot5^4$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.

We use the power law for multiplication within parentheses:

$$$(x\cdot y)^n=x^n\cdot y^n$We apply it to the problem:

$(4\cdot7\cdot3)^2=4^2\cdot7^2\cdot3^2$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.

We begin by using the power rule for parentheses.

$(z\cdot t)^n=z^n\cdot t^n$That is, the power applied to a product inside parentheses is applied to each of the terms within when the parentheses are opened,

We apply the above rule to the given problem:

$(3\cdot2\cdot4\cdot6)^{-4}=3^{-4}\cdot2^{-4}\cdot4^{-4}\cdot6^{-4}$Therefore, the correct answer is option d.

Note:

According to the formula of the power property inside parentheses mentioned above, it might seem as though it refers to only two terms of the product inside of the parentheses, but in reality, it is also valid for the power over a multiplication of many terms inside parentheses, as was seen above.

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.).

We use the formula:

$(a\times b)^n=a^nb^n$

$(5\times x\times3)^3=(15x)^3$

$(15x)^3=(15\times x)^3$

$15^3x^3$

We use the formula:

$(a\times b)^n=a^nb^n$

$(y\times x\times3)^5=y^5x^53^5$

Let us begin by using the law of exponents for a power that is applied to parentheses in which terms are multiplied:

$(x\cdot y)^n=x^n\cdot y^n$We apply the rule to our problem:

$(x\cdot4\cdot3)^3= x^3\cdot4^3\cdot3^3$When we apply the power to the product of the terms within parentheses, we apply the power to each term of the product separately and keep the product,

Therefore, the correct answer is option C.

We use the formula

$(a\times b)^x=a^xb^x$

Therefore, we obtain:

$a^2b^28^2$

We use the formula:

$(a\times b)^x=a^xb^x$

Therefore, we obtain:

$(a\times5\times6\times y)^5=(a\times30\times y)^5$

$a^530^5y^5$

We use the formula:

$(a\times b)^x=a^xb^x$

Therefore, we obtain:

$a^7b^7c^74^7$

We use the power law for a multiplication between parentheses:

$(z\cdot t)^n=z^n\cdot t^n$That is, a power applied to a multiplication between parentheses is applied to each term when the parentheses are opened,

We apply it in the problem:

$(4\cdot9\cdot11)^a=4^a\cdot9^a\cdot11^a$Therefore, the correct answer is option b.

Note:

From the power property formula mentioned, we can understand that it works not only with two terms of the multiplication between parentheses, but also valid with a multiplication between multiple terms in parentheses. As we can see in this problem.

We use the power law for multiplication within parentheses:

$(x\cdot y)^n=x^n\cdot y^n$We apply it in the problem:

$(y\cdot7\cdot3)^4=y^4\cdot7^4\cdot3^4$Therefore, the correct answer is option a.

Note:

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

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

$(x\cdot y)^n=x^n\cdot y^n$We apply the law in the problem:

$(7\cdot4\cdot6\cdot3)^4=7^4\cdot4^4\cdot6^4\cdot3^4$When we apply the exponent to a parentheses with multiplication, we apply the exponent to each term of the multiplication separately, and we keep the multiplication between them.

Therefore, the correct answer is option a.

We begin by applying the power rule to the products within the parentheses:

$(z\cdot t)^n=z^n\cdot t^n$That is, the power applied to a product within parentheses is applied to each of the terms when the parentheses are opened,

We apply the rule to the given problem:

$(8\cdot9\cdot5\cdot3)^{-2}=8^{-2}\cdot9^{-2}\cdot5^{-2}\cdot3^{-2}$Therefore, the correct answer is option c.

Note:

Whilst it could be understood that the above power rule applies only to two terms of the product within parentheses, in reality, it is also valid for the power over a multiplication of multiple terms within parentheses, as was seen in the above problem.

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