Applying Combined Exponents Rules: Variable in the base of the power

$(5\cdot x\cdot3)^3=$

We use the formula:

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

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

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

$15^3x^3$

$15^3\cdot x^3$

Solve the exercise:

$(a^5)^7=$

We use the formula:

$(a^m)^n=a^{m\times n}$

and therefore we obtain:

$(a^5)^7=a^{5\times7}=a^{35}$

$a^{35}$

$(a^4)^6=$

We use the formula

$(a^m)^n=a^{m\times n}$

Therefore, we obtain:

$a^{4\times6}=a^{24}$

$a^{24}$

$(a\times b\times c\times4)^7=$

We use the formula:

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

Therefore, we obtain:

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

$a^7\times b^7\times c^7\times4^7$

$(a\cdot5\cdot6\cdot y)^5=$

We use the formula:

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

Therefore, we obtain:

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

$a^530^5y^5$

$a^5\cdot30^5\cdot y^5$

$(a\cdot b\cdot8)^2=$

We use the formula

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

Therefore, we obtain:

$a^2b^28^2$

$a^2\cdot b^2\cdot8^2$

$(x\cdot4\cdot3)^3=$

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.

$x^3\cdot4^3\cdot3^3$

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

We use the formula:

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

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

$y^5\times x^5\times3^5$

$((y^6)^8)^9=$

We use the power rule of distributing exponents.

$(a^m)^n=a^{m\cdot n}$We apply it in the problem:

$\big((y^6)^8\big)^9=(y^{6\cdot8})^9=y^{6\cdot8\cdot9}=y^{432}$When we use the aforementioned rule twice, the first time for the inner parentheses in the first stage and the second time for the remaining parentheses in the second stage, in the last stage we calculate the result of the multiplication in the power exponent.

Therefore, the correct answer is option b.

$y^{432}$

$((a^2)^3)^{\frac{1}{4}}=$

We use the power rule for exponents.

$(a^m)^n=a^{m\cdot n}$We apply it to the problem:

$\big((a^2)^3\big)^{\frac{1}{4}}=(a^{2\cdot3})^{\frac{1}{4}}=a^{2\cdot3\cdot\frac{1}{4}}=a^{\frac{6}{4}}=a^{\frac{3}{2}}$When we use the previously mentioned rule twice, the first time for the inner parentheses in the first stage and the second time for the remaining parentheses in the second stage, in the third stage we calculate the result of the multiplication in the exponent. While remembering that multiplying by a fraction is actually doubling the numerator of the fraction and, finally, in the last stage we simplify the fraction we obtained in the exponent.

Now remember that -

$\frac{3}{2}=1\frac{1}{2}=1.5$

Therefore, the correct answer is option a.

$a^{1.5}$

$((b^3)^6)^2=$

We use the formula

$(a^m)^n=a^{m\times n}$

Therefore, we obtain:

$((b^3)^6)^2=(b^{3\times6})^2=(b^{18})^2=b^{18\times2}=b^{36}$

$b^{36}$

$(y\times7\times3)^4=$

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.

$y^4\times7^4\times3^4$

Solve the exercise:

$(x^2\times3)^2=$

We have an exponent raised to another exponent with a multiplication between parentheses:

$(z\cdot t)^n=z^n\cdot t^n$This says that in a case where a power is applied to a multiplication between parentheses,the power is applied to each term of the multiplication when the parentheses are opened,

We apply it in the problem:

$(3x^2)^2=3^2(x^2)^2$With the second term of the multiplication we proceed carefully, since it is already in a power (that's why we use parentheses). The term will be raised using the power law for an exponent raised to another exponent:

$(a^m)^n=a^{m\cdot n}$and we apply it in the problem:

$3^2(x^2)^2=9x^{2\cdot2}=9x^4$In the first step we raise the number to the power, and in the second step we multiply the exponent.

Therefore, the correct answer is option a.

$9x^4$

Solve the exercise:

$Y^2+Y^6-Y^5\cdot Y=$

We use the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$We apply it in the problem:

$Y^2+Y^6-Y^5\cdot Y=Y^2+Y^6-Y^{5+1}=Y^2+Y^6-Y^6=Y^2$When we apply the previous property to the third expression from the left in the sum, and then simplify the total expression by adding like terms.

Therefore, the correct answer is option D.

$Y^2$

Solve the exercise:

$a^2:a+a^3\cdot a^5=$

First we rewrite the first expression on the left of the problem as a fraction:

$\frac{a^2}{a}+a^3\cdot a^5$Then we use two properties of exponentiation, to multiply and divide terms with identical bases:

A.

$b^m\cdot b^n=b^{m+n}$2.

$\frac{b^m}{b^n}=b^{m-n}$Returning to the problem and applying the two properties of exponentiation mentioned earlier:

$\frac{a^2}{a}+a^3\cdot a^5=a^{2-1}+a^{3+5}=a^1+a^8=a+a^8$

Later on, keep in mind that we need to factor the expression we obtained in the last step by extracting the common factor,

Therefore, we extract from outside the parentheses the greatest common divisor to the two terms which are:

$a$ We obtain the expression:

$a+a^8=a(1+a^7)$when we use the property of exponentiation mentioned earlier in A.

$$$a^8=a^{1+7}=a^1\cdot a^7=a\cdot a^7$

Summarizing the solution to the problem and all the steps, we obtained the following:

$\frac{a^2}{a}+a^3\cdot a^5=a(1+a^{7)}$Therefore, the correct answer is option b.

$a(1+a^7)$

Simplify the expression:

$a^3\cdot a^2\cdot b^4\cdot b^5=$

In the exercise of multiplying powers, we will add up all the powers of the same product, in this case the terms a, b

We use the formula:

$a^n\times a^m=a^{n+m}$

We are going to focus on the term a:

$a^3\times a^2=a^{3+2}=a^5$

We are going to focus on the term b:

$b^4\times b^5=b^{4+5}=b^9$

Therefore, the exercise that will be obtained after simplification is:

$a^5\times b^9$

$a^5\cdot b^9$

$k^2\cdot t^4\cdot k^6\cdot t^2=$

Using the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$It is important to note that this law is only valid for terms with identical bases,

We notice that in the problem there are two types of terms. First, for the sake of order, we will use the substitution property to rearrange the expression so that the two terms with the same base are grouped together. The, we will proceed to solve:

$k^2t^4k^6t^2=k^2k^6t^4t^2$Next, we apply the power property to each different type of term separately,

$k^2k^6t^4t^2=k^{2+6}t^{4+2}=k^8t^6$We apply the property separately - for the terms whose bases are$k$and for the terms whose bases are$t$We add the powers in the exponent when we multiply all the terms with the same base.

The correct answer then is option b.

$k^8\cdot t^6$

$a\cdot b\cdot a\cdot b\cdot a^2$

We use the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$It is important to note that this property is only valid for terms with identical bases,

We return to the problem

We notice that in the problem there are two types of terms with different bases. First, for the sake of order, we will use the substitution property of multiplication to rearrange the expression so that the two terms with the same base are grouped together. Then, we will proceed to work:

$a\cdot b\operatorname{\cdot}a\operatorname{\cdot}b\operatorname{\cdot}a^2=a\cdot a\cdot a^2\cdot b\cdot b$Next, we apply the power property for each type of term separately,

$a\cdot a\cdot a^2\cdot b\cdot b=a^{1+1+2}\cdot b^{1+1}=a^4\cdot b^2$

$$We apply the power property separately - for the terms whose bases are$a$and then for the terms whose bases are$b$and we add the exponents and simplify the terms.

$$Therefore, the correct answer is option c.

Note:

We use the fact that:

$a=a^1$and the same for $b$.

$a^4\cdot b^2$

$E^6\cdot F^{-4}\cdot E^0\cdot F^7\cdot E=$

We use the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$It should be noted that this property is only valid for terms with identical bases,

We return to the problem

We notice that in the problem there are two types of terms with different bases. First, for the sake of order, we will use the substitution property of multiplication to rearrange the expression so that the two terms with the same base are grouped together. Then, we will proceed to work:

$E^6\cdot F^{-4}\cdot E^0\cdot F^7\cdot E=E^6\cdot E^0\cdot E\cdot F^{-4}\cdot F^7$Next, we apply the power property for each type of term separately,

$E^6\cdot E^0\cdot E\cdot F^{-4}\cdot F^7=E^{6+0+1}\cdot F^{-4+7}=E^7\cdot F^3$

$$We apply the power property separately - for the terms whose bases are$E$and for the terms whose bases are$F$and we add the exponents and simplify the terms with the same base.

The correct answer is then option d.

Note:

We use the fact that:

$E=E^1$.

$E^7\cdot F^3$

$(x^2\times a^3)^{\frac{1}{4}}=$

$x^{\frac{1}{2}}\times a^{\frac{3}{4}}$