Multiplication of Powers: Variable in the base of the power

$a^3\times a^4=$

Here, we will need to calculate a multiplication between terms with identical bases, therefore we will use the appropriate power property:

$b^m\cdot b^n=b^{m+n}$Note that this property can only be used to calculate the multiplication between terms with identical bases,

We apply it to the problem:

$a^3\cdot a^4=a^{3+4}=a^7$Therefore, the correct answer is option b.

$a^7$

$x^2\times x^5=$

Here we will have to to multiply terms with identical bases, therefore we use the appropriate power property:

$b^m\cdot b^n=b^{m+n}$Note that this property can only be used to calculate the multiplication between terms with identical bases,

From now on we no longer write the multiplication sign, but use the accepted form of writing in which placing terms next to each other means multiplication.

We apply it in the problem:

$x^2x^5=x^{2+5}=x^7$Therefore, the correct answer is D.

$x^7$

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:

$E=E^1$.

$E^7\cdot F^3$

Choose the expression that is equal to the following:

$a^4\cdot a^5$

$a^9$

Solve the exercise:

$X^3\cdot X^2:X^5+X^4$

$1+X^4$

Simplify the following:

$\lbrack\frac{a^4}{a^3}\times\frac{a^8}{a^7}\rbrack:\frac{a^{10}}{a^8}$

$1$

Simplify the following:

$\frac{a^{20b}}{a^{15b}}\times\frac{a^{3b}}{a^{2b}}=$

$a^{6b}$

Simplify the following:

$\frac{b^{22}}{b^{20}}\times\frac{b^{30}}{b^{20}}=$

$b^{12}$

$c^{-1}\cdot d^6\cdot d^{-2}\cdot c^3\cdot c^2=$

$c^4\cdot d^4$