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

We use the formula:

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

and therefore we obtain:

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

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:

$a^5\times b^9$

Let's start with multiplying the fractions, remembering that multiplication of fractions is performed by multiplying numerator by numerator and denominator by denominator:

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

Next, we'll notice that both in the numerator and denominator, multiplication occurs between terms with identical bases, so we'll use the power law for multiplying terms with identical bases:

$c^m\cdot c^n=c^{m+n}$

We emphasize that this law can only be used when multiplication is performed between terms with identical bases.

From here on, we will no longer indicate the multiplication sign, but use the conventional writing method where placing terms next to each other means multiplication.
Let's return to the problem and apply the above power law separately to the fraction's numerator and denominator:

$\frac{b^{22}b^{30}}{b^{20}b^{20}}=\frac{b^{22+30}}{b^{20+20}}=\frac{b^{52}}{b^{40}}$

where in the final step we calculated the sum of the exponents in the numerator and denominator.

Now we notice that division is required between two terms with identical bases, so we'll use the power law for division between terms with identical bases:

$\frac{c^m}{c^n}=c^{m-n}$

We emphasize that this law can only be used when division is performed between terms with identical bases.

Let's return to the problem and apply the above power law:

$\frac{b^{52}}{b^{40}}=b^{52-40}=b^{12}$

where in the final step we calculated the subtraction between the exponents.

We have obtained the most simplified expression and therefore we are done.

Therefore, the correct answer is C.

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:

1.

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

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.

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

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

Therefore, we obtain:

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

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:

$a^2b^28^2$

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 $b$.

We use the formula

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

Therefore, we obtain:

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

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.

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 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.

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 formula:

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

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

First, simplify the numerator and the denominator separately:
Numerator: $X^4 \cdot X^3 = X^{4+3} = X^7$
Denominator: $X^5 \cdot X^2 = X^{5+2} = X^7$

Now, combine the simplified numerator and denominator:

$\frac{X^7}{X^7}$

Since any number divided by itself is 1, we have:

$\frac{X^7}{X^7} = 1$

Therefore, the correct answer is:

$1$

First, let's write the problem in an organized way and use fraction notation for the first term:$\frac{}{}\frac{X^3\cdot X^2}{X^5}+X^4$

Let's continue and refer to the first term in the above sum:

$\frac{X^3\cdot X^2}{X^5}$

Let's deal with the numerator, first using the law of exponents for multiplying terms with identical bases:

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

and we get:

$\frac{X^3\cdot X^2}{X^5}=\frac{X^{3+2}}{X^5}=\frac{X^5}{X^5}$

Now let's use the law of exponents for division between terms with identical bases:

$a^m:a^n=\frac{a^m}{a^n}=a^{m-n}$

When in the first stage of the above formula we just wrote the same thing in fraction notation instead of using division (:), let's apply the law of exponents to the problem and calculate the result for the first term we got above:

$\frac{X^5}{X^5}=X^{5-5}=X^0$

Now let's use the law of exponents:

$a^0=1$

We can notice that this rule is actually just the understanding that dividing a number by itself will always give the result 1. Let's return to the problem and we get that the result of the first term in the exercise (meaning - the result of calculating the fraction) is:

$X^0=1$,

let's return to the complete exercise and summarize everything said so far, we got:

$\frac{X^3\cdot X^2}{X^5}+X^4=\frac{X^5}{X^5}+X^4=X^0+X^4=1+X^4$

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.

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.

We will use the power rule for multiplying terms with identical bases:

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

Let's emphasize that this rule is valid only for terms with identical bases,

Here in the problem there are also terms with negative exponents, but this doesn't pose an issue regarding the use of the aforementioned power rule. In fact, this power rule is valid in all cases for numerical terms with different powers, including negative powers, rational number powers, and even irrational number powers, etc.

Let's return to the problem,

Let's note that there are two types of terms in the problem that differ from each other with different bases. First, for good order, we'll use the commutative law of multiplication to arrange the expression so that all terms with the same base are adjacent, let's get to work:

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

Then we'll apply the aforementioned power rule separately to each different type of term,

$c^{-1}\cdot c^3\cdot c^2\cdot d^6\cdot d^{-2}=c^{-1+3+2}\cdot d^{6+(-2)}=c^{-1+3+2}\cdot d^{6-2}=c^4\cdot d^4$

$$When we actually applied the mentioned rule separately - for terms with base $c$and for terms with base $d$ and combined the powers in the exponent when we grouped all terms with the same base together.

Therefore, the correct answer is B.