Applying Combined Exponents Rules: Applying the formula

We use the zero exponent rule.

$X^0=1$We obtain

$112^0=1$Therefore, the correct answer is option C.

To solve the exercise we use the power property:$(a^n)^m=a^{n\cdot m}$

We use the property with our exercise and solve:

$(3^5)^4=3^{5\times4}=3^{20}$

We use the formula:

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

Therefore, we obtain:

$6^{2\times13}=6^{26}$

Let's keep in mind that the numerator and denominator of the fraction have terms with the same base, therefore we use the property of powers to divide between terms with the same base:

$\frac{b^m}{b^n}=b^{m-n}$We apply it in the problem:

$\frac{2^4}{2^3}=2^{4-3}=2^1$Remember that any number raised to the 1st power is equal to the number itself, meaning that:

$b^1=b$Therefore, in the problem we obtain:

$2^1=2$Therefore, the correct answer is option a.

Using the quotient rule for exponents: $\frac{a^m}{a^n} = a^{m-n}$.

Here, we have $\frac{3^5}{3^2} = 3^{5-2}$

Simplifying, we get $3^3$

Using the quotient rule for exponents: $\frac{a^m}{a^n} = a^{m-n}$.

Here, we have $\frac{5^6}{5^4} = 5^{6-4}$. Simplifying, we get $5^2$.

Note that in the fraction and its denominator, there are terms with the same base, so we will use the law of exponents for division between terms with the same base:

$\frac{b^m}{b^n}=b^{m-n}$ Let's apply it to the problem:

$\frac{9^9}{9^3}=9^{9-3}=9^6$$$Therefore, the correct answer is b.

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:

$a^5\times b^9$

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 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 formula:

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

$(4^2)^3+(g^3)^4=4^{2\times3}+g^{3\times4}=4^6+g^{12}$

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 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:

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

Therefore, we obtain:

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

$(\frac{4^2}{7^4})^2=\frac{4^{2\times2}}{7^{4\times2}}=\frac{4^4}{7^8}$

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.

We use the formula:

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

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

We use the formula:

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

$(2^2)^3+(3^3)^4+(9^2)^6=2^{2\times3}+3^{3\times4}+9^{2\times6}=2^6+3^{12}+9^{12}$

To solve the problem$(2\times7\times5)^3$, we need to apply the Power of a Product rule of exponents. This rule states that when you raise a product to a power, it's the same as raising each factor to that power. In mathematical terms, if you have $(abc)^n$, it is equivalent to $a^n \times b^n \times c^n$.

Let's apply this rule step by step:

Our original expression is $(2 \times 7 \times 5)^3$.

We identify the factors inside the parentheses as $2$, $7$, and $5$.

According to the Power of a Product rule, we can distribute the exponent$3$ to each factor:

First, raise $2$ to the power of $3$ to get $2^3$.

Then, raise $7$ to the power of $3$ to get $7^3$.

Finally, raise $5$ to the power of $3$ to get $5^3$.

Therefore, the expression $(2 \times 7 \times 5)^3$ simplifies to $2^3 \times 7^3 \times 5^3$.

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.).