Applying Combined Exponents Rules: Applying the formula

$112^0=\text{?}$

We use the zero exponent rule.

$X^0=1$We obtain

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

1

$\frac{2^4}{2^3}=$

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.

$2$

$(9\times2\times5)^3=$

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.

$9^3\times2^3\times5^3$

$\frac{81}{3^2}=$

First, we recognize that 81 is a power of the number 3, which means that:

$3^4=81$We replace in the problem:

$\frac{81}{3^2}=\frac{3^4}{3^2}$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{3^4}{3^2}=3^{4-2}=3^2$$$Therefore, the correct answer is option b.

$3^2$$$

$(3^5)^4=$

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:

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

$3^{20}$

$(6^2)^{13}=$

We use the formula:

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

Therefore, we obtain:

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

$6^{26}$

$\frac{9^9}{9^3}=$

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.

$9^6$

$(\frac{2}{6})^3=$

We use the formula:

$(\frac{a}{b})^n=\frac{a^n}{b^n}$

$(\frac{2}{6})^3=(\frac{2}{2\times3})^3$

We simplify:

$(\frac{1}{3})^3=\frac{1^3}{3^3}$

$\frac{1\times1\times1}{3\times3\times3}=\frac{1}{27}$

$\frac{1}{27}$

$(\frac{1}{4})^{-1}$

We use the power property for a negative exponent:

$a^{-n}=\frac{1}{a^n}$We will write the fraction in parentheses as a negative power with the help of the previously mentioned power:

$\frac{1}{4}=\frac{1}{4^1}=4^{-1}$We return to the problem, where we obtained:

$\big(\frac{1}{4}\big)^{-1}=(4^{-1})^{-1}$We continue and use the power property of an exponent raised to another exponent:

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

$(4^{-1})^{-1}=4^{-1\cdot-1}=4^1=4$Therefore, the correct answer is option d.

$4$

$5^{-2}$

We use the property of powers of a negative exponent:

$a^{-n}=\frac{1}{a^n}$We apply it to the problem:

$5^{-2}=\frac{1}{5^2}=\frac{1}{25}$

Therefore, the correct answer is option d.

$\frac{1}{25}$

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

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

$\frac{4^4}{7^8}$

$(3\times4\times5)^4=$

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.

$3^44^45^4$

$(3\times2\times4\times6)^{-4}=$

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.).

$3^{-4}\times2^{-4}\times4^{-4}\times6^{-4}$

$(4\times7\times3)^2=$

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.

$4^2\times7^2\times3^2$

$(4^2)^3+(g^3)^4=$

We use the formula:

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

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

$4^6+g^{12}$

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

We use the formula:

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

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

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

$15^3x^3$

$15^3\cdot x^3$

$[(\frac{1}{7})^{-1}]^4=$

We use the power property of a negative exponent:

$a^{-n}=\frac{1}{a^n}$We will rewrite the fraction in parentheses as a negative power:

$\frac{1}{7}=7^{-1}$Let's return to the problem, where we had:

$\bigg( \big( \frac{1}{7}\big)^{-1}\bigg)^4=\big((7^{-1})^{-1} \big)^4$We continue and use the power property of an exponent raised to another exponent:

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

$\big((7^{-1})^{-1} \big)^4 =(7^{-1\cdot-1})^4=(7^1)^4=7^{1\cdot4}=7^4$Therefore, the correct answer is option c

$7^4$

Solve the exercise:

$(a^5)^7=$

We use the formula:

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

and therefore we obtain:

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

$a^{35}$

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

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:

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

$a^530^5y^5$

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