Applying Combined Exponents Rules: Monomial

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.

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.

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:

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

We use the zero exponent rule.

$X^0=1$We obtain

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

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 formula:

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

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

We use the formula

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

Therefore, we obtain:

$a^2b^28^2$

We use the formula:

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

Therefore, we obtain:

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

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.

We use the formula:

$a^0=1$

$\frac{9\times3}{8^0}=\frac{9\times3}{1}=9\times3$

We know that:

$9=3^2$

Therefore, we obtain:

$3^2\times3=3^2\times3^1$

We use the formula:

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

$3^2\times3^1=3^{2+1}=3^3$

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.

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.

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

We begin by using the power rule of negative exponents.

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

$$$4^{-1}=\frac{1}{4^1}=\frac{1}{4}$We can therefore deduce that the correct answer is option B.

We begin by using the power rule of negative exponents.

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

$$$2^{-5}=\frac{1}{2^5}=\frac{1}{32}$We can therefore deduce that the correct answer is option A.

We begin by using the power property for a negative exponent:

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

$(-7)^{-3}=\frac{1}{(-7)^3}$We then subsequently notice that each whole number inside the parentheses is raised to a negative power (that is, the number and its negative coefficient together) When using the previously mentioned power property: We are careful to take this into account,

We then continue by simplifying the expression in the denominator of the fraction, remembering the exponentiation property for the power of terms in multiplication:

$(a^m)^n=a^{m\cdot n}$We apply the resulting expression

$\frac{1}{(-7)^3}=\frac{1}{(-1\cdot7)^3}=\frac{1}{(-1)^3\cdot7^3}=\frac{1}{-1\cdot7^3}=\frac{1}{-7^3}=-\frac{1}{7^3}$

In summary we are able to deduce that the solution to the problem is as follows:

$(-7)^{-3}=\frac{1}{(-7)^3}=\frac{1}{-7^3}=-\frac{1}{7^3}$

$$$$Therefore, the correct answer is option B.

Using the rules of negative exponents: how to raise a number to a negative exponent:

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

$$$$$7^{-24}=\frac{1}{7^{24}}$Therefore, the correct answer is option D.

In order to solve the exercise, we use the negative exponent rule.

$a^{-n}=\frac{1}{a^n}$

We apply the rule to the given exercise:

$19^{-2}=\frac{1}{19^2}$

We can then continue and calculate the exponent.

$\frac{1}{19^2}=\frac{1}{361}$

We use the negative exponent rule.

$b^{-n}=\frac{1}{b^n}$

We apply it to the problem in the opposite sense.:

$$$$$\frac{1}{8^3}=8^{-3}$

Therefore, the correct answer is option A.