Applying Combined Exponents Rules: Calculating powers with negative exponents

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.).

We begin by applying the power rule to the products within the parentheses:

$(z\cdot t)^n=z^n\cdot t^n$That is, the power applied to a product within parentheses is applied to each of the terms when the parentheses are opened,

We apply the rule to the given problem:

$(8\cdot9\cdot5\cdot3)^{-2}=8^{-2}\cdot9^{-2}\cdot5^{-2}\cdot3^{-2}$Therefore, the correct answer is option c.

Note:

Whilst it could be understood that the above power rule applies only to two terms of the product within parentheses, in reality, it is also valid for the power over a multiplication of multiple terms within parentheses, as was seen in the above problem.

A good exercise is to demonstrate that if the previous property is valid for a power over a product of two terms within parentheses (as formulated above), then it is also valid for a power over several terms of the product within parentheses (for example - three terms, etc.).

We begin by using the negative exponent rule:

$b^{-n}=\frac{1}{b^n}$We apply it to the given expression and obtain the following:

$(3a)^{-2}=\frac{1}{(3a)^2}$We then use the power rule for parentheses:

$(x\cdot y)^n=x^n\cdot y^n$We apply it to the denominator of the expression and obtain the following:

$\frac{1}{(3a)^2}=\frac{1}{3^2a^2}=\frac{1}{9a^2}$Let's summarize the solution to the problem:

$(3a)^{-2}=\frac{1}{(3a)^2} =\frac{1}{9a^2}$

Therefore, the correct answer is option A.

We begin by using the rule for multiplying exponents. (the multiplication between terms with identical bases):

$a^m\cdot a^n=a^{m+n}$We then apply it to the problem:

$7^5\cdot7^{-6}=7^{5+(-6)}=7^{5-6}=7^{-1}$When in a first stage we begin by applying the aforementioned rule and then continue on to simplify the expression in the exponent,

Next, we use the negative exponent rule:

$a^{-n}=\frac{1}{a^n}$We apply it to the expression obtained in the previous step:

$7^{-1}=\frac{1}{7^1}=\frac{1}{7}$We then summarise the solution to the problem: $7^5\cdot7^{-6}=7^{-1}=\frac{1}{7}$Therefore, the correct answer is option B.

We begin by using the power rule of exponents; for the multiplication of terms with identical bases:

$a^m\cdot a^n=a^{m+n}$We apply it to the given problem:

$12^4\cdot12^{-6}=12^{4+(-6)}=12^{4-6}=12^{-2}$When in a first stage we apply the aforementioned rule and then simplify the subsequent expression in the exponent,

Next, we use the negative exponent rule:

$a^{-n}=\frac{1}{a^n}$We apply it to the expression that we obtained in the previous step:

$12^{-2}=\frac{1}{12^2}=\frac{1}{144}$Lastly we summarise the solution to the problem: $12^4\cdot12^{-6}=12^{-2} =\frac{1}{144}$Therefore, the correct answer is option A.

Use the power property for a power in parentheses where there is a multiplication of its terms:

$(x\cdot y)^n=x^n\cdot y^n$We apply this law to the problem expression:

$(5\cdot12\cdot4\cdot6)^{a+3bx}=5^{a+3bx}12^{a+3bx}4^{a+3bx}6^{a+3bx}$When we apply a power to parentheses where its terms are multiplied, we do it separately and keep the multiplication.

Therefore, the correct answer is option d.

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.

We begin by using the negative exponent rule.

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

$$$$$a^{-4}=\frac{1}{a^4}$Therefore, the correct answer is option B.

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.

To begin with we deal with the expression in the denominator of the fraction. Making note of the power rule for exponents (raising an exponent to another exponent):

$(a^m)^n=a^{m\cdot n}$We obtain the following:

$(-2)^7=(-1\cdot2)^7=(-1)^7\cdot2^7=-1\cdot2^7=-2^7$

We then return to the initial problem and apply the above information:

$\frac{1}{(-2)^7}=\frac{1}{-2^7}=\frac{1}{-1}\cdot\frac{1}{2^7}=-\frac{1}{2^7}$$$

In the last step we remember that:

$\frac{1}{-1}=-1$

Next, we remember the Negative Exponent rule ( raising exponents to a negative power)

$a^{-n}=\frac{1}{a^n}$We apply it to the expression we obtained in the last step:

$-\frac{1}{2^7}=-2^{-7}$Let's summarize the steps of the solution:

$\frac{1}{(-2)^7}=-\frac{1}{2^7} = -2^{-7}$

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

We use the power property for a negative exponent:

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

$\frac{1}{2^9}=2^{-9}$

Therefore, the correct answer is option A.

To begin with, we must remind ourselves of the Negative Exponent rule:

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

$\frac{1}{12^3}=12^{-3}$Therefore, the correct answer is option A.

We must first remind ourselves of the negative exponent rule:

$a^{-n}=\frac{1}{a^n}$When applied to given the expression we obtain the following:

$7^{-4}=\frac{1}{7^4}=\frac{1}{2401}$

Therefore, the correct answer is option C.

We use the formula:

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

Therefore, we obtain:

$(\frac{3}{2})^4$

We use the formula:

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

Therefore, we obtain:

$\frac{3^4}{2^4}=\frac{3\times3\times3\times3}{2\times2\times2\times2}=\frac{81}{16}$

First, let's recall the negative exponent rule:

$b^{-n}=\frac{1}{b^n}$We'll apply it to the expression we received:

$10^{-5}=\frac{1}{10^5}=\frac{1}{100000}=0.00001$In the final steps, we performed the exponentiation in the numerator and then wrote the answer as a decimal.

Therefore, the correct answer is option A.

We use the formula:

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

We decompose the fraction inside of the parentheses:

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

We obtain:

$7^4\times8^3\times\frac{1^4}{7^4}$

We simplify the powers: $7^4$

We obtain:

$8^3\times1^4$

Remember that the number 1 in any power is equal to 1, thus we obtain:

$8^3\times1=8^3$

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.