Applying Combined Exponents Rules: Calculating powers with negative exponents

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 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.).

Let's handle each expression in the problem separately:

a. We'll start with the leftmost expression, first calculating the result of the multiplication in parentheses, and then use the power rule for power to a power:

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

Let's apply this to the problem for the first expression from the left:

$((7\cdot3)^2)^6=(21^2)^6=21^{2\cdot6}=21^{12}$

where in the final step we calculated the result of multiplication in the power expression,

We're done with this expression, let's move on to the next expression from the left.

b. Let's continue with the second expression from the left, using the power rule for power to a power that we mentioned above and apply it separately to each factor in this expression:

$(3^{-1})^3\cdot(2^3)^4=3^{-1\cdot3}\cdot2^{3\cdot4}=3^{-3}\cdot2^{12}$

Note that the multiplication factors we got have different bases, so we cannot further simplify this expression,

Therefore, let's combine parts a and b above in the result of the original problem:

$((7\cdot3)^2)^6+(3^{-1})^3\cdot(2^3)^4=21^{12}+3^{-3}\cdot2^{12}$

Therefore, the correct answer is answer d.

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 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.

We use the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$It should be noted 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:

$E^6\cdot F^{-4}\cdot E^0\cdot F^7\cdot E=E^6\cdot E^0\cdot E\cdot F^{-4}\cdot F^7$Next, we apply the power property for each type of term separately,

$E^6\cdot E^0\cdot E\cdot F^{-4}\cdot F^7=E^{6+0+1}\cdot F^{-4+7}=E^7\cdot F^3$

$$We apply the power property separately - for the terms whose bases are$E$and for the terms whose bases are$F$and we add the exponents and simplify the terms with the same base.

The correct answer is then option d.

Note:

We use the fact that:

$E=E^1$.

First, let's convert the decimal fraction in the problem to a simple fraction:

$0.25=\frac{25}{100}=\frac{1}{4}$

where we remembered that 0.25 is 25 hundredths, meaning:

$25\cdot\frac{1}{100}=\frac{25}{100}$

If so, let's write the problem:

$(0.25)^{-2}=\big(\frac{1}{4}\big)^{-2}=\text{?}$

Now we'll use the negative exponent law:

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

and deal with the fraction expression inside the parentheses:

$\big(\frac{1}{4}\big)^{-2}=(4^{-1})^{-2}$

when we applied the above exponent law to the expression inside the parentheses,

Next, we'll recall the power of a power law:

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

and we'll apply this law to the expression we got in the last step:

$(4^{-1})^{-2}=4^{(-1)\cdot(-2)}=4^2=16$

where in the first step we carefully applied the above law and used parentheses in the exponent to perform the multiplication between the powers, then we simplified the resulting expression, and finally calculated the numerical result from the last step.

Let's summarize the solution steps:

$(0.25)^{-2}=\big(\frac{1}{4}\big)^{-2}=4^{(-1)\cdot(-2)}=16$

Therefore, the correct answer is answer B.

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.

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 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 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'll use the law of exponents for negative exponents, but in the opposite direction:

$\frac{1}{a^n} =a^{-n}$Let's apply this law to the problem:

$4^5-4^6\cdot\frac{1}{4}= 4^5-4^6\cdot4^{-1}$When we apply the above law to the second term from the left in the sum, and convert the fraction to a term with a negative exponent,

Next, we'll use the law of exponents for multiplying terms with identical bases:

$a^m\cdot a^n=a^{m+n}$Let's apply this law to the expression we got in the last step:

$4^5-4^6\cdot4^{-1} =4^5-4^{6+(-1)}=4^5-4^{6-1}=4^5-4^{5}=0$When we apply the above law of exponents to the second term from the left in the expression we got in the last step, then we'll simplify the resulting expression,

Let's summarize the solution steps:

$4^5-4^6\cdot\frac{1}{4}= 4^5-4^6\cdot4^{-1} =4^5-4^{5}=0$

We got that the answer is 0,

Therefore the correct answer is answer A.

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'll use the power rule for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$and we'll simplify the second term on the left in the equation using it:
$5^3+5^{-3}\cdot5^3=5^3+5^{-3+3}=5^3+5^0=5^3+1$ where in the first stage we applied the mentioned rule to the second term on the left, then we simplified the expression with the exponent, and in the final stage we used the fact that any number raised to the power of 0 equals 1,

We didn't touch the first term of course since it was already simplified,

Therefore the correct answer is answer C.

First let's recall the negative exponent rule:

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

$(-5)^{-3}=\frac{1}{(-5)^3}$Next let's recall the power rule for expressions in parentheses:

$(x\cdot y)^n=x^n\cdot y^n$And we'll apply it to the denominator of the expression we received:

$\frac{1}{(-5)^3}=\frac{1}{(-1\cdot5)^3}=\frac{1}{(-1)^3\cdot5^3}=\frac{1}{-1\cdot5^3}=-\frac{1}{5^3}=-\frac{1}{125}$In the first step, we expressed the negative number inside the parentheses in the denominator as a multiplication between a positive number and negative one, and then we used the power rule for expressions in parentheses to expand the parentheses, and then we simplified the expression.

Let's summarize the solution to the problem:

$(-5)^{-3}=\frac{1}{(-5)^3} =\frac{1}{-5^3}=-\frac{1}{125}$

Therefore, the correct answer is answer 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 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.

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 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 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.