Applying Combined Exponents Rules: Calculating powers with negative exponents

$10^{-5}=?$

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.

$0.00001$

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

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.).

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

$7^{-4}=\text{?}$

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.

$\frac{1}{2401}$

Simplify:

$(5\cdot12\cdot4\cdot6)^{a+3bx}$

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.

$5^{a+3bx}12^{a+3bx}4^{a+3bx}6^{a+3bx}$

$7^5\cdot7^{-6}=\text{?}$

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.

$$$\frac{1}{7}$

$(8\times9\times5\times3)^{-2}=$

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.).

$8^{-2}\times9^{-2}\times5^{-2}\times3^{-2}$

$12^4\cdot12^{-6}=\text{?}$

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.

$\frac{1}{144}$

$(\frac{2}{3})^{-4}=\text{?}$

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:

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

$\frac{81}{16}$

$(3a)^{-2}=\text{?}$

$a\ne0$

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.

$\frac{1}{9a^2}$

$10\cdot10^2\cdot10^{-4}\cdot10^{10}=$

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

$a^m\cdot a^n=a^{m+n}$Keep in mind that this property is also valid for several terms in the multiplication and not just for two, for example for the multiplication of three terms with the same base we obtain:

$a^m\cdot a^n\cdot a^k=a^{m+n}\cdot a^k=a^{m+n+k}$When we use the mentioned power property twice, we could also perform the same calculation for four terms of the multiplication of five, etc.,

Let's return to the problem:

First keep in mind that:

$10=10^1$Keep in mind that all the terms of the multiplication have the same base, so we will use the previous property:

$10^1\cdot10^2\cdot10^{-4}\cdot10^{10}=10^{1+2-4+10}=10^9$

$$Therefore, the correct answer is option c.

$$$10^9$

$5^{-3}\cdot5^0\cdot5^2\cdot5^5=$

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

$a^m\cdot a^n=a^{m+n}$Keep in mind that this property is also valid for several terms in the multiplication and not just for two, for example for the multiplication of three terms with the same base we obtain:

$a^m\cdot a^n\cdot a^k=a^{m+n}\cdot a^k=a^{m+n+k}$When we use the mentioned power property twice, we could also perform the same calculation for four terms of the multiplication of five, etc.,

Let's return to the problem:

$$$$Keep in mind that all the terms of the multiplication have the same base, so we will use the previous property:

$5^{-3}\cdot5^0\cdot5^2\cdot5^5=5^{-3+0+2+5}=5^4$Therefore, the correct answer is option c.

Note:

Keep in mind that $5^0=1$

$5^4$

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

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:

$E=E^1$.

$E^7\cdot F^3$

$(0.25)^{-2}=\text{?}$

$16$

$(-5)^{-3}=\text{?}$

$-\frac{1}{125}$

$\frac{1}{4^{-3}}=?$

$64$

$9^{300}\cdot\frac{1}{9^{-252}}\cdot9^{-549}=\text{?}$

$\frac{1}{9^{-3}}$

$\frac{1}{-3}\cdot3^{-4}\cdot5^3=\text{?}$

$-\frac{5^3}{3^5}$

$45^{-80}\cdot\frac{1}{45^{-81}}\cdot49\cdot7^{-5}=\text{?}$

$\frac{45}{7^3}$

Solve the following:

$\frac{y^3}{y^6}\times\frac{y^4}{y^{-2}}\times\frac{y^{12}}{y^7}=$

$y^8$

$4^{2x}\cdot\frac{1}{4}\cdot4^{-2}=\text{?}$

$\frac{1}{4^{3-2x}}$