Examples with solutions for Applying Combined Exponents Rules: A power law

Exercise #1

Solve the exercise:

(a5)7= (a^5)^7=

Video Solution

Step-by-Step Solution

We use the formula:

(am)n=am×n (a^m)^n=a^{m\times n}

and therefore we obtain:

(a5)7=a5×7=a35 (a^5)^7=a^{5\times7}=a^{35}

Answer

a35 a^{35}

Exercise #2

1120=? 112^0=\text{?}

Video Solution

Step-by-Step Solution

We use the zero exponent rule.

X0=1 X^0=1 We obtain

1120=1 112^0=1 Therefore, the correct answer is option C.

Answer

1

Exercise #3

(35)4= (3^5)^4=

Video Solution

Step-by-Step Solution

To solve the exercise we use the power property:(an)m=anm (a^n)^m=a^{n\cdot m}

We use the property with our exercise and solve:

(35)4=35×4=320 (3^5)^4=3^{5\times4}=3^{20}

Answer

320 3^{20}

Exercise #4

(62)13= (6^2)^{13}=

Video Solution

Step-by-Step Solution

We use the formula:

(an)m=an×m (a^n)^m=a^{n\times m}

Therefore, we obtain:

62×13=626 6^{2\times13}=6^{26}

Answer

626 6^{26}

Exercise #5

2423= \frac{2^4}{2^3}=

Video Solution

Step-by-Step Solution

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:

bmbn=bmn \frac{b^m}{b^n}=b^{m-n} We apply it in the problem:

2423=243=21 \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:

b1=b b^1=b Therefore, in the problem we obtain:

21=2 2^1=2 Therefore, the correct answer is option a.

Answer

2 2

Exercise #6

3532= \frac{3^5}{3^2}=

Step-by-Step Solution

Using the quotient rule for exponents: aman=amn \frac{a^m}{a^n} = a^{m-n} .

Here, we have 3532=352 \frac{3^5}{3^2} = 3^{5-2}

Simplifying, we get 33 3^3

Answer

33 3^3

Exercise #7

5654= \frac{5^6}{5^4}=

Step-by-Step Solution

Using the quotient rule for exponents: aman=amn \frac{a^m}{a^n} = a^{m-n} .

Here, we have 5654=564 \frac{5^6}{5^4} = 5^{6-4} . Simplifying, we get 52 5^2 .

Answer

52 5^2

Exercise #8

9993= \frac{9^9}{9^3}=

Video Solution

Step-by-Step Solution

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:

bmbn=bmn \frac{b^m}{b^n}=b^{m-n} Let's apply it to the problem:

9993=993=96 \frac{9^9}{9^3}=9^{9-3}=9^6 Therefore, the correct answer is b.

Answer

96 9^6

Exercise #9

Simplify the expression:

a3a2b4b5= a^3\cdot a^2\cdot b^4\cdot b^5=

Video Solution

Step-by-Step Solution

In the exercise of multiplying powers, we will add up all the powers of the same product, in this case the terms a, b

We use the formula:

an×am=an+m a^n\times a^m=a^{n+m}

We are going to focus on the term a:

a3×a2=a3+2=a5 a^3\times a^2=a^{3+2}=a^5

We are going to focus on the term b:

b4×b5=b4+5=b9 b^4\times b^5=b^{4+5}=b^9

Therefore, the exercise that will be obtained after simplification is:

a5×b9 a^5\times b^9

Answer

a5b9 a^5\cdot b^9

Exercise #10

(2×8×7)2= (2\times8\times7)^2=

Video Solution

Step-by-Step Solution

We begin by using the power rule for parentheses:

(zt)n=zntn (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 then apply the above rule to the problem:

(287)2=228272 (2\cdot8\cdot7)^2=2^2\cdot8^2\cdot7^2

Therefore, the correct answer is option d.

Note:

From 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.).

Answer

228272 2^2\cdot8^2\cdot7^2

Exercise #11

(3×4×5)4= (3\times4\times5)^4=

Video Solution

Step-by-Step Solution

We use the power law for multiplication within parentheses:

(xy)n=xnyn (x\cdot y)^n=x^n\cdot y^n

We apply it to the problem:

(345)4=344454 (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.

Answer

34×44×54 3^4\times4^4\times5^4

Exercise #12

(42)3+(g3)4= (4^2)^3+(g^3)^4=

Video Solution

Step-by-Step Solution

We use the formula:

(am)n=am×n (a^m)^n=a^{m\times n}

(42)3+(g3)4=42×3+g3×4=46+g12 (4^2)^3+(g^3)^4=4^{2\times3}+g^{3\times4}=4^6+g^{12}

Answer

46+g12 4^6+g^{12}

Exercise #13

(4×7×3)2= (4\times7\times3)^2=

Video Solution

Step-by-Step Solution

We use the power law for multiplication within parentheses:

(xy)n=xnyn (x\cdot y)^n=x^n\cdot y^n

We apply it to the problem:

(473)2=427232 (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.

Answer

42×72×32 4^2\times7^2\times3^2

Exercise #14

(5x3)3= (5\cdot x\cdot3)^3=

Video Solution

Step-by-Step Solution

We use the formula:

(a×b)n=anbn (a\times b)^n=a^nb^n

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

(15x)3=(15×x)3 (15x)^3=(15\times x)^3

153x3 15^3x^3

Answer

153x3 15^3\cdot x^3

Exercise #15

(7463)4=? (7\cdot4\cdot6\cdot3)^4= \text{?}

Video Solution

Step-by-Step Solution

We use the power property for an exponent that is applied to a set parentheses in which the terms are multiplied:

(xy)n=xnyn (x\cdot y)^n=x^n\cdot y^n

We apply the law in the problem:

(7463)4=74446434 (7\cdot4\cdot6\cdot3)^4=7^4\cdot4^4\cdot6^4\cdot3^4

When we apply the exponent to a parentheses with multiplication, we apply the exponent to each term of the multiplication separately, and we keep the multiplication between them.

Therefore, the correct answer is option a.

Answer

74446434 7^4\cdot4^4\cdot6^4\cdot3^4

Exercise #16

(8×9×5×3)2= (8\times9\times5\times3)^{-2}=

Video Solution

Step-by-Step Solution

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

(zt)n=zntn (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:

(8953)2=82925232 (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.).

Answer

82×92×52×32 8^{-2}\times9^{-2}\times5^{-2}\times3^{-2}

Exercise #17

(a4)6= (a^4)^6=

Video Solution

Step-by-Step Solution

We use the formula

(am)n=am×n (a^m)^n=a^{m\times n}

Therefore, we obtain:

a4×6=a24 a^{4\times6}=a^{24}

Answer

a24 a^{24}

Exercise #18

(a56y)5= (a\cdot5\cdot6\cdot y)^5=

Video Solution

Step-by-Step Solution

We use the formula:

(a×b)x=axbx (a\times b)^x=a^xb^x

Therefore, we obtain:

(a×5×6×y)5=(a×30×y)5 (a\times5\times6\times y)^5=(a\times30\times y)^5

a5305y5 a^530^5y^5

Answer

a5305y5 a^5\cdot30^5\cdot y^5

Exercise #19

(ab8)2= (a\cdot b\cdot8)^2=

Video Solution

Step-by-Step Solution

We use the formula

(a×b)x=axbx (a\times b)^x=a^xb^x

Therefore, we obtain:

a2b282 a^2b^28^2

Answer

a2b282 a^2\cdot b^2\cdot8^2

Exercise #20

ababa2 a\cdot b\cdot a\cdot b\cdot a^2

Video Solution

Step-by-Step Solution

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

aman=am+n a^m\cdot a^n=a^{m+n} It is important to note 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:

ababa2=aaa2bb a\cdot b\operatorname{\cdot}a\operatorname{\cdot}b\operatorname{\cdot}a^2=a\cdot a\cdot a^2\cdot b\cdot b Next, we apply the power property for each type of term separately,

aaa2bb=a1+1+2b1+1=a4b2 a\cdot a\cdot a^2\cdot b\cdot b=a^{1+1+2}\cdot b^{1+1}=a^4\cdot b^2

We apply the power property separately - for the terms whose bases area a and then for the terms whose bases areb b and we add the exponents and simplify the terms.

Therefore, the correct answer is option c.

Note:

We use the fact that:

a=a1 a=a^1 and the same for b b .

Answer

a4b2 a^4\cdot b^2