## Engage NY Eureka Math 8th Grade Module 1 Lesson 3 Answer Key

### Eureka Math Grade 8 Module 1 Lesson 3 Example Answer Key

Examples 1–2

Work through Examples 1 and 2 in the same manner. (Supplement with additional examples if needed.) Have students calculate the resulting exponent; however, emphasis should be placed on the step leading to the resulting exponent, which is the product of the exponents.

Example 1.

(7^{2} )^{6}=

Answer:

Example 2.

(1.3)^{3} )^{10}=

Answer:

(1.3×1.3×1.3)^{10}

### Eureka Math Grade 8 Module 1 Lesson 3 Exercise Answer Key

Exercise 1.

(15^{3})^{9}=

Answer:

(15)^{9×3}

Exercise 2.

((-2)^{5} )^{8}=

Answer:

(-2)^{8×5}

Exercise 3.

(3.4^{17})^{4}=

Answer:

3.4^{4×17}

Exercise 4.

Let s be a number.

Answer:

(s^{17} )^{4}=

Answer:

s^{4×17}

Exercise 5.

Sarah wrote (3^{5} )^{7}=3^{12}. Correct her mistake. Write an exponential expression using a base of 3 and exponents of 5, 7, and 12 that would make her answer correct.

Answer:

Correct way: (3^{5} )^{7}=3^{35}; Rewritten Problem: 3^{5}×3^{7}=3^{5+7}=3^{12}.

Exercise 6.

A number y satisfies y^{24}-256=0. What equation does the number x=y^{4} satisfy?

Answer:

Since x=y^{4}, then (x)^{6}=(y^{4} )^{6}. Therefore, x=y^{4} would satisfy the equation x^{6}-256=0.

Exercises 7–13 (10 minutes)

Have students complete Exercises 17–12 independently and then check their answers.

Exercise 7.

(11×4)^{9}=

Answer:

11^{9×1}×4^{9×1}

Exercise 8.

(3^{2}×7^{4} )^{5}=

Ans:

3^{5×2}×7^{5×4}

Exercise 9.

Let a, b, and c be numbers.

(3^{2} a^{4} )^{5}=

Answer:

3^{5×2} a^{5×4}

Exercise 10.

Let x be a number.

(5x)^{7}=

Ans:

5^{7×1} ∙x^{7×1}

Exercise 11.

Let x and y be numbers.

(5xy^{2} )^{7}=

Ans:

5^{7×1} ∙x^{7×1}∙y^{7×2}

Exercise 12.

Let a, b, and c be numbers.

(a^{2} bc^{3} )^{4}=

Ans:

a^{4×2} ∙b^{4×1}∙c^{4×3}

Exercise 13.

Let x and y be numbers, y≠0, and let n be a positive integer. How is (\(\frac{x}{y}\))^{n} related to x^{n} and y^{n}?

Answer:

(\(\frac{x}{y}\))^{n}=\(\frac{x^{n}}{y^{n}}\)

Because

### Eureka Math Grade 8 Module 1 Lesson 3 Problem Set Answer Key

Question 1.

Show (prove) in detail why (2∙3∙7)^{4}=2^{4} 3^{4} 7^{4}.

Answer:

(2∙3∙7)^{4}=(2∙3∙7)(2∙3∙7)(2∙3∙7)(2∙3∙7)

=(2∙2∙2∙2)(3∙3∙3∙3)(7∙7∙7∙7)

By repeated use of the commutative and associative properties

=2^{4} 3^{4} 7^{4} By definition

Question 2.

Show (prove) in detail why (xyz)^{4}=x^{4} y^{4} z^{4} for any numbers x,y,z.

Answer:

The left side of the equation (xyz)^{4} means (xyz) (xyz) (xyz) (xyz). Using the commutative and associative properties of multiplication, we can write (xyz) (xyz) (xyz) (xyz) as (xxxx)(yyyy) (zzzz), which in turn can be written as x^{4} y^{4} z^{4}, which is what the right side of the equation states.

Question 3.

Show (prove) in detail why (xyz)^{n}=x^{n} y^{n} z^{n} for any numbers x, y, and z and for any positive integer n.

Ans:

Beginning with the left side of the equation, (xyz)^{n} means . Using the commutative and associative properties of multiplication, can be rewritten as and, finally, x^{n} y^{n} z^{n}, which is what the right side of the equation states. We can also prove this equality by a different method, as follows. Beginning with the right side x^{n} y^{n} z^{n} means which by the commutative property of multiplication can be rewritten as . Using exponential notation, can be rewritten as (xyz)^{n}, which is what the left side of the equation states.

### Eureka Math Grade 8 Module 1 Lesson 3 Exit Ticket Answer Key

Write each expression as a base raised to a power or as the product of bases raised to powers that is equivalent to the given expression.

Question 1.

(9^{3} )^{6}=

Ans:

(9^{3} )^{6}=9^{6×3}=9^{18}

Question 2.

(113^{2}×37×51^{4} )^{3}=

Ans:

(113^{2}×37×51^{4} )^{3}=((113^{2}×37)×51^{4} )^{3} By associative law

= (113^{2}×37)^{3}×(51^{4} )^{3} Because (xy)^{n}=x^{n} y^{n} for all numbers x, y

= (113^{2})^{3}×37^{3}×(51^{4})^{3} Because (xy)^{n}=x^{n} y^{n} for all numbers x, y

= 113^{6}×37^{3}×51^{12} Because (x^{m})^{n}=x^{mn} for all numbers x

Question 3.

Let x,y,z be numbers. (x^{2} yz^{4} )^{3}=

Answer:

(x^{2}yz^{4} )^{3}=((x^{2}×y)×z^{4} )^{3} By associative law

= (x^{2}×y)^{3}×(z^{4} )^{3} Because (xy)^{n}=x^{n} y^{n} for all numbers x, y

=(x^{2} )^{3}×y^{3}×(z^{4} )^{3} Because (xy)^{n}=x^{n} y^{n} for all numbers x, y

=x^{6}×y^{3}×z^{12} Because (x^{m})^{n}=x^{mn} for all numbers x

= x^{6} y^{3} z^{12}

Question 4.

Let x,y,z be numbers and let m,n,p,q be positive integers. (x^{m} y^{n} z^{p})^{q}=

Ans:

(x^{m} y^{n} z^{p} )^{q}=((x^{m}×y^{n} )×z^{p})^{q} By associative law

=(x^{m}×y^{n} )^{q}×(z^{p} )^{q} Because (xy)^{n}=x^{n} y^{n} for all numbers x, y

=(x^{m} )^{q}×(y^{n} )^{q}×(z^{p})^{q} Because (xy)^{n}=x^{n} y^{n} for all numbers x, y

= x^{mp}×y^{nq}×z^{pq} Because (x^{m} )^{n}=x^{mn} for all numbers x

=x^{mq}y^{nq}z^{pq}

Question 5.

\(\frac{4^{8}}{5^{8}}\) =

Answer:

\(\frac{4^{8}}{5^{8}}\) = \(\left(\frac{4}{5}\right)^{8}\)