## Engage NY Eureka Math 8th Grade Module 1 Lesson 6 Answer Key

### Eureka Math Grade 8 Module 1 Lesson 6 Exercise Answer Key

Exercise 1.
Show that (C) is implied by equation (5) of Lesson 4 when m>0, and explain why (C) continues to hold even when
m=0.
Equation (5) says for any numbers x, y, (y≠0) and any positive integer n, the following holds: ($$\frac{x}{y}$$)n=$$\frac{x^{n}}{y^{n}}$$. So,
($$\frac{1}{x}$$)m = $$\frac{1^{m}}{x^{m}}$$ By ($$\frac{x}{y}$$)n= $$\frac{x^{n}}{y^{n}}$$ for positive integer n and nonzero y (5)
= $$\frac{1}{x^{m}}$$ Because 1m =1

If m= 0, then the left side is
($$\frac{1}{x}$$)m =($$\frac{1}{x}$$)0
=1 By definition of x0,
and the right side is
$$\frac{1}{x^{m}}$$ = $$\frac{1}{x^{0}}$$
= $$\frac{1}{1}$$ By definition of x0
=1.

Exercise 2.
Show that (B) is in fact a special case of (11) by rewriting it as (xm)-1 = x(-1m) for any whole number m, so that if b=m (where m is a whole number) and a=-1, (11) becomes (B).
(B) says x-m = $$\frac{1}{x^{m}}$$.
The left side of (B), x-m is equal to x(-1)m.
The right side of (B), $$\frac{1}{x^{m}}$$, is equal to (xm)-1 by the definition of (xm)-1 in Lesson 5.
Therefore, (B) says exactly that (xm)-1 = x(-1)m.

Exercise 3.
Show that (C) is a special case of (11) by rewriting (C) as (x-1)m = xm(-1) for any whole number m. Thus, (C) is the special case of (11) when b=-1 and a=m, where m is a whole number.
(C) says ($$\frac{1}{x}$$)m = $$\frac{1}{x^{m}}$$ for any whole number m.
The left side of (C) is equal to
($$\frac{1}{x}$$)m =(x-1)mBy definition of x-1 ,
and the right side of (C) is equal to
$$\frac{1}{x^{m}}$$ =x-m By definition of x-m, and the latter is equal to xm(-1). Therefore, (C) says (x-1)m) =xm(-1) for any whole number m.

Exercise 4.
Proof of Case (iii): Show that when a<0 and b≥0, (xb )a=xab is still valid. Let a=-c for some positive integer c. Show that the left and right sides of (xb )a=xab are equal.
The left side is
(xb )a=(xb )-c
= $$\frac{1}{\left(x^{b}\right)^{c}}$$ By x-m = $$\frac{1}{x^{m}}$$ for any whole number m (B)
= $$\frac{1}{x^{c b}}$$ . By (xm)n=xmn for all whole numbers m and n (A)
The right side is
xab = x(-c)b
=x-(cb)
= $$\frac{1}{x^{c b}}$$ . By x-m = $$\frac{1}{x^{m}}$$ for any whole number m (B)
So, the two sides are equal.

### Eureka Math Grade 8 Module 1 Lesson 6 Problem Set Answer Key

Question 1.
You sent a photo of you and your family on vacation to seven Facebook friends. If each of them sends it to five of their friends, and each of those friends sends it to five of their friends, and those friends send it to five more, how many people (not counting yourself) will see your photo? No friend received the photo twice. Express your answer in exponential notation.

The total number of people who viewed the photo is (50+51+52+53 )×7.

Question 2.
Show directly, without using (11), that (1.27-36 )85= 1.27-36∙85.
(1.27-36 )85= ($$\frac{1}{1.27^{36}}$$)85 By definition
= $$\frac{1}{\left(1.27^{36}\right)^{85}}$$By ($$\frac{1}{x}$$)m =$$\frac{1}{x^{m}}$$ for any whole number m (C)
= $$\frac{1}{1.27^{36 \cdot 85}}$$By (xm)n=xmn for whole numbers m and n (7)
= 1.27-36∙85 By x-m = $$\frac{1}{x^{m}}$$ for any whole number m (B)

Question 3.
Show directly that ($$\frac{2}{13}$$)-127∙($$\frac{2}{13}$$)-56=($$\frac{2}{13}$$)-183.

Question 4.
Prove for any nonzero number x, x-127∙x-56=x-183.
x-127∙x-56 =$$\frac{1}{x^{127}}$$ ∙$$\frac{1}{x^{56}}$$ By definition
= $$\frac{1}{x^{127} \cdot x^{56}}$$ By the product formula for complex fractions
=$$\frac{1}{x^{127+56}}$$ By xm ∙xn=xm+n for whole numbers m and n (6)
= $$\frac{1}{x^{183}}$$
= x-183 By x-m = $$\frac{1}{x^{m}}$$ for any whole number m (B)

Question 5.
Prove for any nonzero number x, x-m ∙x-n=x-m-n for positive integers m and n.
x-m∙x-n= $$\frac{1}{x^{m}}$$∙$$\frac{1}{x^{n}}$$ By definition
= $$\frac{1}{x^{m} \cdot x^{n}}$$ By the product formula for complex fractions
= $$\frac{1}{x^{m+n}}$$ By xm ∙xn=xm+n for whole numbers m and n (6)
=x-(m+n) By x-m = $$\frac{1}{x^{m}}$$ for any whole number m (B)
=x-m-n

Question 6.
Which of the preceding four problems did you find easiest to do? Explain.
Students will likely say that x-m ∙x-n=x-m-n (Problem 5) was the easiest problem to do. It requires the least amount of writing because the symbols are easier to write than decimal or fraction numbers.

Question 7.
Use the properties of exponents to write an equivalent expression that is a product of distinct primes, each raised to an integer power.

### Eureka Math Grade 8 Module 1 Lesson 6 Exit Ticket Answer Key

Question 1.
Show directly that for any nonzero integer x, x-5∙x-7 = x-12.
x-5∙x-7 =$$\frac{1}{x^{5}}$$ ∙$$\frac{1}{x^{7}}$$ By x-m = $$\frac{1}{x^{m}}$$ for any whole number m (B)
=$$\frac{1}{x^{5} \cdot x^{7}}$$ By the product formula for complex fractions
=$$\frac{1}{x^{5+7}}$$ By xm ∙xn=xm+n for whole numbers m and n (6)
=$$\frac{1}{x^{12}}$$
= x-12 By x-m = $$\frac{1}{x^{m}}$$ for any whole number m (B)
(x-2)-3 = $$\frac{1}{\left(x^{-2}\right)^{3}}$$ By x-m = $$\frac{1}{x^{m}}$$ for any whole number m (B)
= $$\frac{1}{x^{-(2 \cdot 3)}}$$ By case (ii) of (11)
=$$\frac{1}{x^{-6}}$$
= x6 By x-m = $$\frac{1}{x^{m}}$$ for any whole number m (B)