# Eureka Math Algebra 2 Module 3 Lesson 8 Answer Key

## Engage NY Eureka Math Algebra 2 Module 3 Lesson 8 Answer Key

### Eureka Math Algebra 2 Module 3 Lesson 8 Opening Exercise Answer Key

Opening Exercise:

a. Evaluate each expression. The first two have been completed for you.

i. WhatPower2(8) = 3
3, because 23 = 8

ii. WhatPower3(9) = 2
2, because 32 = 9

iii. WhatPower6(36) = ______
2, because 62 = 36

iv. WhatPower2(32) = _____
5, because 25 = 32

v. WhatPower10(1000) = _____
3, because 103 = 1000

vi. WhatPower10(1000000) = _____
6, because 106 = 1,000,000

vii. WhatPower100(1000000) = _____
3, because 1003 = 1,000,000

viii. WhatPower4(64) = _____
3, because 43 = 64

ix. WhatPower2(64) = _____
6, because 26 = 64

x. WhatPower9(3) = _____
$$\frac{1}{2}$$ because 9$$\frac{1}{2}$$ = 3

xi. WhatPower5(âˆš5) =_____
$$\frac{1}{2}$$ because 5$$\frac{1}{2}$$ = âˆš5

xii. WhatPower$$\frac{1}{2}$$($$\frac{1}{8}$$) = ___
3, because $$\left(\frac{1}{2}\right)^{3}$$ = $$\frac{1}{2}$$

xiii. WhatPower42(1) = ____
0, because 420 = 1

xiv. WhatPower100(0.01) = ___
-1, because 100-1 = 0.01

xv. WhatPower2($$\frac{1}{4}$$)
-2, because 2-2 = $$\frac{1}{4}$$

xvi. WhatPower$$\frac{1}{4}$$(2) = _____
$$\frac{1}{4}$$ because ($$\frac{1}{4}$$)$$-\frac{1}{2}$$ = 4$$\frac{1}{2}$$ = 2

b. With your group members, write a definition for the function WhatPowerb, where b is a number.
The value of WhatPowerb is the number you need to raise b to in order to get x. That is, if bL = x, then L = WhatPowerb(x).

### Eureka Math Algebra 2 Module 3 Lesson 8 Exercise Answer Key

Exercises 1 – 9:

Exercise 1.
WhatPower7(49)
WhatPower7(49) = 2 because 72 = 49

Exercise 2.
WhatPower0(7)
WhatPower0(7) does not make sense because there is no power of 0 that will produce 7.

Exercise 3.
WhatPower5(1)
WhatPower5(1) = 0 because 50 = 1.

Exercise 4.
WhatPower1(5)
WhatPower1 (5) does not exist because for any exponent L, 1L = 1, so there is no power of 1 that will produce 5.

Exercise 5.
WhatPower-2(16)
WhatPower-2(16) = 4 because (-2)4 = 16.

Exercise 6.
WhatPower-2(32)
WhatPower-2(32) does not make sense because there is no power of -2 that will produce 32.

Exercise 7.
WhatPower$$\frac{1}{3}$$(9)
WhatPower$$\frac{1}{3}$$(9) = -2 because ($$\frac{1}{3}$$)-2 = 9

Exercise 8.
WhatPower$$-\frac{1}{3}$$(27)
WhatPower$$-\frac{1}{3}$$(27) does not make sense because there is no power of –$$\frac{1}{3}$$ that will produce 27.

Exercise 9.
Describe the allowable values of b in the expression WhatPowerb(x). When can we define a function f(x) = WhatPowerb(x)? Explain how you know.
1f b = 0 or b = 1, then the expression WhatPowerb(x) does not make sense. If b < 0, then the expression WhatPowerb(x) makes sense for some values of x but not for others, so we cannot define a function f(x) = WhatPowerb(x) if b < 0. Thus, we can define the function f(x) = WhatPowerb(x) if b > 0 and b â‰  1.

Examples:

Example 1.
log2(8) = 3
3, because 23 = 8

Example 2.
log3(9) = 2
2, because 32 = 9

Example 3.
log6(36) = ___
2, because 62 = 36

Example 4.
log2(32) = ___
5, because 25 = 32

Example 5.
log10(1000) = ___
3, because 103 = 1000

Example 6.
log42(1) = ___
0, because 420 = 1

Example 7.
log100(0.01) = ___
-1, because 100-1 = 0.01

Example 8.
log2($$\frac{1}{4}$$) = ___
-2, because 2-2 = $$\frac{1}{4}$$

Exercise 10.

10.
Compute the value of each logarithm. Verify your answers using an exponential statement.

a. log2(32) = ___
log2(32) = 5, because 25 = 32

b. log3(81) = ___
log3(81) = 4, because 34 = 81

c. log9(81) = ___
log9(81) = 2, because 92 = 81

d. log5(625) = ___
log5(625) = 4, because 54 = 625

e. log10(1,000,000,000) = ___
log10(1,000,000,000) = 8, because 109 = 1,000,000,000.

f. log1000(1000000000) = ___
log1000(1000000000) = 3, because 10003 = 10000000000

g. log13(13) = ___
log13(13) = 1, because 131 = 13

h. log13(1) = ___
log13(1) = 0, because 130 = 1.

i. log7(âˆš7) = ___
log7(âˆš7) = 0, because 7$$\frac{1}{2}$$ = âˆš7.

j. log9(27) = ___
log9(27) = $$\frac{3}{2}$$, because 9$$\frac{3}{2}$$, = 33 = 27.

k. logâˆš7(7) = ___
logâˆš7(7) = 2, because (âˆš7)2 = 7.

l. logâˆš7($$\frac{1}{49}$$) = ___
logâˆš7($$\frac{1}{49}$$) = -4, because âˆš7-4 = $$\frac{1}{(\sqrt{7})^{4}}=\frac{1}{49}$$.

m. logx(x2) = ___
logx(x2) = 2, because (x)2 = x2.

### Eureka Math Algebra 2 Module 3 Lesson 8 Problem Set Answer Key

Question 1.
Rewrite each of the following in the form WhatPowerb(x) = L.

a. 35 = 243
WhatPower3(243) = 5

b. 6-3 = $$\frac{1}{216}$$
WhatPower6($$\frac{1}{216}$$) = -3

c. 90 = 1
WhatPower9(1) = 0

Question 2.
Rewrite each of the following in the form log(x) = L.

a. 16$$\frac{1}{4}$$ = 2
log16(2) = $$\frac{1}{4}$$

b. 103 = 1,000
log10(1,000) = 3

c. bk = r
logb(r) = k

Question 3.
Rewrite each of the following in the form bL = x.

a. log5(625) = 4
54 = 625

b. log10(0.1) = -1
10-1 = 0.1

c. log279 = $$\frac{2}{3}$$
27$$\frac{2}{3}$$ = 9

Question 4.
Consider the logarithms base 2. For each logarithmic expression below, either calculate the value of the expression or explain why the expression does not make sense.

a. log2(1024)
10

b. log2(128)
7

c. log2(âˆš8)
$$\frac{3}{2}$$

d. log2($$\frac{1}{16}$$)
-4

e. log2 (0)
This does not make sense. There is no value of L so that 2L = 0

f. log2($$-\frac{1}{32}$$)
This does not make sense. There is no value of L so that 2L is negative.

Question 5.
Consider the logarithms base 3. For each logarithmic expression below, either calculate the value of the expression or explain why the expression does not make sense.

a. log3(243)
5

b. log3(27)
3

c. log3(1)
0

d. log3($$\frac{1}{3}$$)
-1

e. log3(0)
This does not make sense. There is no value of L so that 3L = 0.

f. log3($$-\frac{1}{3}$$)
This does not make sense. There is no value of L so that 3L < 0.

Question 6.
Consider the logarithms base 5. Fo.â€™ each logarithmic expression below, either calculate the value of the expression or explain why the expression does not make sense.

a. log5(3125)
5

b. log5(25)
2

c. log5(1)
o

d. log5($$\frac{1}{25}$$)
-2

e. log5(0)
This does not make sense. There is no value of L so that 5L = o

f. log5(-$$\frac{1}{25}$$)
This does not make sense. There is no value of L so that L is negative.

Question 7.
Is there any positive number b so that the expression log(0) makes sense? Explain how you know.
No, there is no value of L so that bL = 0. I know b has to be a positive number. A positive number raised to an exponent never equals 0.

Question 8.
Is there any positive number b so that the expression logb(-1) makes sense? Explain how you know.
No. Since b is positive, there is no value of L so that bL is negative. A positive number raised to an exponent never has a negative value.

Question 9.
Verify each of the following by evaluating the logarithms.

a. log2(8) + log2(4) = log2(32)
3 + 2 = 5

b. log3(9) + log3(9) = log3(81)
2 + 2 = 4

c. log4(4) + log4(16) = log4(64)
1 + 2 = 3

d. log10(103) + log10(104) = log10(107)
3 + 4 = 7

Question 10.
Looking at the results from Problem 9, do you notice a trend or pattern? Can you make a general statement about the value of log(x) + log(y)?
The sum of two logarithms of the same base is found by multiplying the input values,
logb(x) + logb(y) = logb(xy). (Note to teacher: Do not evaluate this answer harshly. This is just a preview of a property that students learn later in the module.)

Question 11.
To evaluate log2(3), Autumn reasoned that since log2(2) = 1 and log2(4) = 2, log2(3) must be the average of 1 and 2 and therefore log2 (3) = 1. 5. Use the definition of logarithm to show that log2(3) cannot be 1. 5. Why is her thinking not valid?
According to the definition of logarithm, log2(3) = 1.5 only if 21.5 = 3. According to the calculator, 21.5 â‰ˆ 2.828, so log2(3) cannot be 1.5. Autumn was assuming that the outputs would follow a linear pattern, but since the outputs are exponents, the relationship is not linear.

Question 12.
Find the value of each of the following.

a. If x = log2(8) and y = 2x, find the value of y.
y = 8

b. If log2(x) = 6, find the value of x.
x = 64

c. If r = 26 and s = log2(r), find the value of s.
s = 6

### Eureka Math Algebra 2 Module 3 Lesson 8 Exit Ticket Answer Key

Question 1.
Explain why we need to specify 0 < b < 1 and b > 1 as valid values for the base b in the expression logb(x).
If b = 0, then log0(x) = L means that 0L = x, which cannot be true if x â‰  0.
If b = 1, then log1(x) = L means that 1L = x, which cannot be true if x â‰  1.
If b < 0, then logb(x) = L makes sense for some but not all values of x > 0; for example, if b = -2 and x = 32, there is no power of -2 that would produce 32, so log-2(32) does not makes sense.
Thus, if b â‰¤ 0 or b = 1, then for many values of x, the expression logb(x) does not make sense.

Question 2.
Calculate the following logarithms.

a. log5(25)
b. log10($$\frac{1}{100}$$)
log10($$\frac{1}{100}$$) = -2
log9(3) = $$\frac{1}{2}$$
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