# Eureka Math Algebra 2 Module 3 Lesson 11 Answer Key

## Engage NY Eureka Math Algebra 2 Module 3 Lesson 11 Answer Key

### Eureka Math Algebra 2 Module 3 Lesson 11 Opening Exercise Answer Key

Opening Exercise:

Use the logarithm table below to calculate the specified logarithms.

a. log(80)
log(80) = log(101 . 8) = 1 + log(8) ≈ 1.9031

b. log(7000)
log(7000) = log(103 . 7) = 3 + log(7) ≈ 3.8451

c. log(0. 00006)
log(0. 00006) = log(10-5 . 6) = -5 + log(6) ≈ -4.2218

d. log(3.0 × 1027)
log(3.0 × 1027) = log(1027 . 3) = 27 + log(3) ≈ 27.4771

e. log(9.0 × 10k) for an integer k
log(9.0 × 10k) = log(10k . 9) = k + log(9) ≈ k + 0.9542

### Eureka Math Algebra 2 Module 3 Lesson 11 Exercise Answer Key

Exercises 1 – 5:

Exercise 1.
Use your calculator to complete the following table. Round the logarithms to four decimal places.

Exercise 2.
Calculate the following values. Do they appear anywhere else in the table?

a. log(2) + log(4)
We see that log(2) + log(4) ≈ 0.9031, which is approximately log(8).

b. log(2) + log(6)
We see that log(2) + log(6) ≈ 1.0792, which ¡s approximately log(12).

c. log(3) + log(4)
We see that log(3) + log(4) ≈ 1.0792, which ¡s approximately log(12).

d. log(6)+log(6)
We see that log(6) + log(6) ≈ 1. 5663, which ¡s approximately log(36).

e. log(2) + log(18)
We see that log(2) + log(18) ≈ 1.5663, which logs approximately log(36).

f. log(3)+log(12)
We see that log(3) + log(12) ≈ 1.5664, which is approximately log(36).

Exercise 3.
What pattern(s) can you see in Exercise 2 and the table from Exercise 1? Write them using logarithmic notation.
I found the pattern log(xy) = log(x) + log(y).

Exercise 4.
What pattern would you expect to find for log(x2)? Make a conjecture, and test ¡t to see whether or not it appears to be valid.
I would expect that log(x2) = log(x) + log(x) = 2 log(x). This is verified by the fact that log(4) ≈ 0.6021 ≈ 2 log(2), log(9) ≈ 0.9542 ≈ 2 log(3), log(16) ≈ 1.2041 ≈ 2 log(4), and log(25) ≈ 1.3980 ≈ 2 log(5).

Exercise 5.
Make a conjecture for a logarithm of the form log(xyz), where x, y, and z are positive real numbers. Provide evidence that your conjecture ¡s valid.
It appears that log(xyz) = log(x) + log(y) + log(z). This is due to applying the property from Exercise 3 twice.
log(xyz) = log(xy . z)
= log(xy) + log(z)
= log(x) + log(y) + log(z)
OR
It appears that log(xyz) = log(x) + log(y) + log(z). We can see that
log(18) ≈ 1.2553 ≈ 0.3010 + 0.3010 + 0.4771 ≈ log(2) + log(2) + log(3),
log(20) ≈ 1.3010 ≈ 0.3010 + 0.3010 + 0.6990 ≈ log(2) + log(2) + log(5), and
log(36) ≈ 1.5563 ≈ 0.3010 + 0.4771 + 0.7782 ≈ log(2) + log(3) + log(6).

Example 1:

Use die logarithm table from Exercise to approximate the following logarithms.

a. log(14)
log(14) = log(2) + log(7) ≈ 0.3010 + 0.8451, so log(14) ≈ 1.1461.

b. log(35)
log(35) = log(5) + log(7) ≈ 0.6990 + 0.845 1, so log(35) ≈ 1.5441.

c. log(72)
log(72) = log(8) + log(9) ≈ 0.9031 + 0.9542,50 log(72) ≈ 1.8573.

d. log(121)
log(121) = log(11) + log(11), but we do not have a value for log(11) ¡n the table, so we cannot evaluate log(121).

Exercises 6 – 8:

Exercise 6.
Use your calculator to complete the following table. Round the logarithms to four decimal places.

Exercise 7.
What pattern(s) can you see in the table from Exercise 6? Write a conjecture using logarithmic notation.
For any real number x > 0, log ($$\frac{1}{x}$$) = -log(x).

Exercise 8.
Use the definition of logarithm to justify the conjecture you found in Exercise 7.
If log ($$\frac{1}{x}$$) = a for some number a, then 10a = $$\frac{1}{x}$$. Then, 10-a = x, and thus, log(x) = -a. We then have log($$\frac{1}{x}$$) = -log(x).

Example 2:

Use the logarithm tables and the rules we have discovered to estimate the following logarithms to four decimal places:

a. log(2100)
log(2100) = log(102. 21)
= 2 + log(21)
= 2 + log(3) + log(7)
≈ 2 + 0.4771 + 0.8451
≈ 3.3222

b. log(0.00049)
log(0.00049) = log(10-5 . 49)
= -5 + log(49)
= -5 + log(7) + log(7)
≈ -5 + 0.8451 + 0.8451
≈ -3. 3098

c. log(42000000)
log(42000000) = log(106 . 42)
= 6 + log(42)
= 6 + log(6) + log(7)
≈ 6 + 0.7782 + 0.8451
≈ 7.6233

d. log($$\frac{1}{640}$$)
log($$\frac{1}{640}$$) = -log(640)
= -(log(10.64))
= -(1 + log(64))
= -(1 + log(8) + log(8))
≈ -(1 + 0.9031 + 0.9031)
≈ -2. 8062

### Eureka Math Algebra 2 Module 3 Lesson 11 Problem Set Answer Key

Question 1.
Use the table of logarithms to the right to estimate the value of the logarithms in parts (a) – (t).

a. log(25)
1.40

b. log(27)
1.44

c. log(33)
1.52

d. log(55)
1.74

e. log(63)
1.81

f. log(75)
1.88

g. log(81)
1.92

h. log(99)
Ans:
2.00

i. log(350)
2.55

j. log(0.0014)
-2.85

k. log(0.077)
-1.11

l. log(49000)
4.70

m. log(1.69)
0.22

n. log(6.5)
0.81

o. log($$\frac{1}{30}$$)
-1.48

p. log($$\frac{1}{35}$$)
-1.55

q. log($$\frac{1}{40}$$)
-1.60

r. log($$\frac{1}{42}$$)
-1.63

s. log ($$\frac{1}{50}$$)
Ans:
-1.70

t. log($$\frac{1}{64}$$)
Ans:
-1.80

Question 2.
Reduce each expression to a single logarithm of the form log(x).

a. log(5) + log(7)
Ans:
log(35)

b. log(3) + log(9)
Ans:
log(27)

c. log(15) – log(5)
Ans:
log(3)

d. log(8) + log($$\frac{1}{4}$$)
Ans:
log(2)

Question 3.
Use properties of logarithms to write the following expressions involving logarithms of only prime numbers:

a. log(2500)
Ans:
2 + 2 log(5)

b. log(0.00063)
Ans:
-5 + 2 log(3) + log(7)

c. log(1250)
Ans:
1 + 3 log(5)

d. log(26000000)
Ans:
6 + log(2) + log(13)

Question 4.
Use properties of logarithms to show that log(2) – log ($$\frac{1}{13}$$) = log(26).
Ans:
log(2) – log($$\frac{1}{13}$$) = log(2) – log(13-1)
= log(2) + log(13)
= log(26)

Question 5.
Use properties of logarithms to show that log(3) + log(4) + log(5) – log(6) = 1.
Ans:
There are multiple ways to solve this problem.
log(3) + log(4) + log(5) – log(6) = log(3) + log(4) + log(5) + log($$\frac{1}{6}$$)
= log(3 . 4 . 5. $$\frac{1}{6}$$)
= log(10)
= 1 (OR)

log(3) + log(4) + log(5) = log(60)
= log(10 . 6)
= log(10) + log(6)
= 1 + log(6)
log(3) + log(4) + log(5) – log(6) = 1

Question 6.
Use properties of logarithms to show that log ($$\frac{1}{2}$$ – $$\frac{1}{3}$$) + log(2) = -log(3).
Ans:
log ($$\frac{1}{2}$$ – $$\frac{1}{3}$$) + log(2) = log ($$\frac{1}{6}$$) + log(2)
= -log(6) + log(2)
= -(log(2) + log(3)) + log(2)
= -log(3)

Question 7.
Use properties of logarithms to show that log($$\frac{1}{3$$ – $$\frac{1}{4}$$) + (log ($$\frac{1}{3}$$) – log ($$\frac{1}{4}$$)) = -2 log(3).
Ans:
log ($$\frac{1}{3}$$ – $$\frac{1}{4}$$) + (log ($$\frac{1}{3}$$) – log ($$\frac{1}{4}$$)) = log($$\frac{1}{12}$$) + log ($$\frac{1}{3}$$) – log ($$\frac{1}{4}$$)
= -log(12) – log(3) + log(4)
= -(log(3) + log(4)) – log(3) + log(4)
= -2 log(3)

### Eureka Math Algebra 2 Module 3 Lesson 11 Exit Ticket Answer Key

Question 1.
Use the table below to approximate the following logarithms to four decimal places. Do not use a calculator.

a. log(9)
Ans:
log(9) = log(3) + log(3)
≈ 0.4771 + 0.4771
≈0.9542

b. log ($$\frac{1}{15}$$)
Ans:
log($$\frac{1}{15}$$) = -log(15)
= -(log(3) + log(5))
≈ -(0.4771 + 0.6990)
≈ -1. 1761

c. log(45000)
Ans:
log(45 000) = log(103 . 45)
= 3 + log(45)
= 3 + log(5) + log(9)
≈ 3 +0.6990+0.9542
≈ . 6532

Question 2.
Suppose that k is an integer, a is a positive real number, and you know the value of log(a). Explain how to find the value of log(10k . a2).
Ans:
Applying the rule for the logarithm of a number multiplied by a power of 10 and then the rule for the logarithm of a product, we have
log(10k . a2) = k + log(a2)
= k + log(a) + log(a)
= k + 2 log(a).

Scroll to Top