Eureka Math Algebra 2 Module 3 Lesson 11 Answer Key

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

Opening Exercise:

Use the logarithm table below to calculate the specified logarithms.

Eureka Math Algebra 2 Module 3 Lesson 11 Opening Exercise Answer Key 1

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

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

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

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

e. log(9.0 × 10k) for an integer k
Answer:
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.

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

Answer:

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

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

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

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

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

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

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

f. log(3)+log(12)
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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)
Answer:
log(14) = log(2) + log(7) ≈ 0.3010 + 0.8451, so log(14) ≈ 1.1461.

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

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

d. log(121)
Answer:
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.

Eureka Math Algebra 2 Module 3 Lesson 11 Exercise Answer Key 4

Answer:

Eureka Math Algebra 2 Module 3 Lesson 11 Exercise Answer Key 5

Exercise 7.
What pattern(s) can you see in the table from Exercise 6? Write a conjecture using logarithmic notation.
Answer:
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.
Answer:
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)
Answer:
log(2100) = log(102. 21)
= 2 + log(21)
= 2 + log(3) + log(7)
≈ 2 + 0.4771 + 0.8451
≈ 3.3222

b. log(0.00049)
Answer:
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)
Answer:
log(42000000) = log(106 . 42)
= 6 + log(42)
= 6 + log(6) + log(7)
≈ 6 + 0.7782 + 0.8451
≈ 7.6233

d. log(\(\frac{1}{640}\))
Answer:
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).

Eureka Math Algebra 2 Module 3 Lesson 11 Problem Set Answer Key 6

a. log(25)
Answer:
1.40

b. log(27)
Answer:
1.44

c. log(33)
Answer:
1.52

d. log(55)
Answer:
1.74

e. log(63)
Answer:
1.81

f. log(75)
Answer:
1.88

g. log(81)
Answer:
1.92

h. log(99)
Ans:
2.00

i. log(350)
Answer:
2.55

j. log(0.0014)
Answer:
-2.85

k. log(0.077)
Answer:
-1.11

l. log(49000)
Answer:
4.70

m. log(1.69)
Answer:
0.22

n. log(6.5)
Answer:
0.81

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

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

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

r. log(\(\frac{1}{42}\))
Answer:
-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.

Eureka Math Algebra 2 Module 3 Lesson 11 Exit Ticket Answer Key 7

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).

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