## 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)

Answer:

log(80) = log(10^{1} . 8) = 1 + log(8) ≈ 1.9031

b. log(7000)

Answer:

log(7000) = log(10^{3} . 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 × 10^{27}) = log(10^{27} . 3) = 27 + log(3) ≈ 27.4771

e. log(9.0 × 10^{k}) for an integer k

Answer:

log(9.0 × 10^{k}) = log(10^{k} . 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.

Answer:

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(x^{2})? Make a conjecture, and test ¡t to see whether or not it appears to be valid.

Answer:

I would expect that log(x^{2}) = 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.

Answer:

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 10^{a} = \(\frac{1}{x}\). 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(10^{2}. 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(10^{6} . 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).

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.

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(10^{3} . 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(10^{k} . a^{2}).

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(10^{k} . a^{2}) = k + log(a^{2})

= k + log(a) + log(a)

= k + 2 log(a).