## Engage NY Eureka Math Algebra 2 Module 3 Lesson 10 Answer Key

### Eureka Math Algebra 2 Module 3 Lesson 10 Opening Exercise Answer Key

Opening Exercise:

Find the value of the following expressions without using a calculator.

WhatPower_{10}(1000) =

Answer:

3

log_{10}(1000) = Answer:

3

WhatPower_{10}(100) =

Answer:

2

log_{10}(100) =

Answer:

2

WhatPower_{10}(10) =

Answer:

1

log_{10}(10) =

Answer:

1

WhatPower_{10}(1) =

Answer:

0

log_{10}(1) =

Answer:

0

WhatPower_{10}(\(\frac{1}{10}\)) =

Answer:

-1

log_{10}(\(\frac{1}{10}\)) =

Answer:

-1

WhatPower_{10}(\(\frac{1}{100}\)) =

Answer:

-2

log_{10}(\(\frac{1}{100}\)) =

Answer:

-2

Formulate a rule based on your results above: If k is an integer, then log_{10}(10^{k}) = ______

Answer:

1og_{10}(10^{k}) = k

### Eureka Math Algebra 2 Module 3 Lesson 10 Example Answer Key

Example 1:

(100,000,000,000,000) (1,000,000) = 100,000,000,000,000,000,000, so the second star is 100 quintillion miles away from Earth.

→ How did you arrive at that figure?

I counted the zeros; there are 14 zeros in 100,000,000,000,000 and 6 zeros in 1,000,000, so there must be 20 zeros in the product.

→ Can we restate that in terms of exponents?

(10^{14}) (10^{6}) = 10^{20}

→ How are the exponents related?

14 + 6 = 20

→ What are log(10^{14}), log(10^{6}), and log(10^{20})?

14, 6, and 20

→ In this case, can we state an equivalent expression for log(10^{14} . 10^{6})?

log(10^{14} . 10^{6}) = log(10^{14}) + log(10^{6})

→ Why is this equation true?

log(10^{14} 10^{6}) = log(10^{20}) = 20 = 14 + 6 = log(10^{14}) + log(10^{6})

→ Generalize to find an equivalent expression for log(lOm . 109 for integers m and n. Why is this equation true?

log(10^{m} . 10^{n}) = log(10^{m + n}) = m + n = log(10^{m}) + log(10^{n})

This equation is true because when we multiply powers of 10 together, the resulting product is a power of 10 whose exponent is the sum of the exponents of the factors.

→ Keep this result in mind as we progress through the lesson.

### Eureka Math Algebra 2 Module 3 Lesson 10 Exercise Answer Key

Exercises:

Exercise 1.

Find two consecutive powers of 10 so that 30 is between them. That is, find an integer exponent k so that

10^{k} < 30 < 10^{k + 1}

Answer:

Since 10 < 30 < 100, we have k = 1.

Exercise 2.

From your result in Exercise 1, log(30) is between which two integers?

Answer:

Since 30 is some power of 10 between 1 and 2, 1 < log(30) < 2.

Exercise 3.

Find a number k to one decimal place so that 10^{k} < 30 < 10^{k + 0.1}, and use that to find under and overestimates for log(30).

Answer:

Since 10^{1.4} ≈ 25.1189 and 10^{1.5} ≈ 31.6228, we have 10^{1.4} < 30 < 10^{1.5}. Then 1.4 < log(30) < 1.5, and k ≈ 1.4.

Exercise 4.

Find a number k to two decimal places so that 10^{k} < 30 < 10^{k + 0.01}, and use that to find under and overestimates for log(30).

Answer:

Since 10^{1.47} ≈ 29.5121, and 10^{1.48} ≈ 30.1995, we have 10^{1.47} < 30 < 10^{1.48} so that 1.47 < log(30) < 1.48. So, k ≈ 1.47.

Exercise 5.

Repeat this process to approximate the value of log(30) to 4 decimal places.

Answer:

Since 10^{1.477} ≈ 29.9916, and 10^{1.478} ≈ 30.0608, we have 10^{1.477} < 30 < 10^{1.478} so that 1.477 < log(30) < 1.478.

Since 10^{1.4771} ≈ 29.9985, and 10^{1.4772} ≈ 30.0054, we have 10^{1.4771} < 30 < 10^{1.4772} so that 1.4771 < log(30) < 1.4772.

Since 10^{1.47712} ≈ 29.9999, and 10^{1.47713} ≈ 30.0006, we have 10^{1.47712} < 30 < ^{1.47713} so that 1.47712 < log(30) < 1.47713.

Thus, to four decimal places, log(30) ≈ 1.4771.

Exercise 6.

Verify your result on your calculator, using the LOG button.

Answer:

The calculator gives log(30) ≈ 1.477121255.

Exercise 7.

Use your calculator to complete the following table. Round the logarithms to 4 decimal places.

Answer:

Exercise 8.

What pattern(s) can you see in the table from Exercise 7 as x is multiplied by 10? Write the pattern(s) using logarithmic notation.

Answer:

I found the patterns log(10x) = 1 + log(x) and log(100x) = 2 + log(x). I also noticed that log(100x) = 1 + log(10x).

Exercise 9.

What pattern would you expect to find for log(l000x)? Make a conjecture, and test it to see whether or not it appears to be valid.

Answer:

I would guess that the values of log(1000x) will all start with 3. That is, log(1000x) = 3 + log(x). This appears to be the case since log(2000) ≈ 3.3010, log(5000) ≈ 3.6990, and log(8000) ≈ 3.9031.

Exercise 10.

Use your results from Exercises 8 and 9 to make a conjecture about the value of log(10” x) for any positive integer k.

Answer:

It appears that log(10^{k} . x) = k + log(x), for any positive integer k.

Exercise 11.

Use your calculator to complete the following table. Round the logarithms to 4 decimal places.

Answer:

Exercise 12.

What pattern(s) can you see In the table from Exercise 11? Write them using logarithmic notation.

Answer:

I found the patterns log(x) – log (\(\frac{x}{10}\)) = 1, which can be written as log (\(\frac{x}{10}\)) = -1 + log(x), and log (\(\frac{x}{100}\)) = -2 + log(x).

Exercise 13.

What pattern would you expect to find for log (\(\frac{x}{1000}\))? Make a conjecture, and test it to see whether or not It appears to be valid.

Answer:

I would guess that the values of log(\(\frac{x}{1000}\)) will all start with -2 and that log (\(\frac{x}{1000}\)) = -3 + log(x). This appears to be the case since log(0.002) ≈ -2.6990, and -2.6990 = -3 + 0.3010; log(0.005) ≈ -2.3010, and -2.3010 = -3 + 0.6990; log(0.008) ≈ -2.0969, and -2.0969 = -3 + 0.9031.

Exercise 14.

Combine your results from Exercises 10 and 12 to make a conjecture about the value of the logarithm for a multiple of a power of 10; that is, find a formula for log(10^{k} . x) for any integer k.

Answer:

It appears that log(10^{k} . x) = k + log(x), for any integer k.

### Eureka Math Algebra 2 Module 3 Lesson 10 Problem Set Answer Key

Question 1.

Complete the following table of logarithms without using a calculator; then, answer the questions that follow.

Answer:

a. What is log(1)? How does that follow from the definition of a base-10 logarithm?

Answer:

Since 10^{0} = 1, we know that log(1) = 0.

b. What is log(10^{k}) for an integer k? How does that follow from the definition of a base-10 logarithm?

Answer:

By the definition of the logarithm, we know that log(10^{k}) = k.

c. What happens to the value of log(x) as x gets really large?

Answer:

For any x> 1, there exists k > 0 so that 10^{k} < x < 10^{k + 1} As x gets really large, k gets large. Since k ≤ log(x) < k + 1, as k gets large, log(x) gets large. d. For x > 0, what happens to the value of log(x) as x gets really close to zero?

Answer:

For any 0 < x < 1, there exists k > 0 so that 10^{-k} < x < 10^{k + 1}. Then -k ≤ x < -k + 1. As x gets closer to zero, k gets larger. Thus, log(x) is negative, and |log(x)| gets large as the positive number x gets close to zero.

Question 2.

Use the table of logarithms below to estimate the values of the logarithms in parts (a) – (h).

a. log(70000)

Answer:

4.8451

b. log(O. 0011)

Answer:

-2.9586

c. log(20)

Answer:

1.3010

d. log(0.00005)

Answer:

-4.3010

e. log(130 000)

Answer:

5.1139

f. log(3000)

Answer:

3.4771

g. log(0.07)

Answer:

-1.1549

h. log(11000000)

Answer:

7.0414

Question 3.

If log(n) = 0.6, find the value of log(10n).

Answer:

log(10n) = 1.6

Question 4.

If m is a positive integer and log(m) ≈ 3.8, how many digits are there in m? Explain how you know.

Answer:

Since 3 < log(m) < 4, we know 1,000 < m < 10,000; therefore, m has 4 digits.

Question 5.

If m is a positive integer and log(m) ≈ 9. 6, how many digits are there in m? Explain how you know.

Answer:

Since 9 < log(m) < 10, we know 10^{9} < m < 10^{10}; therefore, m has 10 digits.

Question 6.

Vivian says log(452000) = 5 + log(4.52), while her sister Lillian says that log(452000) = 6 + log(0.452). Which sister is correct? Explain how you know.

Answer:

Both sisters are correct. Since 452,000 = 4.52 . 10^{5}, we can write log(452000) = 5 + log(4.52). However, we could also write 452,000 = 0.452 . 10^{6}, so log(452000) = 6 + log(0.452). Both calculations give log(452,000) ≈ 5.65514.

Question 7.

Write the base-10 logarithm of each number in the form k + log(x), where k is the exponent from the scientific notation, and x is a positive real number.

a. 24902 × 10^{4}

Answer:

4+ log(2.4902)

b. 2.58 × 10^{13}

Answer:

13+ log(2.58)

c. 9.109 × 1o^{-31}

Answer:

-31 + log(9.109)

Question 8.

For each of the following statements, write the number in scientific notation, and then write the logarithm base 10 of that number in the form k + log(x), where k is the exponent from the scientific notation, and x is a positive real number.

a. The speed of sound is 1116ft/s.

Answer:

1116 = 1.116 × 10^{3}, so log(1116) = 3 + log(1.116).

b. The distance from Earth to the sun is 93 million miles.

Answer:

93,000,000 = 9.3 × 10^{7}, so log(93000000) = 7 + log(9.3).

c. The speed of light is 29,980,000,000 cm/s.

Answer:

29,980,000,000 = 2.998 × 10^{10}, so log(29, 980,000, 000) = 10 + log(2.998).

d. The weight of Earth is 5,972, 000,000,000, 000,000,000, 000 kg.

Answer:

5, 972, 000, 000, 000, 000, 000, 000, 000 = 5.972 × 10^{24}, so

log(5,972,000,000,000, 000,000,000,000) = 24 + log(5.972).

e. The diameter of the nucleus of a hydrogen atom is 0. 00000000000000175 m.

Answer:

0.00000000000000175 = 1.75 × 10^{-15}, so log(0. 00000000000000175) = -15 + log(1.75).

f. For each part (a) – (e), you have written each logarithm in the form k + log(x), for integers k and positive real numbers x. Use a calculator to find the values of the expressions log(x). Why are all of these values between 0 and 1?

Answer:

log(1.116) ≈ 0.047664

log(9.3) ≈ 0.968483

log(2.998) ≈ 0.476832

log(5.972) ≈ 0.77612

log(1.75) ≈ 0.243038

These values are all between 0 and 1 because x is between 1 and 10. We can rewrite 1 < x < 10 as 100 < x < 10^{1}. If we write x = 10^{L} for some exponent L, then 10^{0 }< 10^{L} < 10^{1}, so 0 < L < 1. This exponent L is the base 10 logarithm of X.

### Eureka Math Algebra 2 Module 3 Lesson 10 Exit Ticket Answer Key

Question 1.

Use the logarithm table below to approximate the specified logarithms to four decimal places. Do not use a calculator.

a. log( 500)

Answer:

log(500) = log(10^{2} . 5)

= 2 + log(5)

≈ 2.6990

b. log(0.0005)

Answer:

log(0.0005) = log(10^{-4} . 5)

= -4 + log(5)

≈ -3.3010

Question 2.

Suppose that A is a number with log(A) = 1.352.

a. What is the value of log(1000A)?

Answer:

log(1000A) = log(10^{3}A) = 3 + log(A) = 4.352

b. Which one of the following statements is true? Explain how you know.

i. A < 0

ii. 0 < A < 10

iii. 10 < A < 100

iv. 100 < A < 1000 v. A >1000

Answer:

Because log(A) = 1.352 = 1 + 0.352, A is greater than 10 and less than 100. Thus, (iii) is true. In fact, from the table above, we can see that A is between 20 and 30 because of log(20) ≈ 1.3010, and log(30) ≈ 1. 4771.