Eureka Math Algebra 2 Module 3 Lesson 3 Answer Key

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

Opening Exercise:

a. What is the value of 2^{\(\frac{1}{2}\)}? Justify your answer.
Answer:
A possible student response follows: I think it will be around 1.5 because 2⁰ = 1 and 2¹ = 2.

b. Graph f(x) = 2^x for each integer x from x = -2 to x = 5. Connect the points on your graph with a smooth curve.

Answer:

c. The graph on the right shows a close-up view of f(x) 2^x for -0. 5 < x < 1.5.

c. Find two consecutive integers that are under and over estimates of the value of 2^{\(\frac{1}{2}\)}.
Answer:
2⁰ < 2^{\(\frac{1}{2}\)} < 2¹

1 < 2^{\(\frac{1}{2}\)} < 2

d. Does it appear that 2^{\(\frac{1}{2}\)} is halfway between the integers you specified in Exercise 1?
Answer:
No. It looks like 2^{\(\frac{1}{2}\)} is a little less than halfway between 1 and 2.

e. Use the graph of f(x) = 2 to estimate the value of 2^{\(\frac{1}{2}\)}.
Answer:
2^{\(\frac{1}{2}\)} ≈ 1.4

f. Use the graph of f(x) = 2^x to estimate the value of 2^{\(\frac{1}{3}\)}.
Answer:
2^{\(\frac{1}{3}\)} ≈ 1.25

Eureka Math Algebra 2 Module 3 Lesson 3 Example Answer Key

a. What is the 4th root of 16?
Answer:
x⁴ = 16 when x = 2 because 2⁴ = 16. Thus, \(\sqrt[4]{16}\) = 2.

b. What is the cube root of 125?
Answer:
x³ = 125 when x = 5 because 5³ = 125. Thus, \(\sqrt[3]{125}\) = 5.

c. What is the 5th root of 100,000?
Answer:
x⁵ = 100,000 when x = 10 because 10⁵ = 100000. Thus, \(\sqrt[5]{100,000}\) = 10.

Eureka Math Algebra 2 Module 3 Lesson 3 Exercise Answer Key

Exercise 1:

Exercise 1.
Evaluate each expression.

a. \(\sqrt[4]{16}\)
Answer:
3

b. \(\sqrt[5]{32}\)
Answer:
2

c. \(\sqrt[3]{9} \cdot \sqrt[3]{3}\)
Answer:
\(\sqrt[3]{27}\) = 3

d. \(\sqrt[4]{25} \cdot \sqrt[4]{100} \cdot \sqrt[4]{4}\)
Answer:
\(\sqrt[4]{10,000}\) = 10

Discussion:

1.
If 2^{\(\frac{1}{2}\)} = √2 and 2^{\(\frac{1}{3}\)} = \(\sqrt[3]{2}\) what does 2^{\(\frac{3}{4}\)} equal? Explain your reasoning.
Answer:
Student solutions and explanations will vary. One possible solution would be 2^{\(\frac{3}{4}\)} = (2^{\(\frac{1}{4}\)})³ , soit must mean that 2^{\(\frac{3}{4}\)} = \((\sqrt[4]{2})^{3}\). Since the properties of exponents and the meaning of an exponent made sense with integers and now for rational numbers in the form \(\frac{1}{n}\), it would make sense that they would work for all rational numbers, too.

Exercises 2 – 12:

Rewrite each exponential expression as a radical expression.

Exercise 2.
3^{\(\frac{1}{2}\)}
Answer:
3^{\(\frac{1}{2}\)} = √3

Exercise 3.
11^{\(\frac{1}{5}\)}
Answer:
11^{\(\frac{1}{5}\)} = \(\sqrt[5]{11}\)

Exercise 4.
\(\left(\frac{1}{4}\right)^{\frac{1}{5}}\)
Answer:
\(\left(\frac{1}{4}\right)^{\frac{1}{5}}\) = \(\sqrt[5]{\frac{1}{4}}\)

Exercise 5.
6^{\(\frac{1}{10}\)}
Answer:
6^{\(\frac{1}{10}\)} = \(\sqrt[10]{6}\)

Rewrite the following exponential expressions as equivalent radical expressions. If the number is rational, write it without radicals or exponents.

Exercise 6.
2^{\(\frac{3}{2}\)}
Answer:
2^{\(\frac{3}{2}\)} = \(\sqrt{2^{3}}\) = 2√2

Exercise 7.
4^{\(\frac{5}{2}\)}
Answer:
4^{\(\frac{5}{2}\)} = \(\sqrt{4^{5}}\) = (√4)⁵ = 2⁵ = 32

Exercise 8.
\(\left(\frac{1}{8}\right)^{\frac{5}{3}}\)
Answer:
\(\left(\frac{1}{8}\right)^{\frac{5}{3}}\) = \(\sqrt[3]{\left(\frac{1}{8}\right)^{5}}\)
= \(\left(\sqrt[3]{\frac{1}{8}}\right)^{5}\) = (\(\frac{1}{2}\))⁵
= \(\frac{1}{32}\)

Exercise 9.
Show why the following statement is true:
\(2^{-\frac{1}{2}}=\frac{1}{2^{\frac{1}{2}}}\)
Answer:
Student solutions and explanations will vary. One possible solution would be

Rewrite the following exponential expressions as equivalent radical expressions. If the number is rational, write it without radicals or exponents.

Exercise 10.
4^{\(-\frac{3}{2}\)}
Answer:
4^{\(-\frac{3}{2}\)} = \(\frac{1}{\sqrt{4^{3}}}=\frac{1}{(\sqrt{4})^{3}}=\frac{1}{8}\)

Exercise 11.
27^{\(-\frac{2}{3}\)}
Answer:
27^{\(-\frac{2}{3}\)} = \(=\frac{1}{27^{\frac{2}{3}}}=\frac{1}{(\sqrt[3]{27})^{2}}=\frac{1}{3^{2}}=\frac{1}{9}\)

Exercise 12.
\(\left(\frac{1}{4}\right)^{-\frac{1}{2}}\)
Answer:
we have \(\left(\frac{1}{4}\right)^{-\frac{1}{2}}\) = \(\left(\sqrt{\frac{1}{4}}\right)^{-1}\) = \(\left(\frac{1}{2}\right)^{-1}\) = 2,

Alternatively, \(\left(\frac{1}{4}\right)^{-\frac{1}{2}}\) = \(\left(\left(\frac{1}{4}\right)^{-1}\right)^{\frac{1}{2}}\)
= (4)^{\(\frac{1}{2}\)} = √4 = 2.

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

Question 1.
Select the expression from (A), (B), and (C) that correctly completes the statement.

a. x^{\(\frac{1}{3}\)} is equivalent to ____.
(A) \(\frac{1}{3}\)x
(B) \(\sqrt[3]{x}\)
(C) \(\frac{3}{x}\)
Answer:
(B)

b. x^{\(\frac{2}{3}\)}
(A) \(\frac{2}{3}\)x
(B) \(\sqrt[3]{x^{2}}\)
(C) \((\sqrt{x})^{3}\)
Answer:
(B)

c \(x^{-\frac{1}{4}}\)
(A) –\(\frac{1}{4}\)x
(B) \(\frac{4}{x}\)
(C) \(\frac{1}{\sqrt[4]{x}}\)
Answer:
(C)

d. \(\left(\frac{4}{x}\right)^{\frac{1}{2}}\)
(A) \(\frac{2}{x}\)
(B) \(\frac{4}{x^{2}}\)
(C) \(\frac{2}{\sqrt{x}}\)
Answer:
(C)

Question 2.
Identify which of the expressions (A), (B), and (C) are equivalent to the given expression.

a. \(16^{\frac{1}{2}}\)
(A) \(\left(\frac{1}{16}\right)^{-\frac{1}{2}}\)
(B) \(8^{\frac{2}{3}}\)
(C) \(64^{\frac{3}{2}}\)
Answer:
(A) and (B)

b. \(\left(\frac{2}{3}\right)^{-1}\)
(A) \(-\frac{3}{2}\)
(B) \(\left(\frac{9}{4}\right)^{\frac{1}{2}}\)
(C) \(\frac{27^{\frac{1}{3}}}{6}\)
Answer:
(B) only

Question 3.
Rewrite in radical form. If the number is rational, write it without using radicals.

a. \(6^{\frac{3}{2}}\)
Answer:
√216

b. \(\left(\frac{1}{2}\right)^{\frac{1}{4}}\)
Answer:
\(\sqrt[4]{\frac{1}{2}}\)

c. 3\((8)^{\frac{1}{3}}\)
Answer:
\(3 \sqrt[3]{8}\) = 6

d. \(\left(\frac{64}{125}\right)^{-\frac{2}{3}}\)
Answer:
\(\left(\sqrt[3]{\frac{125}{64}}\right)^{2}=\frac{25}{16}\)

e. \(81^{-\frac{1}{4}}\)
Answer:
\(\frac{1}{\sqrt[4]{81}}=\frac{1}{3}\)

Question 4.
Rewrite the following expressions in exponent form.

a. √5
Answer:
5^{\(\frac{1}{2}\)}

b. \(\sqrt[3]{5^{2}}\)
Answer:
5^{\(\frac{2}{3}\)}

c. \(\sqrt{5^{3}}\)
Answer:
5^{\(\frac{3}{2}\)}

d. \((\sqrt[3]{5})^{2}\)
Answer:
5^{\(\frac{2}{3}\)}

Question 5.
Use the graph of f(x) = 2X shown to the right to estimate the following powers of 2.

a. 2^{\(\frac{1}{4}\)}
Answer:
≈ 1.2

b. 2^{\(\frac{2}{3}\)}
Answer:
≈ 1.6

c. 2^{\(\frac{3}{4}\)}
Answer:
≈ 1.7

d. 2^0.2
Answer:
≈ 1.15

e. 2^1.2
Answer:
≈ 2.3

f. 2^{–\(\frac{1}{5}\)}
Answer:
≈ 0.85

Question 6.
Rewrite each expression in the form kxⁿ, where k is a real number, x is a positive real number, and n is rational.

a. \(\sqrt[4]{16 x^{3}}\)
Answer:
\(2 x^{\frac{3}{4}}\)

b.\(\frac{5}{\sqrt{x}}\)
Answer:
\(5 x^{-\frac{1}{2}}\)

c. \(\sqrt[3]{ x}\)
Answer:
\(x^{-\frac{4}{3}}\)

d.\(\frac{4}{\sqrt[3]{8 x^{3}}}\)
Answer:
\(2 x^{-1}\)

e. \(\frac{27}{\sqrt{9 x^{4}}}\)
Answer:
\(9 x^{-2}\)

f. \(\left(\frac{125}{x^{2}}\right)^{-\frac{1}{3}}\)
Answer:
\(\frac{1}{5} x^{\frac{2}{3}}\)

Question 7.
Find the value of x for which 2\(x^{\frac{1}{2}}\) = 32
Answer:
256

Question 8.
Find the value of x for which x\(x^{\frac{4}{3}}\) = 81
Answer:
27

Question 9.
If \(x^{\frac{3}{2}}\) = 64, find the value of 4\(x^{\frac{3}{4}}\).
Answer:
x = 16, so 4\((16)^{-\frac{3}{4}}\) = 4(2)^-3 = \(\frac{4}{8}\) = \(\frac{1}{2}\)

Question 10.
Evaluate the following expressions when b = \(\frac{1}{9}\)

a. \(b^{-\frac{1}{2}}\)
Answer:
\(\left(\frac{1}{9}\right)^{-\frac{1}{2}} = 9^{\frac{1}{2}}\) = 3

b. \(b^{\frac{5}{2}}\)
Answer:
\(\left(\frac{1}{9}\right)^{\frac{5}{2}}=\left(\frac{1}{3}\right)^{5}=\frac{1}{243}\)

c. \(\sqrt[3]{3 b^{-1}}\)
Answer:
\(\sqrt[3]{3\left(\frac{1}{9}\right)^{-1}}=\sqrt[3]{27}\) = 3

Question 11.
Show that each expression is equivalent to 2x. Assume x is a positive real number.

a. \(\sqrt[4]{16 x^{4}}\)
Answer:
\(\sqrt[4]{16} \cdot \sqrt[4]{x^{4}}\) = 2x

b. \(\frac{\left(\sqrt[3]{8 x^{3}}\right)^{2}}{\sqrt{4 x^{2}}}\)
Answer:
\(\frac{(2 x)^{2}}{\left(4 x^{2}\right)^{\frac{1}{2}}}=\frac{4 x^{2}}{2 x}\) = 2x

c. \(\frac{6 x^{3}}{\sqrt[3]{27 x^{6}}}\)
Answer:
\(\frac{6 x^{3}}{3 x^{\frac{6}{3}}}=\frac{6 x^{3}}{3 x^{2}}\) = 2x

Question 12.
Yoshiko said that 164 = 4 because 4 is one-fourth of 16. Use properties of exponents to explain why she is or is not correct.
Answer:
Yoshiko’s reasoning is not correct. By our exponent properties, \(\left(16^{\frac{1}{4}}\right)^{4}\) = \(16^{\left(\frac{1}{4}\right) \cdot 4}\) = 16¹ = 16, but 4⁴ = 256.
Since \(\left(16^{\frac{1}{4}}\right)^{4}\) ≠ 4⁴, we know that \(16^{\frac{1}{4}}\) ≠ 4.

Question 13.
Jefferson said that \(8^{\frac{4}{3}}\) = 16 because \(8^{\frac{1}{3}}\) = 2 and 2⁴ = 16. Use properties of exponents to explain why he is or is not correct.
Answer:
Jefferson’s reasoning is correct. We know that \(8^{\frac{4}{3}}\) = \(\left(8^{\frac{1}{3}}\right)^{4}\), so \(8^{\frac{4}{3}}\) = 2⁴, and thus \(8^{\frac{4}{3}}\) = 16.

Question 14.
Rita said that \(8^{\frac{2}{3}}\) = 128 because \(8^{\frac{2}{3}}\) = 8² \(8^{\frac{1}{3}}\), so \(8^{\frac{2}{3}}\) = 64 . 2, and then \(8^{\frac{2}{3}}\) = 128. Use properties of exponents to explain why she is or is not correct.
Answer:
Rita’s reasoning is not correct because she did not apply the properties of exponents correctly. She should also realize that raising 8 to a positive power less than 1 would produce a number less than 8. The correct calculation is below.

\(8^{\frac{2}{3}}\) = \(\left(8^{\frac{1}{3}}\right)^{2}\) = 2² = 4

Question 15.
Suppose for some positive real number a that \(\left(a^{\frac{1}{4}} \cdot a^{\frac{1}{2}} \cdot a^{\frac{1}{4}}\right)^{2}\) = 3.

a. What is the value of a?
Answer:

b. Which exponent properties did you use to find your answer to part (a)?
Answer:
We used the properties bⁿ . b^m = b^{m + n} and b\(\left(\boldsymbol{b}^{m}\right)^{n}\) = b^mn

Question 16.
In the lesson, you made the following argument:

Since \(\sqrt[3]{2}\) is a number so that (\(\sqrt[3]{2}\))³ = 2 and \(2^{\frac{1}{3}}\) is a number so that \(\left(2^{\frac{1}{3}}\right)^{3}\) = 2, you concluded that \(2^{\frac{1}{3}}\) = \(\sqrt[3]{2}\). Which exponent property was used to make this argument?
Answer:
We used the property bⁿ . b^m = b^{m + n}. (Students may also mention the uniqueness of nth roots.)

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

Question 1.
Rewrite the following exponential expressions as equivalent radical expressions.

a. \(2^{\frac{1}{2}}\)
Answer:
\(2^{\frac{1}{2}}\) = √2

b. \(2^{\frac{3}{4}}\)
Answer:
\(2^{\frac{3}{4}}\) = \(\sqrt[4]{2^{3}}\) = \(\sqrt[4]{8}\)

c. \(3^{-\frac{2}{3}}\)
Answer:
\(3^{-\frac{2}{3}}\) = \(\frac{1}{\sqrt[3]{3^{2}}}=\frac{1}{\sqrt[3]{9}}\)

Question 2.
Rewrite the following radical expressions as equivalent exponential expressions.

a. √5
Answer:
√5 = \(5^{\frac{1}{2}}\)

b. \(2 \sqrt[4]{3}\)
Answer:
\(2 \sqrt[4]{3}\) = \(\sqrt[4]{2^{4} \cdot 3}=\sqrt[4]{48}=48^{\frac{1}{4}}\)

c. \(\frac{1}{\sqrt[3]{16}}\)
Answer:
\(\frac{1}{\sqrt[3]{16}}=\left(2^{4}\right)^{-\frac{1}{3}}=2^{-\frac{4}{3}}\)
\(\frac{1}{\sqrt[3]{16}}=(16)^{-\frac{1}{3}}\)

Question 3.
Provide a written explanation for each question below.

a. Is it true that \(\left(1000^{\frac{1}{3}}\right)^{3}=\left(1000^{3}\right)^{\frac{1}{3}}\)? Explain how you know.
Answer:
\(\left(1000^{\frac{1}{3}}\right)^{3}=(\sqrt[3]{1000})^{3}\)
= 10³ = 1000
\(\left(1000^{3}\right)^{\frac{1}{3}}=(1000000000)^{\frac{1}{3}}\) = 1000
So, this statement is true.

b. Is it true that \(\left(4^{\frac{1}{2}}\right)^{3}=\left(4^{3}\right)^{\frac{1}{2}}\)? Explain how you know.
Answer:
\(\left(4^{\frac{1}{2}}\right)^{3}\) = (√4)³ = 2³ = 8
\(\left(4^{3}\right)^{\frac{1}{2}}=64^{\frac{1}{2}}\) = √64 = 8
So, this statement is true.

c. Suppose that m and n are positive integers and b is a real number so that the principal nth root of b exists. In general, does \(\left(b^{\frac{1}{n}}\right)^{m}=\left(b^{m}\right)^{\frac{1}{n}}\)? Explain how you know.
Answer:
From the two examples we have seen, it appears that we can extend the property (\(\left(b^{\frac{1}{n}}\right)^{m}=\left(b^{m}\right)^{\frac{1}{n}}\) for integers m and n to rational exponents.

We know that, in general, we have
\(\left(b^{\frac{1}{n}}\right)^{m}=(\sqrt[n]{b})^{m}\)
= \(\sqrt[n]{\boldsymbol{b}^{m}}\)
= \(\left(\boldsymbol{b}^{m}\right)^{\frac{1}{n}}\)