## Engage NY Eureka Math Algebra 2 Module 3 Lesson 14 Answer Key

### Eureka Math Algebra 2 Module 3 Lesson 14 Opening Exercise Answer Key

Opening Exercise:

Convert the following logarithmic equations to equivalent exponential equations.

a. log(10,000) = 4

Answer:

10^{4} = 10,000

b. log(âˆš10) = \(\frac{1}{2}\)

Answer:

10^{\(\frac{1}{2}\)} = âˆš10

c. log_{2}(256) = 8

Answer:

2^{8}= 256

d. log_{4}(256) = 4

Answer:

4^{4} = 256

e. ln(1)=0

Answer:

e^{0} = 1

f. log(x + 2) = 3

Answer:

x + 2 = 10^{3}

### Eureka Math Algebra 2 Module 3 Lesson 14 Example Answer Key

Examples:

Write each of the following equations as an equivalent exponential equation, and solve for x.

Example 1.

log(3x+7) = 0

Answer:

log(3x + 7) = 0

10^{0} = 3x + 7

1 = 3x +7

x = -2

Example 2.

log_{2}(x + 5) = 4

Answer:

log_{2}(x + 5) = 4

2^{4} = x +5

16 = x +5

x = 11

Example 3.

log(x + 2) + log(x + 5) = 1

Answer:

log(x + 2) + log(x + 5) = 1

log((x + 2)(x + 5)) = 1

(x + 2)(x + 5)= 10^{1}

x^{2} + 7x + 10 = 10

x^{2} + 7x = 0

x(x +7) = O

x = 0 or x = -7

However, if x = -7, then (x + 2) = -5, and (x + 5) = -2, so both logarithms ln the equation are undefined.

Thus, -7 is an extraneous solution, and only 0 is a valid solution to the equation.

### Eureka Math Algebra 2 Module 3 Lesson 14 Exercise Answer Key

Exercises:

Exercise 1.

Drew said that the equation log_{2}[(x + 1)^{4}] = 8 cannot be solved because he expanded (x + 1)^{4} = x^{4} + 4x^{3} + 6x^{2} + 4x + 1 and realized that he cannot solve the equation x^{4} + 4x^{3}+ 6x^{2} + 4x + 1 = 2^{8}. Is he correct? Explain how you know.

Answer:

If we apply the logarithmic properties, this equation is solvable.

log_{2}[(x+ 1)^{4}] = 8

4log_{2}(x+ 1) = 8

log_{2}(x + 1) = 2

x + 1 = 2^{2}

x = 3

Check: If x = 3, then log_{2}[(3 + 1)^{4}] = 4 log_{2}(4) = 4 . 2 = 8, so 3 is a solution to the original equation.

Solve the equations in Exercises 2 – 4 for x.

Exercise 2.

ln((4x)^{5}) = 15

ln(4x) = 15

ln(4x) = 3

e^{3} = 4x

x = \(\frac{e^{3}}{4}\)

Check: Since 4(\(\frac{e^{3}}{4}\)) > 0, we know that ln \(\left(\left(4 \cdot \frac{e^{3}}{5}\right)^{5}\right)\) Â¡s defined. Thus, \(\frac{e^{3}}{4}\) is the solution to the equation.

Exercise 3.

log((2x + 5)^{2}) = 4

Answer:

2 log(2x + 5) = 4

log(2x + 5) = 2

10^{2}= 2x + 5

100 = 2x + 5

95 = 2x

x = \(\frac{95}{2}\)

Check: Since 2(\(\frac{95}{2}\)) + 5 â‰ 0, we know that log ((2 . \(\frac{95}{2}\) + 5)^{2}) is defined.

Thus, \(\frac{95}{2}\) is the solution to the equation.

Exercise 4.

log_{2}((5x + 7)^{19}) = 57

Answer:

19 . log_{2}(5x + 7) = 57

log_{2}(5x + 7) = 3

2^{3} = 5x + 7

8 = 5x + 7

1 = 5x

x = \(\frac{1}{5}\)

Check:

Since 5(\(\frac{1}{5}\)) + 7 > 0, we know that log_{2}(5 . \(\frac{1}{5}\) + 7) is defined.

Thus, \(\frac{1}{5}\) is the solution to this equation.

Solve the logarithmic equations ln Exercises 5 – 9, and identify any extraneous solutions.

Exercise 5.

log(x^{2} + 7x + 12) – log(x +4) = 0

Answer:

log\(\left(\frac{x^{2}+7 x+12}{x+4}\right)\) = 0

\(\frac{x^{2}+7 x+12}{x+4}\) = 10^{0}

\(\frac{x^{2}+7 x+12}{x+4}\) = 1

x^{2} + 7x + 12 = x + 4

0 = x^{2} + 6x + 8

0 = (x + 4)(x + 2)

x = -4 or x = -2

Check:

If x = -4, then log(x + 4) = log(0), which is undefined. Thus, -4 is an extraneous solution. Therefore, the only solution is -2.

Exercise 6.

log_{2}(3x) + log_{2}(4) = 4

Answer:

log_{2}(3x) + 2 = 4

log_{2}(3x) = 2

2^{2} = 3x

4 = 3x

x = \(_{2}\)

Check:

Since \(\frac{4}{3}\) > 0, log_{2}(3 . \(\frac{4}{3}\)) is defined.

Therefore, \(\frac{4}{3}\) is a valid solution.

Exercise 7.

2ln(x + 2) – ln(-x) = 0

Answer:

ln((x+2)^{2}) – ln(-x) = 0

ln\(\left(\frac{(x+2)^{2}}{-x}\right)\) = 0

1 = \(\frac{(x+2)^{2}}{-x}\)

-x = x^{2} + 4x + 4

0 = x^{2} + 5x + 4

0 = (x + 4) (x + 1)

x = -4 or x = -1

Check: Thus, we get x = -4 or x = -1 as solutions to the quadratic equation. However, if x = -4, then ln(x + 2) = ln(-2), so -4 is an extraneous solution. Therefore, the only solution is -1.

Exercise 8.

log(x) = 2 – log(x)

Answer:

log(x) + log(x) = 2

2 . log(x) = 2

log(x) = 1

x = 10

Check: Since 10 > 0, log(10) is defined.

Therefore, 10 is a valid solution to this equation.

Exercise 9.

ln(x + 2) = ln(12) – ln(x+3)

Answer:

ln(x + 2) + ln(x + 3) = ln(12)

ln((x + 2)(x + 3)) = ln(12)

(x + 2)(x + 3) = 12

x^{2} + 5x +6= 12

x^{2} + 5x – 6 = 0

x = -1 or x = -6

Check:

If x = -6, then the expression ln(x + 2) and ln(x + 3) are undefined.

Therefore, the only valid solution to the original solution is 1.

### Eureka Math Algebra 2 Module 3 Lesson 14 Problem Set Answer Key

Question 1.

Solve the following logarithmic equations.

a. log(x) = \(\frac{5}{2}\)

Answer:

log(x) = \(\frac{5}{2}\)

x = 10^{\(\frac{5}{2}\)}

x = 100âˆš10

Check: Since 100âˆš10 > 0, we know log(100âˆš10) is defined.

Therefore, the solution to this equation is 100âˆš10.

b. 5log(x + 4) = 10

Answer:

log(x + 4) = 2

x + 4 = 10^{2}

x + 4 = 100

x = 96

Check: Since 96 + 4 > 0, we know log(96 + 4) is defined.

Therefore, the solution to this equation is 96.

c. log_{2}(1 – x) = 4

Answer:

1 – x = 2^{4}

x = -15

Check: Since 1 – (-15) > 0, we know log_{2}(1 – (-15)) is defined.

Therefore, the solution to this equation is -15.

d. log_{2}(49x^{2}) = 4

Answer:

log_{2}[(7x)^{2}] = 4

2 log_{2}(7x) = 4

log_{2}(7x) = 2

7x = 2^{2}

x = \(\frac{4}{7}\)

Check: Since 49(\(\frac{4}{7}\))^{2} > 0, we know log_{2}(49(\(\frac{4}{7}\))^{2}) is defined.

Therefore, the solution to this equation is \(\frac{4}{7}\).

e. log_{2}(9x^{2} + 30x + 25) = 8

Answer:

log_{2}[(3x + 5)^{2}] = 8

2 . log_{2}(3x + 5) = 8

log_{2}(3x + 5) = 4

3x + 5 = 2^{4}

3x + 5 = 16

3x = 11

x = \(\frac{11}{3}\)

Check:

Since 9\(\left(\frac{11}{3}\right)^{2}\) + 30\(\frac{11}{3}\) + 25 = 256, and 256 > 0, log_{2}(9\(\left(\frac{11}{3}\right)^{2}\) + 30\(\frac{11}{3}\) + 25) is defined.

Therefore, the solution to this equation is \(\frac{11}{3}\).

Question 2.

Solve the following logarithmic equations.

a. ln(x^{6}) = 36

Answer:

6 . ln(x) = 36

ln(x) = 6

x = e6

Check:

Since e^{6} > 0, we know ln((e^{6})^{6}) is defined.

Therefore, the only solution to this equation is e^{6}.

b. log[(2x^{2} + 45x – 25)^{5}] = 10

Answer:

5 . log(2x^{2} + 45x – 25) = 10

log(2x^{2} + 45x – 25) = 2

2x^{2} + 45x – 25 = 10^{2}

2x^{2} + 45x – 125 = 0

2x^{2} +50x – 5x – 125 = 0

2x(x + 25) – 5(x + 25) = 0

(2x – 5)(x + 25) = 0

Check: Since 2x^{2} + 45x – 25 > 0 for both x = -25 and x = \(\frac{5}{2}\) we know the left side of the equation is defined at these values.

Therefore, the two solutions to this equation are -25 and \(\frac{5}{2}\)

c. log[(x^{2} + 2x – 3)^{4}] = 0

Answer:

4 log(x^{2} + 2x – 3) = 0

log(x^{2} + 2x – 3) = 0

x^{2} + 2x – 3 = 100

x^{2} + 2x – 3 = 1

x^{2} + 2x – 4=0

x = \(\frac{-2 \pm \sqrt{4+16}}{2}\) = -1 Â± âˆš5

Check:

Since x^{2} + 2x – 3 = 1 when x = -1 + âˆš5 or x = -1 – âˆš5, we know the logarithm is defined for these values of x.

Therefore, the two solutions to the equation are -1 +âˆš5 and -1 – âˆš5.

Question 3.

Solve the following logarithmic equations.

a. log(x) + log(x – 1) = log(3x + 12)

Answer:

log(x) + log(x – 1) = log(3x + 12)

log(x(x – 1)) = log(3x + 12)

x(x – 1) = 3x + 12

x^{2} – 4x – 12 = 0

(x + 2)(x – 6) = 0

Check: Since log(-2) Â¡s undefined, -2 Â¡s an extraneous solution.

Therefore, the only solution to this equation is 6.

b. ln(32x^{2}) – 3 In(2) = 3

Answer:

ln(32x^{2}) – ln(2^{3}) = 3

ln(\(\frac{32 x^{2}}{8}\)) = 3

4x^{2} = e^{3}

x^{2} = \(\frac{e^{3}}{4}\)

x = \(\frac{\sqrt{e^{3}}}{2}\) or x = –\(\frac{\sqrt{e^{3}}}{2}\)

Check:

Since the value of x Â¡n the logarithmic expression is squared, ln(32x^{2}) is defined for any nonzero value of x.

Therefore, both \(\frac{\sqrt{e^{3}}}{2}\) and –\(\frac{\sqrt{e^{3}}}{2}\) are valid solutions to this equation.

c. log(x) + log(-x) = 0

Answer:

log(x(-x)) = o

log(-x^{2}) = 0

-x^{2} = 100

x^{2} = -1

Since there is no real number x so that x^{2} = -1, there is no solution to this equation.

d. log(x + 3) + log(x + 5) = 2

Answer:

log((x + 3)(x + 5)) = 21

(x + 3)(x + 5) = 10^{2}

x^{2} + 8x + 15 – 100 = 0

x^{2} + 8x – 85 = 0

x = \(\frac{-8 \pm \sqrt{64+340}}{2}\)

= -4 Â± âˆš101

Check:

The left side of the equation is not defined for x = -4 -âˆš101, but Â¡t is for x = -4 + âˆš101.

Therefore, the only solution to this equation is x = -4 + âˆš101.

e. log(10x + 5) – 3 = log(x – 5)

Answer:

log(10x + 5) – log(x – 5) = 3

log(\(\frac{10 x+5}{x-5}\)) = 3

\(\frac{10 x+5}{x-5}\) = 10^{3}

\(\frac{10 x+5}{x-5}\) = 1000

10x + 5 = 1000x – 5000

5005 = 990x

x = \(\frac{91}{18}\)

Check:

Both sides of the equation ore defined for x = \(\frac{91}{18}\)

Therefore, the solution to this equation is \(\frac{91}{18}\)

f. log_{2}(x) + log_{2}(2x) + log_{2}(3x) + log_{2}(36) = 6

Answer:

log_{2}(x . 2x . 3x . 36) = 6

log_{2}2(6^{3} x^{3}) = 6

log_{2} [(6x)^{3}] = 6

3 log_{2}(6x) = 6

log_{2}(6x) = 2

6x = 2^{2}

x = \(\frac{2}{3}\)

Check:

Since \(\frac{2}{3}\) > 0, all logarithmic expressions in this equation are defined for x = \(\frac{2}{3}\).

Therefore, the solution to this equation is \(\frac{2}{3}\).

Question 4.

Solve the following equations.

a. log_{2}(x) = 4

Answer:

16

b. log_{6} (x) = 1

Answer:

6

c. log_{3}(x) = -4

Answer:

\(\frac{1}{81}\)

d. log_{âˆš2(x)} = 4

Answer:

4

e. log_{âˆš5(x)} = 3

Answer:

5âˆš5

f. log_{3}(x^{2}) = 4

Answer:

9, -9

g. log_{2}(x^{3}) = 12

Answer:

\(\frac{1}{16}\)

h. log_{3}(8x + 9) = 4

Answer:

9

i. 2 = log_{4}(3x – 2)

Answer:

6

j. log_{5}(3 – 2x)

Answer:

1

k. ln(2x) = 3

Answer:

\(\frac{e^{3}}{2}\)

l. log_{3}(x^{2} – 3x + 5) = 2

Answer:

4, -1

m. log((x^{2} + 4)^{5}) = 10

Answer:

4âˆš6, -4âˆš6

n. log(x) + log(x + 21) = 2

Answer:

4

o. log_{4}(x – 2) + log_{4}(2x) = 2

Answer:

4

p. log(x) – log(x + 3) = -1

Answer:

\(\frac{1}{3}\)

q. log_{4}(x + 3) – log_{4}(x – 5) = 2

Answer:

\(\frac{83}{15}\)

r. log(x) + 1 = log(x + 9)

Answer:

1

s. log_{3}(x^{2} – 9) – log_{3}(x + 3) = 1

Answer:

6

t. 1 – log_{8}(x – 3) = log_{8}(2x)

Answer:

4

u. log_{2}(x^{2} – 16) – log_{2}(x – 4) = 1

Answer:

No Solution

v. log(\(\sqrt{(x+3)^{3}}\)) = \(\frac{3}{2}\)

Answer:

7

w. ln(4x^{2} – 1) = 0

Answer:

\(\frac{\sqrt{2}}{2}\), –\(\frac{\sqrt{2}}{2}\)

x. ln(x + 1) – ln(2) = 1

Answer:

2e – 1

### Eureka Math Algebra 2 Module 3 Lesson 14 Exit Ticket Answer Key

Find all solutions to the following equations. Remember to check for extraneous solutions.

Question 1.

log_{2}(3x + 7) = 4

Answer:

log_{2}(3x + 7) = 4

3x + 7 = 2^{4}

3x = 16 – 7

x = 3

Since 3(3) + 7 > 0, we know 3 is a valid solution to the equation.

Question 2.

log(x – 1) + log(x – 4) = 1

Answer:

log((x – 1)(x – 4)) = 1

log(x^{2} – 5x + 4) = 1

x^{2} – 5x + 4 = 10

x^{2} – 5x – 6 = 0

(x – 6)(x + 1) = 0

x = 6 or x = -1

Check:

Since the left side of the equation is not defined for x = -, this is an extraneous solution. Therefore, the only valid solution is 6.