Eureka Math Algebra 2 Module 3 Lesson 14 Answer Key

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:
10⁴ = 10,000

b. log(√10) = \(\frac{1}{2}\)
Answer:
10^{\(\frac{1}{2}\)} = √10

c. log₂(256) = 8
Answer:
2⁸= 256

d. log₄(256) = 4
Answer:
4⁴ = 256

e. ln(1)=0
Answer:
e⁰ = 1

f. log(x + 2) = 3
Answer:
x + 2 = 10³

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
10⁰ = 3x + 7
1 = 3x +7
x = -2

Example 2.
log₂(x + 5) = 4
Answer:
log₂(x + 5) = 4
2⁴ = 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)= 10¹
x² + 7x + 10 = 10
x² + 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 log₂[(x + 1)⁴] = 8 cannot be solved because he expanded (x + 1)⁴ = x⁴ + 4x³ + 6x² + 4x + 1 and realized that he cannot solve the equation x⁴ + 4x³+ 6x² + 4x + 1 = 2⁸. Is he correct? Explain how you know.
Answer:
If we apply the logarithmic properties, this equation is solvable.
log₂[(x+ 1)⁴] = 8
4log₂(x+ 1) = 8
log₂(x + 1) = 2
x + 1 = 2²
x = 3
Check: If x = 3, then log₂[(3 + 1)⁴] = 4 log₂(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
e³ = 4x
x = \(\frac{e^{3}}{4}\)
Check: Since 4(\(\frac{e^{3}}{4}\)) > 0, we know that ln \(\left(\left(4 \cdot \frac{e^{3}}{5}\right)^{5}\right)\) ¡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
10²= 2x + 5
100 = 2x + 5
95 = 2x
x = \(\frac{95}{2}\)

Check: Since 2(\(\frac{95}{2}\)) + 5 ≠ 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.
log₂((5x + 7)¹⁹) = 57
Answer:
19 . log₂(5x + 7) = 57
log₂(5x + 7) = 3
2³ = 5x + 7
8 = 5x + 7
1 = 5x
x = \(\frac{1}{5}\)
Check:
Since 5(\(\frac{1}{5}\)) + 7 > 0, we know that log₂(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(x² + 7x + 12) – log(x +4) = 0
Answer:
log\(\left(\frac{x^{2}+7 x+12}{x+4}\right)\) = 0
\(\frac{x^{2}+7 x+12}{x+4}\) = 10⁰
\(\frac{x^{2}+7 x+12}{x+4}\) = 1
x² + 7x + 12 = x + 4
0 = x² + 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.
log₂(3x) + log₂(4) = 4
Answer:
log₂(3x) + 2 = 4
log₂(3x) = 2
2² = 3x
4 = 3x
x = \(₂\)
Check:
Since \(\frac{4}{3}\) > 0, log₂(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 = x² + 4x + 4
0 = x² + 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
x² + 5x +6= 12
x² + 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 = 100√10
Check: Since 100√10 > 0, we know log(100√10) is defined.
Therefore, the solution to this equation is 100√10.

b. 5log(x + 4) = 10
Answer:
log(x + 4) = 2
x + 4 = 10²
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. log₂(1 – x) = 4
Answer:
1 – x = 2⁴
x = -15
Check: Since 1 – (-15) > 0, we know log₂(1 – (-15)) is defined.
Therefore, the solution to this equation is -15.

d. log₂(49x²) = 4
Answer:
log₂[(7x)²] = 4
2 log₂(7x) = 4
log₂(7x) = 2
7x = 2²
x = \(\frac{4}{7}\)
Check: Since 49(\(\frac{4}{7}\))² > 0, we know log₂(49(\(\frac{4}{7}\))²) is defined.
Therefore, the solution to this equation is \(\frac{4}{7}\).

e. log₂(9x² + 30x + 25) = 8
Answer:
log₂[(3x + 5)²] = 8
2 . log₂(3x + 5) = 8
log₂(3x + 5) = 4
3x + 5 = 2⁴
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, log₂(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(x⁶) = 36
Answer:
6 . ln(x) = 36
ln(x) = 6
x = e6
Check:
Since e⁶ > 0, we know ln((e⁶)⁶) is defined.
Therefore, the only solution to this equation is e⁶.

b. log[(2x² + 45x – 25)⁵] = 10
Answer:
5 . log(2x² + 45x – 25) = 10
log(2x² + 45x – 25) = 2
2x² + 45x – 25 = 10²
2x² + 45x – 125 = 0
2x² +50x – 5x – 125 = 0
2x(x + 25) – 5(x + 25) = 0
(2x – 5)(x + 25) = 0
Check: Since 2x² + 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[(x² + 2x – 3)⁴] = 0
Answer:
4 log(x² + 2x – 3) = 0
log(x² + 2x – 3) = 0
x² + 2x – 3 = 100
x² + 2x – 3 = 1
x² + 2x – 4=0
x = \(\frac{-2 \pm \sqrt{4+16}}{2}\) = -1 ± √5
Check:
Since x² + 2x – 3 = 1 when x = -1 + √5 or x = -1 – √5, we know the logarithm is defined for these values of x.
Therefore, the two solutions to the equation are -1 +√5 and -1 – √5.

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
x² – 4x – 12 = 0
(x + 2)(x – 6) = 0
Check: Since log(-2) ¡s undefined, -2 ¡s an extraneous solution.
Therefore, the only solution to this equation is 6.

b. ln(32x²) – 3 In(2) = 3
Answer:
ln(32x²) – ln(2³) = 3
ln(\(\frac{32 x^{2}}{8}\)) = 3
4x² = e³
x² = \(\frac{e^{3}}{4}\)
x = \(\frac{\sqrt{e^{3}}}{2}\) or x = –\(\frac{\sqrt{e^{3}}}{2}\)
Check:
Since the value of x ¡n the logarithmic expression is squared, ln(32x²) 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(-x²) = 0
-x² = 100
x² = -1
Since there is no real number x so that x² = -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) = 10²
x² + 8x + 15 – 100 = 0
x² + 8x – 85 = 0
x = \(\frac{-8 \pm \sqrt{64+340}}{2}\)
= -4 ± √101
Check:
The left side of the equation is not defined for x = -4 -√101, but ¡t is for x = -4 + √101.
Therefore, the only solution to this equation is x = -4 + √101.

e. log(10x + 5) – 3 = log(x – 5)
Answer:
log(10x + 5) – log(x – 5) = 3
log(\(\frac{10 x+5}{x-5}\)) = 3
\(\frac{10 x+5}{x-5}\) = 10³
\(\frac{10 x+5}{x-5}\) = 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. log₂(x) + log₂(2x) + log₂(3x) + log₂(36) = 6
Answer:
log₂(x . 2x . 3x . 36) = 6
log₂2(6³ x³) = 6
log₂ [(6x)³] = 6
3 log₂(6x) = 6
log₂(6x) = 2
6x = 2²
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. log₂(x) = 4
Answer:
16

b. log₆ (x) = 1
Answer:
6

c. log₃(x) = -4
Answer:
\(\frac{1}{81}\)

d. log_√2(x) = 4
Answer:
4

e. log_√5(x) = 3
Answer:
5√5

f. log₃(x²) = 4
Answer:
9, -9

g. log₂(x³) = 12
Answer:
\(\frac{1}{16}\)

h. log₃(8x + 9) = 4
Answer:
9

i. 2 = log₄(3x – 2)
Answer:
6

j. log₅(3 – 2x)
Answer:
1

k. ln(2x) = 3
Answer:
\(\frac{e^{3}}{2}\)

l. log₃(x² – 3x + 5) = 2
Answer:
4, -1

m. log((x² + 4)⁵) = 10
Answer:
4√6, -4√6

n. log(x) + log(x + 21) = 2
Answer:
4

o. log₄(x – 2) + log₄(2x) = 2
Answer:
4

p. log(x) – log(x + 3) = -1
Answer:
\(\frac{1}{3}\)

q. log₄(x + 3) – log₄(x – 5) = 2
Answer:
\(\frac{83}{15}\)

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

s. log₃(x² – 9) – log₃(x + 3) = 1
Answer:
6

t. 1 – log₈(x – 3) = log₈(2x)
Answer:
4

u. log₂(x² – 16) – log₂(x – 4) = 1
Answer:
No Solution

v. log(\(\sqrt{(x+3)^{3}}\)) = \(\frac{3}{2}\)
Answer:
7

w. ln(4x² – 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.
log₂(3x + 7) = 4
Answer:
log₂(3x + 7) = 4
3x + 7 = 2⁴
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(x² – 5x + 4) = 1
x² – 5x + 4 = 10
x² – 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.