# 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
104 = 10,000

b. log(âˆš10) = $$\frac{1}{2}$$
10$$\frac{1}{2}$$ = âˆš10

c. log2(256) = 8
28= 256

d. log4(256) = 4
44 = 256

e. ln(1)=0
e0 = 1

f. log(x + 2) = 3
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
log(3x + 7) = 0
100 = 3x + 7
1 = 3x +7
x = -2

Example 2.
log2(x + 5) = 4
log2(x + 5) = 4
24 = x +5
16 = x +5
x = 11

Example 3.
log(x + 2) + log(x + 5) = 1
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.
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 $$\left(\left(4 \cdot \frac{e^{3}}{5}\right)^{5}\right)$$ Â¡s defined. Thus, $$\frac{e^{3}}{4}$$ is the solution to the equation.

Exercise 3.
log((2x + 5)2) = 4
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 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.
log2((5x + 7)19) = 57
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
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
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
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)
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)
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) = $$\frac{5}{2}$$
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
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
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
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
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
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
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
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 Â± âˆš5
Check:
Since x2 + 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)
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) Â¡s undefined, -2 Â¡s an extraneous solution.
Therefore, the only solution to this equation is 6.

b. ln(32x2) – 3 In(2) = 3
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 Â¡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
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
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 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)
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
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
16

b. log6 (x) = 1
6

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

d. logâˆš2(x) = 4
4

e. logâˆš5(x) = 3
5âˆš5

f. log3(x2) = 4
9, -9

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

h. log3(8x + 9) = 4
9

i. 2 = log4(3x – 2)
6

j. log5(3 – 2x)
1

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

l. log3(x2 – 3x + 5) = 2
4, -1

m. log((x2 + 4)5) = 10
4âˆš6, -4âˆš6

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

o. log4(x – 2) + log4(2x) = 2
4

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

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

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

s. log3(x2 – 9) – log3(x + 3) = 1
6

t. 1 – log8(x – 3) = log8(2x)
4

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

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

w. ln(4x2 – 1) = 0
$$\frac{\sqrt{2}}{2}$$, –$$\frac{\sqrt{2}}{2}$$

x. ln(x + 1) – ln(2) = 1
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
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