# Eureka Math Algebra 2 Module 3 Lesson 15 Answer Key

## Engage NY Eureka Math Algebra 2 Module 3 Lesson 15 Answer Key

### Eureka Math Algebra 2 Module 3 Lesson 15 Exercise Answer Key

Exercises:

Exercise 1.
Solve the following equations. Remember to check for extraneous solutions because logarithms are only defined for positive real numbers.

a. log(x2) = log(49)
x2 = 49
x = 7 or x = -7
Check: Both solutions are valid since 72, (-72) are both positive numbers.
The two solutions are 7 and -7.

b. log(x + 1) + log(x + 2) = log(7x – 17)
log((x + 1)(x – 2)) = log(7x – 17)
(x + 1)(x – 2) = 7x – 17
x2 – x – 2 = 7x – 17
x2 – 8x + 15 = 0
(x – 5)(x – 3) = 0
x = 3 or x = 5
Check: Since x + 1, x – 2, and 7x – 17 are all positive for either x = 3 or x = 5, both solutions are valid.
Thus, the solutions to this equation are 3 and 5.

c. log(x2 + 1) = log(x(x – 2))
x2 + 1 = x(x- 2)
= x2 – 2x
1 = -2x
x = –$$\frac{1}{2}$$
Check:
Both (-$$\frac{1}{2}$$)2 + 1 > 0 and –$$\frac{1}{2}$$ (-$$\frac{1}{2}$$ – 2) > 0, so the solution –$$\frac{1}{2}$$ is valid.
Thus, –$$\frac{1}{2}$$ is the only valid solution to this equation.

d. log(x +4) + log(x – 1) = log(3x)
log((x + 4)(x – 1)) = log(3x)
(x + 4)(x – 1) = 3x
x2 + 3x – 4= 3x
x2 – 4 = 0
x = 2 or x = -2
Check:
Since log(3x) is undefined when x = -2, there is an extraneous solution of x = -2. The only valid solution to this equation is 2.

e. log(x2 – x) – log(x – 2) = log(x – 3)
log(x – 2) + log(x – 3) = log(x2 – x)
log((x – 2)(x – 3)) = log(x2 – x)
(x – 2)(x – 3) = x2 – x
x2 – 5x + 6 = x2 – x
4x = 6
x = $$\frac{3}{2}$$
Check: When x = $$\frac{3}{2}$$, we have x – 2 < 0, so log(x – 2), log(x – 3), and log(x2 – x) are all undefined. So, the solution x = $$\frac{3}{2}$$ is extraneous.
There are no valid solutions to this equation.

f. log(x) + log(x – 1) + log(x + 1) = 3 log(x)
log(x(x – 1)(x + 1)) = log(x3)
log(x3 – x) = log(x3)
x3 – x = x3
x = 0.
Since log(0) is undefined, x = 0 is an extraneous solution.
There are no valid solutions to this equation.

g. log(x – 4)= -log(x – 2)
Two possible approaches to solving this equation are shown.
log(x – 4) = log($$\frac{1}{x-2}$$)
x – 4 = $$\frac{1}{x-2}$$
(x – 4)(x – 2) = 1
x2 – 6x + 8 = 1
x2 – 6x + 7 = 0
x = 3 ± √2

log(x – 4) + log(x – 2) = 0
log((x – 4)(x – 2)) = log(1)
(x – 4)(x – 2) = 1
x2 -6x + 8 = 1
x2 – 6x + 7 = 0
x = 3 ± √2
Check:
If x = 3 – √2, then x < 2, so log(x – 2) is undefined. Thus, 3 – √2 is an extraneous solution. The only valid solution to this equation is 3 + √2.

Exercise 2.
How do you know if you need to use the definition of a logarithm to solve an equation involving logarithms as we did in Lesson 15 or if you can use the methods of this lesson?
If the equation involves only logarithmic expressions, then it can be reorganized to be of the form log(X) = log(Y)
and then solved by equating X = Y. If there are constants involved, then the equation can be solved using the definition and properties of logarithms.

### Eureka Math Algebra 2 Module 3 Lesson 15 Problem Set Answer Key

Question 1.
Use the table of logarithms to approximate solutions to the following logarithmic equations:

a. log(x) = 0.5044
In the table, 0.5044 is closest to log(3.19), so log(x) ≈ log(3.19).
Therefore, x ≈ 3. 19.

b. log(x) = -0.5044 (Hint: Begin by writing -0.5044 as [(-0.5044) + 1] – 1.)
log(x) = [(-0.5044) + 1] – 1
= 0.4956 – 1
In the table, 0.4956 is closest to log(3.13), so
log(x) ≈ log(3.13) – 1
≈ log(3.13) – log(10)
≈ log($$\frac{3.13}{10}$$)
≈ log(0. 313).
Therefore, x ≈ 0.313.
Alternatively, -.5044 is the opposite of 0.5044, so x is the reciprocal of the answer in part (a). Thus, x ≈ 3.19-1 ≈ 0.313.

c. log(x) = 35.5044
log(x) = 35 + 0.5044
= log(1035) + 0.5044
≈ log(1035) + log(3.19)
≈ log(3.19 × 1035)
Therefore, x ≈ 3.19 × 1035)

d. log(x) = 4.9201
log(x) = 4 + 0.9201
= log(104) + 0.9201
≈ log(4) + log(8.32)
≈ log(8.32 × 104)
Therefore, x ≈ 83200.

Question 2.
Use logarithms and the logarithm table to evaluate each expression.

a. √2.33
log ((2.33)$$\frac{1}{2}$$) = $$\frac{1}{2}$$log(2.33) ≈ (0.3674) ≈ 0.1837
Thus, log(√2.33) ≈ 0. 1837, and locating 0.1837 in the logarithm table gives a value of approximately 1.53. Therefore, √2.33 ≈ 1.53.

b. 13500 . 3600
log(1 . 35 . 104 . 3 . 6 . 103) = log(1.35) + log(3.6) + 4 + 3
≈ 0.1303 + 0.5563 + 7
≈ 7.6866
Thus, log(13500 . 3600) ≈ 7.6866. Locating 0.6866 in the logarithm table gives a value of 4.86. Therefore, the product is approximately 4.86 × 107

c. $$\frac{7.2 \times 10^{9}}{1.3 \times 10^{5}}$$
log(7.2) + log(109) – log(1.3) – log(105) ≈ 0.8573 + 9 – 0.1139 – 5 ≈ 4.7434
Locating 0.7434 in the logarithm table gives 5.54. So, the quotient is approximately 5.54 × 104.

Question 3.
Solve for x: log(3) + 2 log(x) = log(27).
log(3) + 2 log(x) = log(27)
log(3) + log(x2) = log(27)
log(3x2) = log(27)
3x2 = 27
x2 = 9
x = ±3
Because log(x) is only defined for positive real numbers x, the only solution to the equation is 3.

Question 4.
Solve for x: log(3x) + log(x +4) = log(15).
log(3x2 + 12x) = log(15)
3x2 + 12x = 15
3x2 + 12x – 15 = 0
3x2 + 15x – 3x – 15 = 0
3x(x + 5) – 3(x + 5) = 0
(3x – 3)(x + 5) = 0
Thus, 1 and -5 solve the quadratic equation, but -5 is an extraneous solution to the logarithmic equation. Hence, 1 is the only solution.

Question 5.
Solve for x.

a. log(x) = log(y + z) + log(y – z)
log(x) = log(y + z) + log(y – z)
log(x) = log((y + z)(y – z))
log(x) = log(y2 – z2)
x = y2 – z2

b. log(x) = (log(y) + log(z)) + (log(y) – log(z))
log(x) = (log(y) + log(z)) + (log(y) – log(z))
log(x) = 2 log(y)
log(x) = log(y2)
x = y2

Question 6.
If x and y are positive real numbers, and log(y) = 1 + log(x), express y in terms of x.
Since log(10x) = 1 + log(x), we see that log(y) = log(10x). Then y = 10x.

Question 7.
If x, y, and z are positive real numbers, and log(x) – log(y) = log(y) – log(z), express y in terms of x and z.
log(x) – log(y) = log(y) – log(z)
log(x) + log(z) = 2 log(y)
log(xz) = log(y2)
xz = y2
y = $$\sqrt{x z}$$

Question 8.
If x and y are positive real numbers, and log(x) = y(log(y + 1) – log(y)), express x in terms of y.
log(x) = y(log(y + 1) – log(y))
log(x) = y(log($$\left(\frac{y+1}{y}\right)$$)
log(x) = log(($$\left(\frac{y+1}{y}\right)^{y}$$)
x = $$\left(\frac{y+1}{y}\right)^{y}$$

Question 9.
If x and y are positive real numbers, and log(y) = 3 + 2 log(x), express y in terms of x.
Since log(1000x2) = 3 + log(x2) = 3 + 2 log(x), we see that log(y) = log(1000x2). Thus, y = 1000x2.

Question 10.
If x, y, and z are positive real numbers, and log(z) = log(y) + 2 log(x) – 1, express z in terms of x and y.
Since log ($$\frac{x^{2} y}{10}$$) = log(y) + 2 log(x) – 1, we see that log(z) = log ($$\frac{x^{2} y}{10}$$) Thus, z = $$\frac{x^{2} y}{10}$$.

Question 11.
Solve the following equations.

a. ln(10) – ln(7 – x) = ln(x)
Ans:
ln($$\frac{10}{7-x}$$) =ln(x)
$$\frac{10}{7-x}$$ =x
10 = x(7 – x)
x2 – 7x + 10 = 0
(x – 5)(x – 2) = o
x = 2 or x = 5
Check:
If x = 2 or x = 5, then the expressions x and 7 – x are positive.
Thus, both 2 and 5 are valid solutions to this equation.

b. ln(x + 2) + ln(x – 2) = In(9x – 24)
ln((x + 2)(x – 2)) = ln(9x – 24)
x2 – 4 = 9x – 24
x2 – 9x + 20 = 0
(x – 4)(x – 5) = 0
x = 4 or x = 5
Check:
If x = 4 or x = 5, then the expressions x + 2, x – 2, and 9x – 24 are all positive.
Thus, both 4 and 5 are valid solutions to this equation.

c. ln(x + 2) + ln(x – 2) = ln(-2x – 1)
ln((x + 2)(x – 2)) = ln(-2x – 1)
ln(x2 – 4) = ln(-2x – 1)
x2 – 4 = -2x – 1
x2 + 2x – 3 = 0
(x+ 3)(x – 1) = 0
x = -3 or x = 1
So, x = -3 or x = 1, but x = -3 makes the input to both logarithms on the left-hand side negative, and x = 1 makes the input to the second and third logarithms negative. Thus, there are no solutions to the original equation.

Question 12.
Suppose the formula P = P0(1 + r)t gives the population of a city P growing at an annual percentage rate r, where P0 is the population t years ago.

a. Find the time t it takes this population to double.
Let P = 2P0.
Then,
2P0 = P0(1 + r)t
2 =(1 + r)t
log(2) = log(1 + r)t
log(2) = tlog(l + r)
t = $$\frac{\log (2)}{\log (1+r)}$$.

b. Use the structure of the expression to explain why populations with lower growth rates take a longer time to double.
If r is a decimal between 0 and 1, then the denominator will be a number between 0 and log(2). Thus, the value of t will be large for small values of r and getting closer to 1 as r increases.

c. Use the structure of the expression to explain why the only way to double the population in one year is if there is a 100 percent growth rate.
For the population to double, we need to have t = 1. This happens if log(2) = log( 1 + r), and then we have 2 = 1 + r and r = 1.

Question 13.
lf x > 0, a + b > 0, a > b, and log(x) = log(a+b) + log(a – b),find x interms of a and b.
Applying properties of logarithms, we have
log(x) = log(a + b) + log(a – b)
= log((a + b)(a – b))
= log(a2 – b2).
So, x = a2 – b2.

Question 14.
Jenn claims that because log(1) + log(2) + log (3) = log(6), then log(2) + log(3) + log(4) = log(9).

a. Is she correct? Explain how you know.
Jenn is not correct. Even though log(1) + log(2) + log(3) = log(1 . 2 . 3) = log(6), the logarithm properties give log(2) + log(3) + log(4) = log(2 . 3 . 4) = log(24). Since 9 ≠ 24, we know that log(9) ≠ log(24).

b. If log(a) + log(b) + log(c) = log(a + b + c), express c in terms of a and b. Explain how this result relates to your answer to part (a).
log(a) + log(b) + log(c) = log(a + b + c)
log(abc) = log(a + b + c)
abc = a + b + c
abc – c = a + b
c(ab – 1) = a + b
c = $$\frac{a+b}{a b-1}$$
If log(2) + log(3) + log(4) were equal to log(9), then we would have 4 = $$\frac{2+3}{2 \cdot 3-1}$$ However,
$$\frac{2+3}{2 \cdot 3-1}$$ = $$\frac{5}{5}$$ = 1 ≠ 4, so we know that log(2) + log(3) + log(4) ≠ log(9).

c. Find other values of a, b, and c so that log(a) + log(b) + log(c) = log(a + b + c).
Many answers are possible; in fact, any positive values of a and b where ab ≠ 1 and c = $$\frac{a+b}{a b-1}$$ will satisfy log(a) + log(b) + log(c) = log(a + b + c). One such answer is a = 3, b = 7, and c = $$\frac{1}{2}$$.

Question 15.
In Problem 7 of the Lesson 12 Problem Set, you showed that for x ≥ 1, log(x + $$\sqrt{x^{2}-1}$$) + log(x – $$\sqrt{x^{2}-1}$$) = 0. It follows that log(x + $$\sqrt{x^{2}-1}$$) = -log(x – $$\sqrt{x^{2}-1}$$). What does this tell us about the relationship between the expressions x + $$\sqrt{x^{2}-1}$$ and x – $$\sqrt{x^{2}-1}$$?
Since we know log(x + $$\sqrt{x^{2}-1}$$) = – log(x – $$\sqrt{x^{2}-1}$$), and -log(x – $$\sqrt{x^{2}-1}$$) = log ($$\frac{1}{x-\sqrt{x^{2}-1}}$$), we know that log(x + $$\sqrt{x^{2}-1}$$) = log ( 1 )• Then x + $$\sqrt{x^{2}-1}$$ = $$\frac{1}{x-\sqrt{x^{2}-1}}$$. We can verify that these expressions are reciprocals by multiplying them together:
(x + $$\sqrt{x^{2}-1}$$)(x – $$\sqrt{x^{2}-1}$$) = x2 + x$$\sqrt{x^{2}-1}$$ – x$$\sqrt{x^{2}-1}$$ – ($$\sqrt{x^{2}-1}$$)2

= x2 – (x2 – 1) = 1.

Question 16.
Use the change of base formula to solve the following equations.

a. log(x) = log100(x2 – 2x + 6)
log(x) = $$\frac{\log \left(x^{2}-2 x+6\right)}{\log (100)}$$
log(x) = $$\frac{1}{2}$$log(x2 – 2x + 6)
2log(x)= log(x2 – 2x+6)
log(x2) = log(x2 – 2x +6)
x2 = x2 – 2x +6
2x = 6
x = 3
Since both sides of the equation are defined for x = 3, the only solution to this equation is 3.

b. log(x – 2) = $$\frac{\log (14-x)}{\log (100)}$$
log(x – 2) = log(100)
log(x – 2) = $$\frac{1}{2}$$log(14 – x)
2 log(x – 2) = log( 14 – x)
log((x – 2)2) = log(14 – x)
(x – 2)2 = 14 – x
x2 – 4x +4= 14 – x
x2 – 3x – 10 = 0
(x – 5)(x + 2) = 0
Thus, either x = 5 or x = -2. Since the left side of the equation is undefined when x = -2, but both sides are defined for x = 5, the only solution to the equation is 5.

c. log2(x + 1) = log4(x2 + 3x + 4)
log2(x + 1) = log4(x2 + 3x + 4)
log2(x + 1) = $$\frac{\log _{2}\left(x^{2}+3 x+4\right)}{\log _{2}(4)}$$
2log2(x + 1) = log2(x2 + 3x + 4)
log2((x + 1)2) = log2(x2 + 3x + 4)
(x + 1)2 = x2 + 3x + 4
x2 + 2x + 1 = x2 + 3x + 4
2x + 1 = 3x +4
x = -3
Since the left side of the equation is undefined for x = -3, there is no solution to this equation.

d. log2(x – 1) = log8(x3 – 2x2 – 2x + 5)
log2(x – 1) = $$\frac{\log _{2}\left(x^{3}-2 x^{2}-2 x+5\right)}{\log _{2}(8)}$$
3 log2(x – 1) = log2(x3 -2x2 – 2x + 5)
log2((x – i)) = log2(x3 – 2×2 – 2x + 5)
(x – 1)3 = x3 – 2x2 – 2x + 5
x3 – 3x2 + 3x – 1 = x3 – 2x2 -2x + 5
x2 – 5x + 6 = 0
(x – 3)(x – 2) = 0
Since both sides of the equation are defined for x = 3 and x = 2, 2 and 3 are both valid solutions to this equation.

Question 17.
Solve the following equation: log(9x) = $$\frac{2 \ln (3)+\ln (x)}{\ln (10)}$$
Rewrite the left-hand side using the change of base formula:
log(9x) = $$\frac{\ln (9 x)}{\ln (10)}$$
= $$\frac{\ln (9)+\ln (x)}{\ln (10)}$$
= $$\frac{\ln \left(3^{2}\right)+\ln (x)}{\ln (10)}$$
Thus, the equation is true for all x > 0.

### Eureka Math Algebra 2 Module 3 Lesson 15 Exit Ticket Answer Key

Question 1.
The surface area of Jupiter is 6.14 × 1010 km2, and the surface area of Earth is 5.10 × 108 km2. Without using a calculator but using the table of logarithms, find how many times greater the surface area of Jupiter is than the surface area of Earth.
Let R be the ratio of the two surface areas. Then R = $$\frac{6.14 \times 10^{10}}{5.10 \times 10^{8}}=\frac{6.14}{5.10}$$ . 102.
log(R) = log ($$\left(\frac{6.14}{5.10} \cdot 10^{2}\right)$$)
= 2 + log($$\frac{6.14}{5.10}$$)