Eureka Math Algebra 2 Module 3 Lesson 15 Answer Key

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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)
Answer:
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)
Answer:
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))
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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?
Answer:
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
Answer:
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.)
Answer:
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
Answer:
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
Answer:
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
Answer:
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
Answer:
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. \(\frac{7.2 \times 10^{9}}{1.3 \times 10^{5}}\)
Answer:
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).
Answer:
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).
Answer:
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)
Answer:
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))
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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)
Answer:
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)
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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).
Answer:
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).
Answer:
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 – \(\sqrt{x^{2}-1}\)?
Answer:
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)
Answer:
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) = \(\frac{\log (14-x)}{\log (100)}\)
Answer:
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)
Answer:
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)
Answer:
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) = \(\frac{2 \ln (3)+\ln (x)}{\ln (10)}\)
Answer:
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.
Answer:
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.
Taking the logarithm of each side,
log(R) = log (\(\left(\frac{6.14}{5.10} \cdot 10^{2}\right)\))
= 2 + log(\(\frac{6.14}{5.10}\))
= 2 + log(6.14) – log(5.10)
≈ 2 +0.7882 – 0.7076
≈ 2. 0806.
Find 0.0806 in the table entries to estimate R.
Look up 0.0806, which is closest to log(1.20). Note that 2 + 0.0806 ≈ log(100) + log(1.20), so log(120) ≈ 2.0806. Therefore, the surface area of Jupiter is approximately 120 times that of Earth.

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