## 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(x^{2}) = log(49)

Answer:

x^{2} = 49

x = 7 or x = -7

Check: Both solutions are valid since 7^{2}, (-7^{2}) 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

x^{2} – x – 2 = 7x – 17

x^{2} – 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(x^{2} + 1) = log(x(x – 2))

Answer:

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

= x^{2} – 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

x^{2} + 3x – 4= 3x

x^{2} – 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(x^{2} – x) – log(x – 2) = log(x – 3)

Answer:

log(x – 2) + log(x – 3) = log(x^{2} – x)

log((x – 2)(x – 3)) = log(x^{2} – x)

(x – 2)(x – 3) = x^{2} – x

x^{2} – 5x + 6 = x^{2} – 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(x^{2} – 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(x^{3} – x) = log(x^{3})

x^{3} – x = x^{3}

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

x^{2} – 6x + 8 = 1

x^{2} – 6x + 7 = 0

x = 3 Â± âˆš2

log(x – 4) + log(x – 2) = 0

log((x – 4)(x – 2)) = log(1)

(x – 4)(x – 2) = 1

x^{2} -6x + 8 = 1

x^{2} – 6x + 7 = 0

x = 3 Â± âˆš2

Check:

If x = 3 – âˆš2, then x < 2, so log(x – 2) is undefined. Thus, 3 – âˆš2 is an extraneous solution. The only valid solution to this equation is 3 + âˆš2.

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) â‰ˆ log(3.19).

Therefore, x â‰ˆ 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) â‰ˆ log(3.13) – 1

â‰ˆ log(3.13) – log(10)

â‰ˆ log(\(\frac{3.13}{10}\))

â‰ˆ log(0. 313).

Therefore, x â‰ˆ 0.313.

Alternatively, -.5044 is the opposite of 0.5044, so x is the reciprocal of the answer in part (a). Thus, x â‰ˆ 3.19^{-1} â‰ˆ 0.313.

c. log(x) = 35.5044

Answer:

log(x) = 35 + 0.5044

= log(10^{35}) + 0.5044

â‰ˆ log(10^{35}) + log(3.19)

â‰ˆ log(3.19 Ã— 10^{35})

Therefore, x â‰ˆ 3.19 Ã— 10^{35})

d. log(x) = 4.9201

Answer:

log(x) = 4 + 0.9201

= log(10^{4}) + 0.9201

â‰ˆ log(^{4}) + log(8.32)

â‰ˆ log(8.32 Ã— 10^{4})

Therefore, x â‰ˆ 83200.

Question 2.

Use logarithms and the logarithm table to evaluate each expression.

a. âˆš2.33

Answer:

log ((2.33)^{\(\frac{1}{2}\)}) = \(\frac{1}{2}\)log(2.33) â‰ˆ (0.3674) â‰ˆ 0.1837

Thus, log(âˆš2.33) â‰ˆ 0. 1837, and locating 0.1837 in the logarithm table gives a value of approximately 1.53. Therefore, âˆš2.33 â‰ˆ 1.53.

b. 13500 . 3600

Answer:

log(1 . 35 . 10^{4} . 3 . 6 . 10^{3}) = log(1.35) + log(3.6) + 4 + 3

â‰ˆ 0.1303 + 0.5563 + 7

â‰ˆ 7.6866

Thus, log(13500 . 3600) â‰ˆ 7.6866. Locating 0.6866 in the logarithm table gives a value of 4.86. Therefore, the product is approximately 4.86 Ã— 10^{7}

c. \(\frac{7.2 \times 10^{9}}{1.3 \times 10^{5}}\)

Answer:

log(7.2) + log(10^{9}) – log(1.3) – log(10^{5}) â‰ˆ 0.8573 + 9 – 0.1139 – 5 â‰ˆ 4.7434

Locating 0.7434 in the logarithm table gives 5.54. So, the quotient is approximately 5.54 Ã— 10^{4}.

Question 3.

Solve for x: log(3) + 2 log(x) = log(27).

Answer:

log(3) + 2 log(x) = log(27)

log(3) + log(x^{2}) = log(27)

log(3x^{2}) = log(27)

3x^{2} = 27

x^{2} = 9

x = Â±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(3x^{2} + 12x) = log(15)

3x^{2} + 12x = 15

3x^{2} + 12x – 15 = 0

3x^{2} + 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(y^{2} – z^{2})

x = y^{2} – z^{2}

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(y^{2})

x = y^{2}

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(y^{2})

xz = y^{2}

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(1000x^{2}) = 3 + log(x^{2}) = 3 + 2 log(x), we see that log(y) = log(1000x^{2}). Thus, y = 1000x^{2}.

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)

x^{2} – 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)

x^{2} – 4 = 9x – 24

x^{2} – 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(x^{2} – 4) = ln(-2x – 1)

x^{2} – 4 = -2x – 1

x^{2} + 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 = P_{0}(1 + r)^{t} gives the population of a city P growing at an annual percentage rate r, where P_{0} is the population t years ago.

a. Find the time t it takes this population to double.

Answer:

Let P = 2P_{0}.

Then,

2P_{0} = P_{0}(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 = a^{2} – b^{2}.

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 â‰ 24, we know that log(9) â‰ 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 â‰ 4, so we know that log(2) + log(3) + log(4) â‰ 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 â‰ 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 â‰¥ 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 )â€¢ 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 – \(\sqrt{x^{2}-1}\)) = x^{2} + x\(\sqrt{x^{2}-1}\) – x\(\sqrt{x^{2}-1}\) – (\(\sqrt{x^{2}-1}\))^{2}

= x^{2} – (x^{2} – 1) = 1.

Question 16.

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

a. log(x) = log_{100}(x^{2} – 2x + 6)

Answer:

log(x) = \(\frac{\log \left(x^{2}-2 x+6\right)}{\log (100)}\)

log(x) = \(\frac{1}{2}\)log(x^{2} – 2x + 6)

2log(x)= log(x^{2} – 2x+6)

log(x2) = log(x^{2} – 2x +6)

x^{2} = x^{2} – 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

x^{2} – 4x +4= 14 – x

x^{2} – 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. log_{2}(x + 1) = log4(x^{2} + 3x + 4)

Answer:

log_{2}(x + 1) = log_{4}(x^{2} + 3x + 4)

log_{2}(x + 1) = \(\frac{\log _{2}\left(x^{2}+3 x+4\right)}{\log _{2}(4)}\)

2log_{2}(x + 1) = log_{2}(x^{2} + 3x + 4)

log_{2}((x + 1)2) = log2(x^{2} + 3x + 4)

(x + 1)^{2} = x^{2} + 3x + 4

x^{2} + 2x + 1 = x^{2} + 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. log_{2}(x – 1) = log_{8}(x^{3} – 2x^{2} – 2x + 5)

Answer:

log_{2}(x – 1) = \(\frac{\log _{2}\left(x^{3}-2 x^{2}-2 x+5\right)}{\log _{2}(8)}\)

3 log_{2}(x – 1) = log_{2}(x^{3} -2x^{2} – 2x + 5)

log_{2}((x – i)) = log_{2}(x3 – 2×2 – 2x + 5)

(x – 1)^{3} = x^{3} – 2x^{2} – 2x + 5

x^{3} – 3x^{2} + 3x – 1 = x^{3} – 2x^{2} -2x + 5

x^{2} – 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 Ã— 10^{10} km^{2}, and the surface area of Earth is 5.10 Ã— 10^{8} km^{2}. 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 = \(\frac{6.14 \times 10^{10}}{5.10 \times 10^{8}}=\frac{6.14}{5.10}\) . 10^{2}.

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)

â‰ˆ 2 +0.7882 – 0.7076

â‰ˆ 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 â‰ˆ log(100) + log(1.20), so log(120) â‰ˆ 2.0806. Therefore, the surface area of Jupiter is approximately 120 times that of Earth.