# Eureka Math Algebra 2 Module 3 Lesson 12 Answer Key

## Engage NY Eureka Math Algebra 2 Module 3 Lesson 12 Answer Key

### Eureka Math Algebra 2 Module 3 Lesson 12 Opening Exercise Answer Key

Opening Exercise:

Use the approximation log(2) ≈ 0.3010 to approximate the values of each of the following logarithmic expressions.

a. log(20)
log(20) = log(10 . 2)
= log(10) + log(2)
≈ 1 + 03010
≈ 1.3010

b. log(0.2)
log(0.2) = log(0.12)
= log(0.1) + log(2)
≈ -1 + 0.3010
≈ -0.6990

c. log(24)
log(24) = log(2 . 2 . 2 . 2)
= log(2 . 2) + log(2 . 2)
= log(2) + log(2) + log(2) + log(2)
≈ 4(0.3010)
≈ 1.2040

### Eureka Math Algebra 2 Module 3 Lesson 12 Exercise Answer Key

Exercises:

For Exercises 1 – 6, explain why each statement below is a property of base-10 logarithms.

Exercise 1.
Property 1: log(1) = 0
Because L = log(x) means 10L = x, then when x = 1, L = 0.

Exercise 2.
Property 2: log(10) = 1.
Because L = log(x) means 10L = x, then when x = 10, L = 1.

Exercise 3.
Property 3: For all real numbers r, log(10r) = r.
Because L = log(x) means 10L = x, then when x = 10r, L = r.

Exercise 4.
Property 4: For any x > 0, 10log(x) = x.
Because L = log(x) means 10L = x, then x = 10log(x)

Exercise 5.
Property 5: For any positive real numbers x and y, log(x . y) = log(x) + log(y).
Hint: Use an exponent rule as well as property 4.
By the rule ab . ac = ab + c, 10log(x) . 10log(y) = 10log(x) + log(y)
By property 4, 10log(x) . 10log(y) = x . y
Therefore, x . y = 10log(x) + log(y) Again, by property 4, x . y = 10log(x.y)
Then,
10log(x.y) = 10log(x) + log(y); so, the exponents must be equal, and log(x . y) = log(x) + log(y)

Exercise 6.
Property 6: For any positive real number x and any real number r, log(xr) = r . log(x).
Hint: Use an exponent rule as well as property 4.
By the rule (ab)c = abc, 10k log(x) = (10log(x))k
By property 4, (10log(x))r = xr.
Therefore, xr = 10r log(x) Again, by property 4, xr = 10log(xr)
Then, 10log(xr) = 10r log(x); 50 the exponents must be equal, and log(xr) = r log(x).

Exercise 7.
Apply properties of logarithms to rewrite the following expressions as a single logarithm or number.

a. $$\frac{1}{2}$$log(25) + log(4)
log(5) + log(4) = log(20)

b. $$\frac{1}{3}$$log(8) + log(16)
log(2) + log(24) = log(32)

c. 3 log(5) + log(0.8)
log(125) + log(0.8) = log(100) = 2

Exercise 8.
Apply properties of logarithms to rewrite each expression as a sum of terms involving numbers, log(x), and log(y),
where x and y are positive real numbers.

a. log(3x2y5)
log(3) + 2 log(x) + 5 log(y)

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

Exercise 9.
In mathematical terminology, logarithms are well defined because if X = Y, then log(X) = log(Y) for X, Y > 0. This means that if you want to solve an equation involving exponents, you can apply a logarithm to both sides of the equation, just as you can take the square root of both sides when solving a quadratic equation. You do need to be careful not to take the logarithm of a negative number or zero.

Use the property stated above to solve the following equations.

a. 1010x = 100
log(1010x) = log(100)
10x = 2
x = $$\frac{1}{5}$$

b. 10x – 1 = $$\frac{1}{10^{x+1}}$$
log(10x – 1) = -log(10x + 1)
x – 1 = -(x+ 1)
2x = 0
x = 0

c. 1002x = 103x – 1
log(1002x) = log(103x – 1)
2xlog(100) = (3x -1)
4x = 3x – 1
x = -1

Exercise 10.
Solve the following equations.

a. 10x = 27
log(10x) = log(27)
x = 7log(2)

b. 10x2 + 1 = 15
log(10x2 + 1) = log(15)
x2 + 1 = log(15)
x= ±$$\sqrt{\log (15)-1}$$

c. 4x = 53
log(4x) = log(53)
xlog(4) = 3log(5)
x = $$\frac{3 \log (5)}{\log (4)}$$

Eureka Math Algebra 2 Module 3 Lesson 12 Problem Set Answer Key

Question 1.
Use the approximate logarithm values below to estimate the value of each of the following logarithms. In which properties you used.

log(2) = 0.3010 log(3) = 0.4771
log(5) = 0.6990 log(7) = 0.8451

a. log(6)
Using property 5,
log(6) = log(3) + log(2) ≈ 0.7781.

b. log(15)
Using property 5,
log(15) = log(3) + log(5) ≈ 1.1761.

c. log(12)
Using properties 5 and 6,
log(12) = log(3) + log(22) = log(3) + 2 log(2) ≈ 1,0791.

d. log(107)
Using property 3,
log(107) = 7.

e. log($$\frac{1}{5}$$)
Using property 7,
log($$\frac{1}{5}$$) = -log(5) ≈ -0.6990.

f. log($$\frac{3}{7}$$)
Using property 8,
log($$\frac{3}{7}$$) = log(3) – log(7) ≈ -0.368.

g. log($$\sqrt[4]{2}$$)
Using property 6,
log($$\sqrt[4]{2}$$) = log (2$$\frac{1}{4}$$) = $$\frac{1}{4}$$log(2) ≈ 0.0753.

Question 2.
Let log(X) = r, log(Y) = s, and log(Z) = t. Express each of the following in terms of r, s, and t.

a. log($$\frac{X}{\boldsymbol{Y}}$$)
r – s

b. log(XZ)
s + t

c. log(Xr)
r2

d. log($$\sqrt[3]{Z}$$)
$$\frac{t}{3}$$

e. log(\sqrt[4]{\frac{\boldsymbol{Y}}{\boldsymbol{Z}}})
$$\frac{s-t}{4}$$

f. log(XY2Z3)
r + 2s + 3t

Question 3.
Use the properties of logarithms to rewrite each expression in an equivalent form containing a single logarithm.

a. log($$\frac{13}{5}$$) + log($$\frac{5}{4}$$)
log($$\frac{13}{4}$$)

b. log($$\frac{5}{6}$$) – log($$\frac{2}{3}$$)
log($$\frac{5}{4}$$)

c. $$\frac{1}{2}$$log(16) + log(3) + log ($$\frac{1}{4}$$)
log(3)

Question 4.
Use the properties of logarithms to rewrite each expression in an equivalent form containing a single logarithm.

a. log(√x) + $$\frac{1}{2}$$log($$\frac{1}{x}$$) + 2 log(x)
log(x2)

b. log($$\sqrt[5]{x}$$) + log($$\sqrt[5]{x^{4}}$$)
log(x)

c. log(x) + 2log(y) – $$\frac{1}{2}$$log(z)
log($$\frac{x y^{2}}{\sqrt{z}}$$)

d. $$\frac{1}{3}$$(log(x) – 3 log(y) + log(z))
log($$\sqrt[3]{\frac{x z}{y^{3}}}$$)

e. 2(log(x) – log(3y)) + 3(log(z) – 2 log(x))
$$\log \left(\left(\frac{x}{3 y}\right)^{2}\right)+\log \left(\left(\frac{z}{x^{2}}\right)^{3}\right)=\log \left(\frac{z^{3}}{9 y^{2} x^{4}}\right)$$

Question 5.
In each of the following expressions, x, y, and z represent positive real numbers. Use properties of logarithms to rewrite each expression in an equivalent form containing only log(x), log(y), log(z), and numbers.

a. log($$\frac{3 x^{2} y^{4}}{\sqrt{z}}$$)
log(3) + 2log(x) + 4log(y) – $$\frac{1}{2}$$log(z)

b. log ($$\frac{42 \sqrt[3]{x y^{7}}}{x^{2} z}$$)
log(42) – $$\frac{5}{3}$$log(x) + $$\frac{7}{3}$$log(y) — log(z)

c. log($$\frac{100 x^{2}}{y^{3}}$$)
2 + 2log(x) – 3 log(y)

d. log($$\sqrt{\frac{x^{3} y^{2}}{10 z}}$$)
$$\frac{1}{2}$$(3log(x) + 2 log(y) – 1 – log(z))

e. log($$\frac{1}{10 x^{2} z}$$)
-1 – 2log(x) – log(z)

Question 6.
Express log ($$\frac{1}{x}-\frac{1}{x+1}$$) + (log($$\frac{1}{x}$$) – log($$\frac{1}{x+1}$$)) as a single logarithm for positive numbers x.
log($$\frac{1}{x}-\frac{1}{x+1}$$) + (log($$\frac{1}{x}$$) – log($$\frac{1}{x+1}$$)) = log($$\frac{1}{x(x+1)}$$) + log($$\frac{1}{x}$$) – log($$\frac{1}{x+1}$$)
= – log(x(x + 1)) – log(x) + log(x + 1)
= -log(x) – log(x + 1) – log(x) + log(x + 1)
= -2log(x)

Question 7.
Show that log(x + $$\sqrt{x^{2}-1}$$) + log(x – $$\sqrt{x^{2}-1}$$) = 0 for x ≥ 1.
log(x + $$\sqrt{x^{2}-1}$$) + log(x – $$\sqrt{x^{2}-1}$$) = log ((x + $$\sqrt{x^{2}-1}$$) (x – $$\sqrt{x^{2}-1}$$))
= log(x2 – ($$\sqrt{x^{2}-1}$$)2)
= log(x2 – x2 + 1)
= log(1)
= 0

Question 8.
If xy = 103.67 for some positive real numbers x and y, find the value of log(x) + log(y).
xy = 103.67
3.67 = log(xy)
log(xy) = 3.67
log(x) + log(y) = 3.67

Question 9.
Solve the following exponential equations by taking the logarithm base 10 of both sides. Leave your answers stated in terms of logarithmic expressions.

a. 10x2 = 320
log(10x2) = log(320)
x2 = log(320)
x = ±$$\sqrt{\log (320)}$$

b. 10$$\frac{x}{8}$$ = 300
log (10$$\frac{x}{8}$$) = log(300)
$$\frac{x}{8}$$ = log(102 . 3)
$$\frac{x}{8}$$ = 2 + log(3)
x = 16 + 8 log(3)

c. 103x = 400
log(103x) = log(400)
3x . log(10) = log(102 . 4)
3x . 1 = 2 + log(4)
x = $$\frac{1}{3}$$(2 + log(4))

d. 52x = 200
log(52x) = log(200)
2x . log(5) = log(100) + log(2)
2x = $$\frac{2+\log (2)}{\log (5)}$$
x = $$\frac{2+\log (2)}{2 \log (5)}$$

e. 3x = 7-3x + 2
log(3x) = log (7-3x + 2)
x log(3) = (-3x + 2)log(7)
x log(3) + 3x log(7) = 2 log(7)
x(log(3) +3 log(7)) = 2 log(7)
x = $$\frac{2 \log (7)}{\log (3)+3 \log (7)}=\frac{\log (49)}{\log (3)+\log (343)}=\frac{\log (49)}{\log (1029)}$$
(Any of the three equivalent forms given above are acceptable answers.)

Question 10.
Solve the following exponential equations.

a. 10x = 3
x = log(3)

b. 10y = 30
y = log(30)

c. 10z = 300
z = log(300)

d. Use the properties of logarithms to justify why x, y, and z form an arithmetic sequence whose constant difference is 1.
Since y = log(30), y = log(10 . 3) = 1 + log(3) = 1 + x.
Similarly, z = 2 + log(3) = 2 + x.
Thus, the sequence x, y, z is the sequence log(3), 1 + log(3), 2 + log(3), and these numbers form an arithmetic sequence whose first term is log(3) with constant difference 1.

Question 11.
Without using a calculator, explain why the solution to each equation must be a real number between 1 and 2.

a. 11x = 12
12 is greater thon 111 and less thon 112, so the solution is between 1 and 2.

b. 21x = 30
30 is greater than 211 and less thon 212, so the solution is between 1 and 2.

c. 100x = 2000
1002 = 10000, so 2,000 is between 1001 and 1002, so the solution is between 1 and 2.

d. $$\left(\frac{1}{11}\right)^{x}$$ = 0.01
$$\frac{1}{100}$$ is between $$\frac{1}{11}$$ and $$\frac{1}{121}$$ so the solution is between 1 and 2.

e. $$\left(\frac{2}{3}\right)^{x}$$
($$\left(\frac{2}{3}\right)^{2}$$) = $$\frac{4}{9}$$ and $$\frac{1}{2}$$ is between $$\frac{4}{9}$$ and $$\frac{2}{3}$$ sothe solution is between land 2.

f. 99x = 9000
992 = 9801. Since 9,000 is less than 9,801 and greater than 99, the solution is between 1 and 2.

Question 12.
Express the exact solution to each equation as a base-10 logarithm. Use a calculator to approximate the solution to the nearest 1000th.

a. 11x = 12
log(11x) = log(12)
xlog(11) = log(12)
x = $$\frac{\log (12)}{\log (11)}$$
x ≈ 1.036

b. 21x = 30
x = $$\frac{\log (30)}{\log (21)}$$
x ≈ 1.117

c. 100x = 2000
x = $$\frac{\log (2000)}{\log (100)}$$
x ≈ 1.651

d. $$\left(\frac{1}{11}\right)^{x}$$ = 0.01
x = –$$\frac{2}{\log \left(\frac{1}{11}\right)}$$
x ≈ 1.921

e. $$\left(\frac{2}{3}\right)^{x}$$ = $$\frac{1}{2}$$
x = $$\frac{\log \left(\frac{1}{2}\right)}{\log \left(\frac{2}{3}\right)}$$
x ≈ 1.710

f. 99x = 9000
x = $$\frac{\log (9000)}{\log (99)}$$
x ≈ 1.981

Question 13.
Show that for any real number r, the solution to the equation 10x = 3 . 10r is log(3) + r.
Substituting x = log(3) + r into 10x and using properties of exponents and logarithms gives
10x = 10log(3) + r
= 10log(3) 10r
= 3. 10r
Thus, x = log(3) + r is a solution to the equation 10x = 3 . 10r

Question 14.
Solve each equation. If there is no solution, explain why.

a. 3 . 5x = 21
5x = 7
log(5x) = log(7)
x log(5) = log(7)
x = $$\frac{\log (7)}{\log (5)}$$

b. 10x – 3 = 25
log(10x – 3) = log(25)
x = 3 + log(25)

c. 10x + 10x + 1 = 11
10x(1 + 10) = 11
10x = 1
x = 0.

d. 8 – 2x = 10
-2x = 2
2x = -2
There is no solution because 2x is always positive for all real x.

Question 15.
Solve the following equation for n: A = P(1 + r)n.
A = P(1+r)n
log(A) = log[(P(1 + r)n]
log(A) = log(P) + log(1 + r)n]
log(A) – log(P) = n log(1 + r)
log(A) – log(P)
n = $$\frac{\log (A)-\log (P)}{\log (1+r)}$$
n = $$\frac{\log \left(\frac{A}{P}\right)}{\log (1+r)}$$

16.
In this exercise, we will establish a formula for the logarithm of a sum. Let L = log(x + y), where x, y > 0.

a. Show log(x) + log (1 + $$\frac{y}{x}$$) = L. State as a property of logarithms after showing this is a true statement.
log(x) + log(1 + $$\frac{y}{x}$$) = log(x(1 + $$\frac{y}{x}$$))
= log(x + $$\frac{xy}{x}$$)
= log(x + y)
= L
Therefore, for x,y > 0, log(x + y) = log(x) + log (1 + $$\frac{y}{x}$$).

b. Use part (a) and the fact that log(100) = 2 to rewrite log(365) as a sum.
log(365) = log( 100 + 265)
= log(100) + log (1 + $$\frac{265}{100}$$)
= log(100) + log(3.65)
= 2 + log(3.65)

c. Rewrite 365 in scientific notation, and use properties of logarithms to express log(365) as a sum of an integer and a logarithm of a number between 0 and 10.
365 = 3.65 × 102
log(365) = log(3. 65 × 102)
= log(3.65) + log(102)
= 2 + log(3. 65)

Separating 365 into 100 + 265 and using the formula for the logarithm of a sum is the same as writing 365 in scientific notation and using the formula for the logarithm of a product.

e. Find two integers that are upper and lower estimates of log(365).
Since 1 < 3.65 < 10, we know that 0 < log(3. 65) < 1. This tells us that 2 < 2 + log(3.65) < 3, so 2 < log(365) < 3.

### Eureka Math Algebra 2 Module 3 Lesson 12 Exit Ticket Answer Key

In this lesson, we have established six logarithmic properties for positive real numbers x and y and real numbers r.
1. log(1) = 0
2. log(10) = 1
3. log(10r) = r
4 10log(x) = x
5. log(x . y) = log(x)+log(y)
6. log(xr) = r.log(x)

Question 1.
Use properties 1 – 6 of logarithms to establish property 7: log ($$\frac{1}{x}$$) = -log(x) for all x > 0.
By property 6, log(xk) = k log(x).
Let k = -1; then for x > 0, log(x-1) = (-1) log(x), which is equivalent to log ($$\frac{1}{x}$$) = -log(x).
Thus, for any x > 0, log ($$\frac{1}{x}$$) = -log(x).

Question 2.
Use properties 1 – 6 of logarithms to establish property 8: log ($$\frac{x}{y}$$) = log(x) – log(y) for x > 0 and y > 0.
log($$\frac{x}{y}$$) = log(x) + log($$\frac{1}{y}$$)
Thus, for any x > 0 and y > 0, log ($$\frac{x}{y}$$) = log(x) – log(y).