## 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)

Answer:

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(2^{4})

Answer:

log(2^{4}) = 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

Answer:

Because L = log(x) means 10^{L} = x, then when x = 1, L = 0.

Exercise 2.

Property 2: log(10) = 1.

Answer:

Because L = log(x) means 10^{L} = x, then when x = 10, L = 1.

Exercise 3.

Property 3: For all real numbers r, log(10^{r}) = r.

Answer:

Because L = log(x) means 10^{L} = x, then when x = 10^{r}, L = r.

Exercise 4.

Property 4: For any x > 0, 10^{log(x)} = x.

Answer:

Because L = log(x) means 10^{L} = x, then x = 10^{log(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.

Answer:

By the rule a^{b} . a^{c} = a^{b + c}, 10^{log(x)} . 10^{log(y)} = 10^{log(x) + log(y)}

By property 4, 10^{log(x)} . 10^{log(y)} = x . y

Therefore, x . y = 10^{log(x) + log(y)} Again, by property 4, x . y = 10^{log(x.y)}

Then,

10^{log(x.y)} = 10^{log(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(x^{r}) = r . log(x).

Hint: Use an exponent rule as well as property 4.

Answer:

By the rule (a^{b})^{c} = a^{bc}, 10^{k log(x)} = (10^{log(x)})^{k}

By property 4, (10^{log(x)})^{r} = x^{r}.

Therefore, x^{r} = 10^{r log(x)} Again, by property 4, x^{r} = 10^{log(xr)}

Then, 10^{log(xr)} = 10^{r log(x)}; 50 the exponents must be equal, and log(x^{r}) = 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)

Answer:

log(5) + log(4) = log(20)

b. \(\frac{1}{3}\)log(8) + log(16)

Answer:

log(2) + log(2^{4}) = log(32)

c. 3 log(5) + log(0.8)

Answer:

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(3x^{2}y^{5})

Answer:

log(3) + 2 log(x) + 5 log(y)

b. log(\(\sqrt{x^{7} y^{3}}\))

Answer:

\(\frac{7}{2}\)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. 10^{10x} = 100

Answer:

log(10^{10x}) = log(100)

10x = 2

x = \(\frac{1}{5}\)

b. 10^{x – 1} = \(\frac{1}{10^{x+1}}\)

Answer:

log(10^{x – 1}) = -log(10^{x + 1})

x – 1 = -(x+ 1)

2x = 0

x = 0

c. 100^{2x} = 10^{3x – 1}

Answer:

log(100^{2x}) = log(10^{3x – 1})

2xlog(100) = (3x -1)

4x = 3x – 1

x = -1

Exercise 10.

Solve the following equations.

a. 10^{x} = 2^{7}

Answer:

log(10^{x}) = log(2^{7})

x = 7log(2)

b. 10^{x2 + 1} = 15

Answer:

log(10^{x2 + 1}) = log(15)

x^{2} + 1 = log(15)

x= ±\(\sqrt{\log (15)-1}\)

c. 4^{x} = 5^{3}

Answer:

log(4^{x}) = log(5^{3})

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)

Answer:

Using property 5,

log(6) = log(3) + log(2) ≈ 0.7781.

b. log(15)

Answer:

Using property 5,

log(15) = log(3) + log(5) ≈ 1.1761.

c. log(12)

Answer:

Using properties 5 and 6,

log(12) = log(3) + log(2^{2}) = log(3) + 2 log(2) ≈ 1,0791.

d. log(10^{7})

Answer:

Using property 3,

log(10^{7}) = 7.

e. log(\(\frac{1}{5}\))

Answer:

Using property 7,

log(\(\frac{1}{5}\)) = -log(5) ≈ -0.6990.

f. log(\(\frac{3}{7}\))

Answer:

Using property 8,

log(\(\frac{3}{7}\)) = log(3) – log(7) ≈ -0.368.

g. log(\(\sqrt[4]{2}\))

Answer:

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(\(\frac{X}{\boldsymbol{Y}}\))

Answer:

r – s

b. log(XZ)

Answer:

s + t

c. log(X^{r})

Answer:

r^{2}

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

Answer:

\(\frac{t}{3}\)

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

Answer:

\(\frac{s-t}{4}\)

f. log(XY^{2}Z^{3})

Answer:

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(\(\frac{5}{4}\))

Answer:

log(\(\frac{13}{4}\))

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

Answer:

log(\(\frac{5}{4}\))

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

Answer:

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)

Answer:

log(x^{2})

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

Answer:

log(x)

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

Answer:

log(\(\frac{x y^{2}}{\sqrt{z}}\))

d. \(\frac{1}{3}\)(log(x) – 3 log(y) + log(z))

Answer:

log(\(\sqrt[3]{\frac{x z}{y^{3}}}\))

e. 2(log(x) – log(3y)) + 3(log(z) – 2 log(x))

Answer:

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

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(\(\frac{3 x^{2} y^{4}}{\sqrt{z}}\))

Answer:

log(3) + 2log(x) + 4log(y) – \(\frac{1}{2}\)log(z)

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

Answer:

log(42) – \(\frac{5}{3}\)log(x) + \(\frac{7}{3}\)log(y) — log(z)

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

Answer:

2 + 2log(x) – 3 log(y)

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

Answer:

\(\frac{1}{2}\)(3log(x) + 2 log(y) – 1 – log(z))

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

Answer:

-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.

Answer:

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.

Answer:

log(x + \(\sqrt{x^{2}-1}\)) + log(x – \(\sqrt{x^{2}-1}\)) = log ((x + \(\sqrt{x^{2}-1}\)) (x – \(\sqrt{x^{2}-1}\)))

= log(x^{2} – (\(\sqrt{x^{2}-1}\))2)

= log(x^{2} – x^{2} + 1)

= log(1)

= 0

Question 8.

If xy = 10^{3.67} for some positive real numbers x and y, find the value of log(x) + log(y).

Answer:

xy = 10^{3.67}

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. 10^{x2} = 320

Answer:

log(10^{x2}) = log(320)

x^{2} = log(320)

x = ±\(\sqrt{\log (320)}\)

b. 10^{\(\frac{x}{8}\)} = 300

Answer:

log (10^{\(\frac{x}{8}\)}) = log(300)

\(\frac{x}{8}\) = log(10^{2} . 3)

\(\frac{x}{8}\) = 2 + log(3)

x = 16 + 8 log(3)

c. 10^{3x} = 400

Answer:

log(10^{3x}) = log(400)

3x . log(10) = log(10^{2} . 4)

3x . 1 = 2 + log(4)

x = \(\frac{1}{3}\)(2 + log(4))

d. 5^{2x} = 200

Answer:

log(5^{2x}) = 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. 3^{x} = 7^{-3x + 2}

Answer:

log(3^{x}) = 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. 10^{x} = 3

Answer:

x = log(3)

b. 10^{y} = 30

Answer:

y = log(30)

c. 10^{z} = 300

Answer:

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.

Answer:

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. 11^{x} = 12

Answer:

12 is greater thon 11^{1} and less thon 11^{2}, so the solution is between 1 and 2.

b. 21^{x} = 30

Answer:

30 is greater than 21^{1} and less thon 21^{2}, so the solution is between 1 and 2.

c. 100^{x} = 2000

Answer:

100^{2} = 10000, so 2,000 is between 100^{1} and 100^{2}, so the solution is between 1 and 2.

d. \(\left(\frac{1}{11}\right)^{x}\) = 0.01

Answer:

\(\frac{1}{100}\) is between \(\frac{1}{11}\) and \(\frac{1}{121}\) so the solution is between 1 and 2.

e. \(\left(\frac{2}{3}\right)^{x}\)

Answer:

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

f. 99^{x} = 9000

Answer:

99^{2} = 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. 11^{x} = 12

Answer:

log(11^{x}) = log(12)

xlog(11) = log(12)

x = \(\frac{\log (12)}{\log (11)}\)

x ≈ 1.036

b. 21^{x} = 30

Answer:

x = \(\frac{\log (30)}{\log (21)}\)

x ≈ 1.117

c. 100^{x} = 2000

Answer:

x = \(\frac{\log (2000)}{\log (100)}\)

x ≈ 1.651

d. \(\left(\frac{1}{11}\right)^{x}\) = 0.01

Answer:

x = –\(\frac{2}{\log \left(\frac{1}{11}\right)}\)

x ≈ 1.921

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

Answer:

x = \(\frac{\log \left(\frac{1}{2}\right)}{\log \left(\frac{2}{3}\right)}\)

x ≈ 1.710

f. 99^{x} = 9000

Answer:

x = \(\frac{\log (9000)}{\log (99)}\)

x ≈ 1.981

Question 13.

Show that for any real number r, the solution to the equation 10^{x} = 3 . 10^{r} is log(3) + r.

Answer:

Substituting x = log(3) + r into 10^{x} and using properties of exponents and logarithms gives

10^{x} = 10^{log(3) + r}

= 10^{log(3)} 10^{r}

= 3. 10^{r}

Thus, x = log(3) + r is a solution to the equation 10^{x} = 3 . 10^{r}

Question 14.

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

a. 3 . 5^{x} = 21

Answer:

5^{x} = 7

log(5^{x}) = log(7)

x log(5) = log(7)

x = \(\frac{\log (7)}{\log (5)}\)

b. 10^{x – 3} = 25

Answer:

log(10^{x – 3}) = log(25)

x = 3 + log(25)

c. 10^{x} + 10^{x + 1} = 11

Answer:

10^{x}(1 + 10) = 11

10^{x} = 1

x = 0.

d. 8 – 2^{x} = 10

Answer:

-2^{x} = 2

2^{x} = -2

There is no solution because 2^{x} is always positive for all real x.

Question 15.

Solve the following equation for n: A = P(1 + r)^{n}.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

365 = 3.65 × 10^{2}

log(365) = log(3. 65 × 10^{2})

= log(3.65) + log(10^{2})

= 2 + log(3. 65)

d. What do you notice about your answers to (b) and (c)?

Answer:

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).

Answer:

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(10^{r}) = r

4 10^{log(x)} = x

5. log(x . y) = log(x)+log(y)

6. log(x^{r}) = 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.

Answer:

By property 6, log(x^{k}) = 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.

Answer:

By property 5, log(x . y) = log(x) + log(y).

By Problem 1 above, for y > 0, log(y^{-1}) = (-1) log(y).

Therefore,

log(\(\frac{x}{y}\)) = log(x) + log(\(\frac{1}{y}\))

= log(x) + (-1)log(y)

= log(x) – log(y).

Thus, for any x > 0 and y > 0, log (\(\frac{x}{y}\)) = log(x) – log(y).