# Eureka Math Algebra 2 Module 3 Lesson 13 Answer Key

## Engage NY Eureka Math Algebra 2 Module 3 Lesson 13 Answer Key

### Eureka Math Algebra 2 Module 3 Lesson 13 Exercise Answer Key

Exercises:

Exercise 1.
Assume that x, a, and b are all positive real numbers so that a ≠ 1 and b ≠ 1. What is logb(x) ln terms of loga(x)? The resulting equation allows us to change the base of a logarithm from a to b.
Let L = logb(x). Then bL = x Taklng the logarithm base a of each side, we get
loga(bL) = loga(x)
L . loga(b) = loga(x)
L = $$\frac{\log _{a}(x)}{\log _{a}(b)}$$
Therefore, logb(x) = $$\frac{\log _{a}(x)}{\log _{a}(b)}$$.

Exercise 2.
Approximate each of the following logarithms to four decimal places. Use the LOG key on your calculator rather than logarithm tables, first changing the base of the logarithm to 10 ¡f necessary.

a. log(32)
log(32) = log(9) ≈ 0.9542
Therefore, log(32) ≈ 0.9542.
OR
log(32) = 2 log(3) ≈ 20.4771 ≈ 0.9542
Therefore, log(32) ≈ 0.9542.

b. log3(32)
log3(32) = $$\frac{2 \log (3)}{\log (3)}$$ = 2
Therefore, log3(32) = 2.0000.

c. log2(32)
log2(32) = log2(9) ≈ 3.1699
Therefore, log2(32) ≈ 3.1699.

Exercise 3.
ln Lesson 12, we justified a number of properties of base-10 logarithms. Worklng ln pairs, justify the followlng properties of base-b logarithms:

a. logb(1) = 0
Because L = logb(x) means bL = x, then when x = 1, L = 0.

b. logb(b) = 1
Because L = logb(x) means bL = x, then when x = b, L = 1.

C. logb(br) = r
Because L = logb(x) means bL = x, then when x = br, L = r.

d. blogb(x) = x9
Because L = logb(x) means bL = x, then x =logb(x).

e. logb(x . y) = logb(x) + logb(y)
By the rule aq . ar= aq + r, blogb(x) . blogb(y) = blogb(x) + logb(y).
By property 4, blogb(x) . blogb(y) = x . y
Therefore, x . y = blogb(x) + logb(y) By property 4 agaln, x . y = blogb(x . y)
So, the exponents must be equal, and logb(x . y) = logb(x) + logb(y).

f. logb(xr) = r . logb(x)
By the rule (aq)r = aqr, br logb(x) = (blogb(x))r.
By property 4, (blogb(x))r = xr
Therefore, xr = brlogb(x)
By property 4 agaln, xr = blogb(xr)
So, the exponents must be equal, and logb(xr) = r . logb(x).

Exercise 4.
Use the keys on your calculator to find the value of each logarithm to four decimal places.

a. ln(1)
0.0000

log(1)
0.0000

b. ln(3)
1.0986

log(3)
0.4771

c. ln(10)
2.3026

log(10)
1.0000

d. ln(25)
3.2189

log(25)
1.3979

e. ln(100)
4.6052

log(100)
2.0000

Exercise 5.
Make a conjecture that compares values of log(x) to ln(x) for x ≥ 1.
It appears that for x ≥ 1, log(x) ≤ ln(x).

Exercise 6.
Justify your conjecture ln Exercise 5 uslng the change of base formula.
By the change of base formula, log(x) = $$\frac{\ln (x)}{\ln (10)}$$ Then ln(10) . log(x) = ln(x). Slnce
ln(10) ≈ 2.3, log(x) ≤ ln(10) . log(x), and thus log(x) ≤ ln(x).

Exercise 7.
Write as a single logarithm.

a. ln(4) – 3 ln($$\frac{1}{3}$$) + ln(2)
ln(4) — 3 ln ($$\frac{1}{3}$$) + ln(2) = ln(4) + ln(33) + ln(2)
= ln(4 . 33 . 2)
= ln(216)
= ln(63)
= 3 ln(6)
Any of the last three expressions is an acceptable final answer.

b. ln(5) + ($$\frac{3}{5}$$)ln(32) – ln(4) = ln(5) + ln(32$$\frac{3}{5}$$) – ln(4) = ln(5) + ln (8) – ln(4)
= ln(5) + ln(8) – ln(4)
= ln(5. 8) – ln(4)
= ln($$\frac{40}{4}$$)
= ln(10)
Therefore, ln(5) + $$\frac{3}{5}$$ln(32) – ln(4) = ln(10).

Exercise 8.
Write each expression as a sum of differences of constants and logarithms of simpler terms.

a. ln$$\left(\frac{\sqrt{5 x^{3}}}{e^{2}}\right)$$
ln$$\left(\frac{\sqrt{5 x^{3}}}{e^{2}}\right)$$ = ln(√5) + ln($$\sqrt{x^{3}}$$) + ln(e2)
= $$\frac{1}{2}$$ln(5) + $$\frac{3}{2}$$ln(x) – 2

b. $$Answer: [latex]\ln \left(\frac{(x+y)^{2}}{x^{2}+y^{2}}\right)$$ = ln(x + y)2 – ln(x2 + y2)
= 2 ln(x + y) – ln(x2 + y2)
The point of this simplification is that neither of these terms can be simplified further.

### Eureka Math Algebra 2 Module 3 Lesson 13 Problem Set Answer Key

Question 1.
Evaluate each of the following logarithmic expressions, approximating to four decimal places If necessary. Use the or key on your calculator rather than a table.

a. log8(16)
log8(16) = $$\frac{\log (16)}{\log (8)}$$
= $$\frac{\log \left(2^{4}\right)}{\log \left(2^{3}\right)}$$
= $$\frac{4 \cdot \log (2)}{3 \cdot \log (2)}$$
= $$\frac{4}{3}$$

Therefore, log816) = $$\frac{4}{3}$$.

b. log7(11)
log7(11) = $$\frac{\log (11)}{\log (7)}$$
≈ 1.2323
Therefore, log7(11) ≈ 1.2323.

c. log3(2) + log2(3)
log3(2) + log2(3) = $$\frac{\log (2)}{\log (3)}+\frac{\log (3)}{\log (2)}$$
≈ 2.2159
Therefore, log3(2) + log2(3) ≈ 2.2159

Question 2.
Use logarithmic properties and the fact that ln(2) ≈ 0.69 and ln(3) ≈ 1. 10 to approximate the value of each of the followlng logarithmic expressions. Do not use a calculator.

a. ln(e4)
ln(e4) = 4ln(e)
= 4
Therefore, ln(e4) = 4.

b. ln(6)
ln(6) = ln(2) + ln(3)
≈ 0.69 + 1.10
≈ 1.79
Therefore, ln(6) ≈ 1.79.

c. ln(108)
ln(108) = ln(4.27)
= ln(4) + ln(27)
≈ 2 ln(2) + 3 ln(3)
≈ 1.38 + 3.30
≈ 4.68
Therefore, ln(108) ≈ 4.68.

d. ln($$\frac{8}{3}$$)
ln($$\frac{8}{3}$$ ) = ln(8) – ln(3)
= ln(23) – ln(3)
≈ 3(0.69) – 1.10
≈ 0.97
Therefore, ln ($$\frac{8}{3}$$) ≈ 0.97.

Question 3.
Compare the values of log$$\frac{1}{9}$$(10) and log9($$\frac{1}{10}$$) without using a calculator.
Using the change of base formula,

log$$\frac{1}{9}$$(10) = $$\frac{\log _{9}(10)}{\log _{9}\left(\frac{1}{9}\right)}$$
= $$\frac{\log _{9}(10)}{-1}$$
= -log9(10)
= log9($$\frac{1}{10}$$)
Thus, log$$\frac{1}{9}$$(10) and log9($$\frac{1}{10}$$)

Question 4.
Show that for any positive numbers a and b with a ≠ 1 and b ≠ 1, loga(b) logb(a) = 1.
Using the change of base formula,
loga(b) = $$\frac{\log _{b}(b)}{\log _{b}(a)}=\frac{1}{\log _{b}(a)}$$ Thus,
loga(b) . logb(a) = $$\frac{1}{\log _{b}(a)}$$ . logb(a) = 1.

Question 5.
Express x ln terms of a, e, and y if ln(x) – ln(y) = 2a.
ln(x) – ln(y) = 2a
ln$$\left(\frac{x}{y}\right)$$ = 2a
$$\frac{x}{y}$$ = e2a
x = y e2a

Question 6.
Rewrite each expression ln an equivalent form that only contains one base-10 logarithm.

a. log2(800)

b. logx($$\frac{1}{10}$$), for positive real values of x ≠ 1
logx($$\frac{1}{10}$$) = $$\frac{\log \left(\frac{1}{10}\right)}{\log (x)}=-\frac{1}{\log (x)}$$

c. log5(12500)

d. log3(0.81)

Question 7.
Write each number ln terms of natural logarithms, and then use the properties of logarithms to show that it is a rational number.

a. log9(√27)
$$\frac{\ln (\sqrt{27})}{\ln (9)}=\frac{\ln \left(3^{\frac{3}{2}}\right)}{\ln \left(3^{2}\right)}$$ = $$\frac{\frac{3}{2} \ln (3)}{2 \ln (3)}=\frac{3}{4}$$

b. log8(32)
$$\frac{\ln (32)}{\ln (8)}=\frac{\ln \left(2^{5}\right)}{\ln \left(2^{3}\right)}=\frac{5}{3}$$

c. log4($$\frac{1}{8}$$)
$$\frac{\ln \left(\frac{1}{8}\right)}{\ln (4)}=\frac{\ln \left(2^{-3}\right)}{\ln \left(2^{2}\right)}=-\frac{3}{2}$$

Question 8.
Write each expression as an equivalent expression with a single logarithm. Assume x, y and z are positive real numbers.

a. ln(x) + 2 ln(y) – 3 ln(z)
ln($$\frac{x y^{2}}{z^{3}}$$)

b. $$\frac{1}{2}$$ln(x+y) – ln(z))
ln($$\sqrt{\frac{x+y}{z}}$$)

c. (x+y) + ln(z)
(x+y)ln(e) + ln(z) = ln(ex + y) + ln(z) = ln(ex + y . z)

Question 9.
Rewrite each expression as sums and differences ln terms of ln(x), ln(y), and ln(z).

a. ln(xyz3)
ln(x) + ln(y) + 3ln(z)

b.
ln$$\left(\frac{e^{3}}{x y z}\right)$$
3 – ln(x) – ln(y) – ln(z)

c. ln$$\left(\sqrt{\frac{x}{y}}\right)$$
$$\frac{1}{2}$$(ln(x) – ln(y))

Question 10.
Use base-5 logarithms to rewrite each exponential equation as a logarithmic equation, and solve the resulting equation. Use the change of base formula to convert to a base-10 logarithm that can be evaluated on a calculator. Give each answer to 4 decimal places. If an equation has no solution, explain why.

a. 52x = 20
2x = log5(20)
x = $$\frac{1}{2}$$ .log5(20)
x = $$\frac{\log (20)}{2 \log (5)}$$

b. 75 = 10 . 5x – 1
7.5 = 5x – 1
x = log5(7.5) + 1
x = $$\frac{\log (7.5)}{\log (5)}$$ + 1
x ≈ 2.2519

c. 52 + x – 5x = 10
5x(52 – 1) = 10
5x = $$\frac{10}{24}$$
x = log5($$\frac{10}{24}$$)
x = $$\frac{\log \left(\frac{10}{24}\right)}{\log (5)}$$
x = $$\frac{\log (10)-\log (24)}{\log (5)}$$
x ≈ -0.5440

d. 5x2 = 0.25
x2 = log5(0.25)
x2 = $$\frac{\log (0.25)}{\log (5)}$$
x2 ≈ -0.8614

This equation has no real solution because x2 cannot be negative for any real number x.

Question 11.
ln Lesson 6, you discovered that log(x . 10k) = k + log(x) by looking at a table of logarithms. Use the properties of logarithms to justify this property for an arbitrary base b > 0 with b ≠ 1. That is, show that logb(x bk) = k + logb(x).
logb(x . bk) = logb(x) + logb(bk)
= k + logb(x)

Question 12.
Larissa argued that since log2(2) = 1 and log2(4) = 2, then it must be true that log2(3) = 1.5. Is she correct? Explain how you know.
Larissa is not correct. If log2(x) = 1.5, then 21.5 = x, so x = 2$$\frac{3}{2}$$ = 2√2. Since 3 ≠ 2√2 , Larissa’s calculation is not correct.

Question 13.
Extension: Suppose that there is some positive number b so that
logb(2) = 0.36
logb(3) = 0.57
logb(5) = 0.84.

a. Use the given values of logb(2), logb(3), and logb(S) to evaluate the following logarithms.

i. logb(6)
logb(6) = logb(2 . 3)
= logb(2) + logb(3)
= 0.36 + 0.57
= 0.93

ii. logb(8)
logb(8) = logb(23)
= 3 . logb(2)
= 3 . 0.36
= 1.08

iii. logb(10)
logb(10) = logb(2.5)
= logb(2) + logb(5)
= 0.36 + 0.84
= 1.20

iv. logb(600)
logb(600) = logb(6 . 100)
= logb(6) + logb(100)
= 0.93 + 2 logb(10)
= 0.93 + 2(1.20)
= 0.93 + 2.40
= 3.33

b. Use the change of base formula to convert logb(10) to base 10, and solve for b. Giver your answer to four decimal places.
From part(iii) above, logb(10) = 1.20. Then,
1.20 = logb(10)
1.20 = $$\frac{\log _{10}(10)}{\log _{10}(b)}$$
1.20 = $$\frac{1}{\log _{10}(b)}$$
$$\frac{1}{1.20}$$ = log10(b)
b = 10$$\frac{1}{1.20}$$
b ≈ 6.8129

Question 14.
Use a logarithm with an appropriate base to solve the following equations.

a. 23x = 16
log2(23x) = log2(16)
3x = 4
x = $$\frac{4}{3}$$

b. 2x + 3 = 43x
log2(2x + 3) = log2(43x)
x + 3 = 3x . 2
5x = 3
x = $$\frac{3}{5}$$

c. 34x – 2 = 27x + 2
log3(34x – 2) = log3(27x + 2)
(4x – 2)log3(3) = (x + 2)log3(27)
4x – 2 = 3(x + 2)
4x – 2 = 3x + 6
x = 8

d. 42x = ($$\frac{1}{4}$$)3x
log4(42x) = log4(($$\frac{1}{4}$$)3x)
2xlog4(4) = 3xlog4($$\frac{1}{4}$$)
2x = 3x(-1)
5x = 0
x = 0

e. 50.2x + 3 = 625
log5(50.2x + 3) = log5(625)
(0.2x + 3)log5(5) = log5(54)
0.2x + 3 = 4
0.2x = 1
x = 5

Question 15.
Solve each exponential equation.

a. 32x = 81
x = 2

b. 63x = 36x + 1
x = 2

c. 625 = 53x
x = $$\frac{4}{3}$$

d. 254 – x = 53x
x = $$\frac{8}{5}$$

e. 32x – 1 = $$\frac{1}{2}$$
x = $$\frac{4}{5}$$

f. $$\frac{4^{2 x}}{2^{x-3}}$$ = 1
x = -1

g. $$\frac{1}{8^{2 x-4}}$$ = 1
x = 2

h. 2x = 81
x = $$\frac{\ln (81)}{\ln (2)}$$

i. 8 = 3x
x = $$\frac{\ln (8)}{\ln (3)}$$

j. 6x + 2 = 12
x = -2 + $$\frac{\log (12)}{\log (6)}$$

k. 10x + 4 = 27
x = -4 + log(27)

l. 2x + 1 = 31 – x
x = $$\frac{\log (3)-\log (2)}{\log (2)+\log (3)}$$

m. 32x – 3 = 2x + 4
x = $$\frac{4 \log (2)+3 \log (3)}{3 \log (3)-\log (2)}$$

n. e2x = 5
x = $$\frac{\ln (5)}{2}$$

o. ex – 1
x = 1 + ln(6)

Question 16.
ln Problem 9(e) of Lesson 12, you solved the equation 3x = 7-3x + 2 using the logarithm base 10.

a. Solve 3x = 7-3x+2 using the logarithm base 3.
log3(3x) = log3(7-3x + 2)
x = (-3x + 2)log3(7)
x = -3x log3 (7) + 2 log3 (7)
x + 3x log3(7) = 2 log3 (7)
x(1 + 3 log3(7)) = 2 log3(7)
x = $$\frac{2 \log _{3}(7)}{1+3 \log _{3}(7)}$$

b. Apply the change of base formula to show that your answer to part (a) agrees with your answer to Problem 9(e) of Lesson 12.
Changing from base 3 to bose 10, we see that
log3(7) = $$\frac{\log (7)}{\log (3)}$$
$$\frac{2 \log _{3}(7)}{1+3 \log _{3}(7)}$$ = $$\frac{2\left(\frac{\log (7)}{\log (3)}\right)}{1+3\left(\frac{\log (7)}{\log (3)}\right)}$$ = $$\frac{2 \log (7)}{\log (3)+3 \log (7)}$$

which was the answer from Problem 9(e) of Lesson 12.

c. Solve 3x = 7-3x+2 using the logarithm base 7.
log7(3’) = log7(7-3x+2)
xlog7(3) = -3x + 2
3x + xlog7(3) =2
x(3 + log7(3)) = 2
x = $$\frac{2}{3+\log _{7}(3)}$$

d. Apply the change of base formula to show that your answer to part (t) also agrees with your answer to Problem 9(e) of Lesson 12.
Changing from base 7 to base 10, we see that
log7(3) = $$\frac{\log (3)}{\log (7)}$$
Then,
$$\frac{2}{3+\log _{7}(3)}$$ = $$\frac{2}{3+\frac{\log (3)}{\log (7)}}$$
= $$\frac{2 \log (7)}{3 \log (7)+\log (3)}$$
which was the answer from Problem 9(e) of Lesson 12.

Question 17.
Pearl solved the equation 2x = 10 as follows:
log(2x) = log(10)
x log(2) = 1
x = $$\frac{1}{\log (2)}$$
Jess solved the equation 2x = 10 as follows:
log2(2x) = log2(10)
xlog2(2) = llog2(10)
x = log2(10).
Is pearl correct? Is Jess correct? Explain how you know.
Both Pearl and Jess are correct. if we take Jess’s solution and apply the change of base formula, we have
x = log2(10)
= $$\frac{\log (10)}{\log (2)}$$
= $$\frac{1}{\log (2)}$$.
Thus, the two solutions are equivalent, and both students are correct.

### Eureka Math Algebra 2 Module 3 Lesson 13 Exit Ticket Answer Key

Question 1.
Are there any properties that hold for base-10 logarithms that would not be valid for the logarithm base e? Why? Are there any properties that hold for base-10 logarithms that would not be valid for some positive base b, such that b ≠ 1?
No. Any property that ¡s true for a base-10 logarithm will be true for a base-e logarithm. The only difference between a common logarithm and a natural logarithm is a scale change because log(x) = $$\frac{\ln (x)}{\ln (10)}$$ and ln(x) = $$\frac{\log (x)}{\log (e)}$$.
Since logb(x) = $$\frac{\log (x)}{\log (b)^{\prime}}$$ we would only encounter a problem if log(b) = 0, but this only happens when b = 1, and 1 is not a valid base for logarithms.

Question 2.
Write each logarithm as an equivalent expression involving only logarithms base 10.

a. log3(25)
log3(25) = $$\frac{\log (25)}{\log (3)}$$

b. log100(x2)
log(x2) = $$=\frac{\log \left(x^{2}\right)}{\log (100)}$$ = $$\frac{2 \log (x)}{2}$$
= log(x)

Question 3.
Rewrite each expression as an equivalent expression containing only one logarithm.

a. 3ln(p + q) – 2ln(q) – 7ln(p)
= ln($$\left(\frac{(\boldsymbol{p}+\boldsymbol{q})^{3}}{\boldsymbol{q}^{2} \boldsymbol{p}^{7}}\right)$$
b. ln(xy) – ln$$\left(\frac{x}{y}\right)$$
ln(xy) – ln$$\left(\frac{x}{y}\right)$$ = ln(x) + ln(y) – ln(x) + ln(y)
Therefore, ln(xy) – ln$$\left(\frac{x}{y}\right)$$ ¡s equivalent to both 2 ln(y) and ln(y2).