Question

UNIT 3
Exponential & Logarithmic Functions
LESSON 13
Exponential & Logarithmic Functions Unit Test

Answer 1

Below is a complete unit test on Exponential & Logarithmic Functions (Unit 3, Lesson 13). It includes instructions, 4 sections (multiple choice, short-answer algebra, word problems, graphing/transformations), a scoring guide, suggested time, and a full answer key with worked solutions. Use the answer key for grading or review.

INSTRUCTIONS
- No calculators unless indicated.
- Show all work for full credit.
- Suggested time: 60 minutes.
- Total points: 100.

SCORING
- Section A (Multiple Choice): 12 questions × 2 pts = 24 pts
- Section B (Short Answer / Algebra): 8 questions × 4 pts = 32 pts
- Section C (Applications / Word Problems): 4 questions × 6 pts = 24 pts
- Section D (Graphing & Transformations): 4 questions × 5 pts = 20 pts

SECTION A — MULTIPLE CHOICE (circle the best answer) — 2 pts each
1. Which of the following is the inverse function of f(x) = 3^x?
A. log_3(x) B. ln(x) C. 3^(-x) D. log(x)

2. Solve for x: 2^(x+1) = 16.
A. 2 B. 3 C. 4 D. 5

3. Which identity is correct?
A. log_a(bc) = log_a b + log_a c
B. log_a(b^c) = c log_b a
C. log_a(b/c) = log_b c - log_b a
D. log_a b^c = log_a b / c

4. Evaluate exactly: ln(e^3)
A. 1 B. e^3 C. 3 D. 0

5. Which graph has a horizontal asymptote y = 0, passes through (0,1), and increases for all x?
A. y = ln(x) B. y = e^x C. y = -e^x D. y = log_2(x)

6. Solve: log_5 (x) = 2
A. x = 2 B. x = 25 C. x = 5^2 + 1 D. x = 1/25

7. Using change-of-base, log_2 10 equals:
A. ln 10 / ln 2 B. log 2 / log 10 C. ln 2 / ln 10 D. log_10 2

8. Which equation represents exponential decay?
A. y = 5(1.03)^t B. y = 120(0.5)^(t/6) C. y = 3(1.2)^t D. y = -4(2)^t

9. Solve for x: e^(2x) = 7
A. x = ln(7)/2 B. x = ln(2)/7 C. x = 2 ln(7) D. x = ln(7^2)

10. Simplify: log_3(9)
A. 1 B. 2 C. 3 D. log 9 / log 3

11. Which is true for the domain and range of y = log_10(x - 2)?
A. domain x > 0, range all real B. domain x > 2, range all real
C. domain all real, range y > 2 D. domain x ≥ 2, range all real

12. If f(x) = a^x with a > 1, then f^-1(x) is:
A. log_a x B. a^(-x) C. ln x D. log x (base 10)

SECTION B — SHORT ANSWER / ALGEBRA (show work) — 4 pts each
13. Solve exactly: 5^(2x - 1) = 125.

14. Solve for x: log(x + 3) - log(x - 1) = log 4. (Assume log = log_10). State any domain restrictions.

15. Solve: 3^(x) = 10. Give an exact answer (in terms of logs) and an approximate value to three decimal places.

16. Expand (using log rules): log[ (x^2 * sqrt(y)) / z^3 ] (assume logs base 10). Write as sum/difference of simpler logs and with exponents in front.

17. Solve for x: ln(x) + ln(2) = 3.

18. Simplify: log_7(49) - log_7(7) + log_7(√7)

19. Solve for t: 200 = 50 e^(0.04 t)

20. Solve: 4^(x+1) = 2^(3x - 2). Give exact solution.

SECTION C — APPLICATION / WORD PROBLEMS — show work — 6 pts each
21. A culture of bacteria grows according to P(t) = 500 * 2^(t/3), where t is hours. How many bacteria after 9 hours? After how many hours will the culture reach 4000 bacteria? Give exact and approximate answers as needed.

22. A radioactive isotope has half-life 12 years. If initial mass is 80 grams, find the formula for mass remaining M(t). How long until 5 grams remain? (Round to nearest tenth of a year.)

23. A certain investment grows continuously at an annual rate of 5%. You invest $10,000. Write the value function and find the value after 7 years. (Give exact expression and approximate to nearest dollar.)

24. The pH of a solution is defined by pH = -log[H+], where [H+] is the hydrogen ion concentration in moles per liter. If pH = 3.7, find [H+]. Express answer in scientific notation to two significant figures.

SECTION D — GRAPHING & TRANSFORMATIONS — show work/explain — 5 pts each
25. Give domain, range, vertical asymptote, x- and y-intercepts (if any), and sketch (describe shape) of y = log_2(x + 4) - 1.

26. Describe the sequence of transformations that converts y = e^x into y = 3 e^(x - 2) + 4. State horizontal shift, vertical stretch/compression and vertical shift.

27. Without graphing technology, determine whether y = -2^x has x-intercepts or y-intercepts. Explain and state any asymptotes.

28. Given the graph of y = a^x passes through (1, 6) and (3, 54), find a and write the equation. Then give the inverse function and its value at x = 6.

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ANSWER KEY & WORKED SOLUTIONS

Section A (answers only)
1. A
2. B (2^(x+1)=16 → 2^(x+1)=2^4 → x+1=4 → x=3)
3. A
4. C (ln(e^3)=3)
5. B
6. B (log_5 x = 2 → x = 5^2 = 25)
7. A
8. B
9. A
10. B (log_3 9 = 2)
11. B (domain x>2, range all real)
12. A

Section B — Solutions

13. 5^(2x - 1) = 125 = 5^3. So 2x - 1 = 3 ⇒ 2x = 4 ⇒ x = 2.

14. log(x + 3) - log(x - 1) = log 4
Use subtraction rule: log[(x+3)/(x-1)] = log 4
Therefore (x+3)/(x-1) = 4
Solve: x + 3 = 4(x - 1) = 4x - 4
3 + 4 = 4x - x ⇒ 7 = 3x ⇒ x = 7/3
Domain: logs require x+3>0 ⇒ x>-3 (okay), x-1>0 ⇒ x>1. So domain x>1; x = 7/3 ≈ 2.333 is valid.

15. 3^x = 10 ⇒ x = log(10)/log(3) = ln 10 / ln 3 (exact)
Approx: ln10 ≈ 2.302585, ln3 ≈ 1.098612 → x ≈ 2.0959 → 2.096 (3 d.p.)

16. log[ (x^2 * sqrt(y)) / z^3 ] = log(x^2) + log(y^(1/2)) - log(z^3)
= 2 log x + (1/2) log y - 3 log z

17. ln(x) + ln(2) = 3 ⇒ ln(2x) = 3 ⇒ 2x = e^3 ⇒ x = e^3 / 2.

18. log_7(49) - log_7(7) + log_7(√7)
49 = 7^2 → log_7 49 = 2
log_7 7 = 1
√7 = 7^(1/2) → log_7 √7 = 1/2
So 2 - 1 + 1/2 = 1 + 1/2 = 1.5 = 3/2.

19. 200 = 50 e^(0.04 t) ⇒ divide by 50: 4 = e^(0.04 t)
Take ln: ln 4 = 0.04 t ⇒ t = ln 4 / 0.04
ln 4 ≈ 1.386294 → t ≈ 34.65735 ⇒ t ≈ 34.66 (or 34.7) units.

20. 4^(x+1) = 2^(3x - 2)
4 = 2^2 so left side: (2^2)^(x+1) = 2^(2x + 2)
So 2^(2x + 2) = 2^(3x - 2) ⇒ 2x + 2 = 3x - 2
Solve: 2 + 2 = 3x - 2x ⇒ 4 = x
So x = 4.

Section C — Applications

21. P(t) = 500 * 2^(t/3).
After 9 hours: P(9) = 500 * 2^(9/3) = 500 * 2^3 = 500 * 8 = 4000.
When will P(t) = 4000? Solve 500 * 2^(t/3) = 4000 → 2^(t/3) = 8 = 2^3 ⇒ t/3 = 3 ⇒ t = 9 hours.
(Exact answer: t = 9; approximate not needed.)

22. Half-life 12 years: decay formula M(t) = M0 * (1/2)^(t/12). With M0 = 80:
M(t) = 80 * (1/2)^(t/12).
Solve for M(t) = 5: 80 * (1/2)^(t/12) = 5 → (1/2)^(t/12) = 5/80 = 1/16
But 1/16 = (1/2)^4. So t/12 = 4 ⇒ t = 48 years.
(If solving with logs: t = 12 * ln(1/16) / ln(1/2) = 12 * (-ln 16)/(-ln 2) = 12*4 = 48.)

23. Continuous compounding: A(t) = P e^(rt). P = 10000, r = 0.05
A(t) = 10000 e^(0.05 t)
After 7 years: A(7) = 10000 e^(0.35) ≈ 10000 * 1.4190675 = $14,190.68 → $14,191 (nearest dollar).

24. pH = 3.7 = -log[H+] ⇒ log[H+] = -3.7
So [H+] = 10^(-3.7).
10^(-3.7) = 10^(-4 + 0.3) = 10^-4 * 10^0.3.
10^0.3 ≈ 1.995262 (since 10^0.3010 = 2). Using calculator: 10^(-3.7) ≈ 2.0 × 10^-4
More precisely: 10^(-3.7) ≈ 1.995 × 10^-4. To two significant figures: 2.0 × 10^-4 M.

Section D — Graphing & Transformations

25. y = log_2(x + 4) - 1
Domain: x + 4 > 0 ⇒ x > -4.
Range: all real numbers (-∞, ∞).
Vertical asymptote: x + 4 = 0 ⇒ x = -4.
x-intercept: set y = 0 → log_2(x + 4) - 1 = 0 ⇒ log_2(x + 4) = 1 ⇒ x + 4 = 2 ⇒ x = -2. So x-intercept (-2, 0).
y-intercept: set x = 0 → y = log_2(4) - 1 = 2 - 1 = 1 → (0,1).
Shape: Standard logarithmic curve shifted left 4 and down 1: passes through (-3, ?), increases slowly, vertical asymptote at x = -4.

26. y = 3 e^(x - 2) + 4 from y = e^x:
- Horizontal shift right by 2 (x replaced by x - 2).
- Vertical stretch by factor 3 (multiply function by 3).
- Vertical shift up by 4 (add 4).

27. y = -2^x (interpretation: negative of 2^x)
- y-intercept: at x = 0 → y = -(2^0) = -1 → y-intercept (0,-1).
- x-intercept: solve -2^x = 0 → 2^x = 0 — no solution (exponential never zero). So no x-intercepts.
- Asymptote: horizontal asymptote y = 0 (x-axis). Because function is negative and approaches 0 from below as x → -∞, and goes to -∞ as x → ∞? Actually 2^x → ∞ as x→∞ so -2^x → -∞ as x→∞, and as x→-∞, 2^x → 0^+ so -2^x → 0^- . So horizontal asymptote y = 0.

28. Find a for y = a^x passing through (1, 6) and (3, 54).
Use (1,6): a^1 = 6 ⇒ a = 6.
Check (3,54): 6^3 = 216, not 54, so a is not 6. Instead solve using both points:
a^1 = 6 ⇒ a = 6 (if both true they'd conflict). Better approach: Suppose a^1 = y1 and a^3 = y2. Take ratio:
(a^3)/(a^1) = a^2 = 54/6 = 9 ⇒ a^2 = 9 ⇒ a = 3 (take positive base >0, a ≠ 0).
Then check a^1 = 3^1 = 3, but that would not give y=6 at x=1. Correct method: Let y = A * b^x (general exponential with coefficient A). The question stated "y = a^x passes through ...", but two points do not fit that form unless consistent. To resolve, we assume the model y = A * b^x. Using that:

Let y = A b^x.
From (1,6): A b^1 = 6 → A b = 6 ...(i)
From (3,54): A b^3 = 54 ...(ii)
Divide (ii)/(i): (A b^3)/(A b) = b^2 = 54/6 = 9 ⇒ b = 3.
Then A b = 6 ⇒ A*3 = 6 ⇒ A = 2.
So function: y = 2 * 3^x.
If the intended function was a^x without coefficient, the two points are inconsistent; thus the correct general form is y = A b^x as above.

Inverse function: For y = 2 * 3^x, first solve for x:
y = 2 * 3^x ⇒ 3^x = y/2 ⇒ x = log_3 (y/2).
So inverse f^-1(x) = log_3 (x/2).
f^-1(6) = log_3 (6/2) = log_3 3 = 1.

(If one insisted on y = a^x, note the two points are inconsistent; the general exponential with coefficient is required.)

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Notes for grading / common pitfalls
- For Section B and C, full credit requires showing algebraic steps (e.g., using log rules, isolating exponentials before taking logs).
- Watch domain restrictions for logs: arguments must be positive.
- In Section D question 28, accept answers that correctly point out inconsistency of y = a^x and then proceed using y = A b^x; award partial credit if student recognizes inconsistency.

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