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This challenging AP-style assessment covers the full Unit 2 syllabus, including arithmetic vs geometric change, exponential growth and decay, function manipulation, data modeling, competing model validation, composition and inverse functions, logarithmic laws, exponential and logarithmic equations and inequalities, and interpreting semi-log plots. The test is designed to push students to mastery level through a combination of advanced multiple-choice and rigorous free-response questions. Students are encouraged to show full working and apply conceptual reasoning, algebraic precision, and real-world modeling skills throughout.
Aviate Learnings
AP Precalculus – Unit 2
Exponential & Logarithmic Functions
Ultra-challenging 1-hour test on exponential & logarithmic functions: sequences, modeling, compositions, inverses, log laws, and semi-log plots.
1-Hour Tough Test
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Test Instructions
Total: 100 marks | Time: 60 min
Section A: 12 multiple-choice questions (4 marks each).
Section B: 6 free-response questions (52 marks total).
Show work in an exam setting. Here, answers are provided in collapsible
panels for self-checking.
Section A – Multiple Choice (4 marks each)
1.
The sequence an = 7(1.03)n + 4n represents a revenue model.
Which statement is TRUE?
Answer
C. 7(1.03)n is geometric and 4n is arithmetic; their sum is a combination of both.
2.
For an exponential function f(x) = a bx, suppose f(2) = 36
and f(5) = 972. Find b.
Answer
B. f(5)/f(2) = b3 = 972/36 = 27, so b = 3.
3.
If g(x) = 5e0.8x, what is g−1(x)?
Answer
A. y = 5e0.8x ⇒ ln(y/5) = 0.8x ⇒ x = (1/0.8) ln(y/5).
4.
Which is equivalent to 3 log(x) − (1/2) log(9x4)?
Answer
D. 3 log x = log(x3) and (1/2) log(9x4) = log(3x2). So the expression is log(x3) − log(3x2) = log(x3 / 3x2).
5.
Interest is compounded continuously. A deposit doubles in 9 years.
What is the interest rate r?
Answer
B. 2P = P e9r ⇒ e9r = 2 ⇒ r = (ln 2)/9.
6.
Which function grows fastest as x → ∞?
Answer
D. xx/2 is super-exponential; it dominates the others as x → ∞.
7.
Let h(x) = log(3x − 4). What is its domain?
Answer
B. Need 3x − 4 > 0 ⇒ x > 4/3.
8.
Solve for x: 2x+1 = 7.
Answer
A. x + 1 = log2 7 ⇒ x = log2 7 − 1.
9.
Composition: If f(x) = 2x and g(x) = ln x, then (f ∘ g)(x) equals:
Answer
B. (f ∘ g)(x) = f(g(x)) = 2ln x.
10.
Bacterial growth: P(t) = 800 e0.32t. After how long will
the population reach 6000?
Answer
A. 6000 = 800 e0.32t ⇒ e0.32t = 6000/800 ⇒ t = ln(6000/800)/0.32.
11.
Which function could generate a straight line on a semi-log plot
(logarithmic y-axis, linear x-axis)?
Answer
B. Exponential models y = a·bx appear as lines when log y is plotted versus x.
12.
In model validation, which indicates that a competing exponential model
fits better?
Answer
B. Smaller RSS means the model’s predictions are closer overall to the observed data.
Section B – Free Response (52 marks total)
Q1. (8 marks) – Sequences & Modeling
The table shows population of an endangered species:
Year t: 0, 1, 2, 3, 4
P(t): 410, 472, 544, 627, 723
(a) Determine whether the data is better modeled by a linear or an exponential function.
(b) Build a suitable exponential model P(t).
(c) Use your model to estimate P(8).
Answer (outline)
Ratios are nearly constant:
472/410 ≈ 1.15, 544/472 ≈ 1.15, 627/544 ≈ 1.15, so an exponential model fits better.
A reasonable model is P(t) ≈ 410(1.15)t (any base around 1.15 is acceptable).
Then P(8) ≈ 410(1.15)8 ≈ 1.25 × 10³
(about 1250–1300 animals, depending on rounding).
Q2. (8 marks) – Exponential Manipulation
Solve for x:
32x+1 − 7·3x = 18.
Answer
Let y = 3x > 0. Then 32x+1 = 3·(3x)² = 3y².
Equation: 3y² − 7y − 18 = 0.
y = [7 ± √(49 + 216)] / 6 = [7 ± √265] / 6.
Only the positive root is valid:
y = (7 + √265) / 6.
Therefore x = log3((7 + √265) / 6).
Q3. (8 marks) – Semi-log Plot & Decay
A medicine concentration decreases exponentially. On a semi-log plot of
concentration vs time, the straight-line equation is:
log10(C(t)) = −0.37 t + 2.81.
(a) Write the exponential function C(t) in standard form.
(b) Find the half-life.
Answer
(a) C(t) = 102.81 − 0.37t = 102.81 · 10−0.37t.
(b) Half-life t1/2 solves C(t1/2) = ½ C(0):
½ = 10−0.37 t1/2.
Taking log10: −0.3010 ≈ −0.37 t1/2 ⇒
t1/2 ≈ 0.3010 / 0.37 ≈ 0.81 time units.
Q4. (10 marks) – Composition & Inverses
Let
f(x) = e2x−3, g(x) = ln(x + 5).
(a) Find f(g(x)).
(b) Find g(f(x)).
(c) Find f−1(x).
Answer
(a) f(g(x)) = e2 ln(x + 5) − 3 = e−3(x + 5)².
(b) g(f(x)) = ln(e2x − 3 + 5).
(c) y = e2x − 3 ⇒ ln y = 2x − 3 ⇒ x = (ln y + 3)/2.
So f−1(x) = (ln x + 3)/2.
Q5. (10 marks) – Logarithmic Inequality
Solve completely:
2 ln(x − 3) + ln(5 − x) ≥ ln 6.
State any domain restrictions.
Answer
Domain: x − 3 > 0 ⇒ x > 3, and 5 − x > 0 ⇒ x < 5, so 3 < x < 5.
Combine logs:
2 ln(x − 3) + ln(5 − x) = ln[(x − 3)²(5 − x)].
So (x − 3)²(5 − x) ≥ 6 on 3 < x < 5.
For 3 < x < 5, we have 0 < (x − 3)² < 4 and 0 < 5 − x < 2,
so the product is < 8. Checking values (e.g. x = 4 gives 1) shows it
never reaches 6 inside the domain.
Therefore there are no real solutions; solution set is ∅.
Q6. (8 marks) – Competing Model Validation
Two schools model fuel consumption (in gallons) of buses:
Month: 1 2 3 4
Actual: 215 228 243 265
Model A:219 239 262 289
Model B:215 229 246 268
(a) Compute the residual sum of squares (RSS) for each model.
(b) State which model is better and why.
Answer
Residuals (Actual − Model):
Model A: −4, −11, −19, −24 ⇒ squares: 16, 121, 361, 576.
RSSA = 16 + 121 + 361 + 576 = 1074.
Model B: 0, −1, −3, −3 ⇒ squares: 0, 1, 9, 9.
RSSB = 0 + 1 + 9 + 9 = 19.
Model B is far better because it has a much smaller residual
sum of squares.