--- title: "AP Precalculus Unit 2 Practice Test" description: "A timed AP Precalculus Unit 2 (Exponential and Logarithmic Functions) practice test in the AP format, drawn from 1,736 multiple-choice questions and 33 FRQs written for the unit. The standard version takes about 85 minutes." canonical: "https://fiveable.me/ap-pre-calc/unit-test/2" type: "unit-test" subject: "AP Precalculus" unit: "Unit 2" --- # AP Precalculus Unit 2 Practice Test ## Overview A timed AP Precalculus Unit 2 (Exponential and Logarithmic Functions) practice test in the AP format, drawn from 1,736 multiple-choice questions and 33 FRQs written for the unit. The standard version takes about 85 minutes. ## Test Format - 20 Multiple Choice Questions (50 minutes) - 1 FRQ – FRQ 1 – Function Concepts (Calculator) (18 minutes) - 1 FRQ – FRQ 2 – Modeling a Non-Periodic Context (Calculator) (18 minutes) ## Unit FRQs by Type - [FRQ 1 – Function Concepts (Calculator)](/ap-pre-calc/frq-practice?type=frq-1-function-concepts-calculator&id=6977ca40384a204905d878e6): 11 questions - [FRQ 2 – Modeling a Non-Periodic Context (Calculator)](/ap-pre-calc/frq-practice?type=frq-2-modeling-a-non-periodic-context-calculator&id=6a72d0738dbb41238f18eb9e): 11 questions - [FRQ 4 – Symbolic Manipulations (No Calculator)](/ap-pre-calc/frq-practice?type=frq-4-symbolic-manipulations-no-calculator&id=6977ce6e55025854818eb1c6): 11 questions ## Practice Questions ### Question 1 The function $$f(x) = \log_2(x)$$ has domain $$(0, \infty)$$ and a vertical asymptote at $$x = 0$$. If $$r(x) = \log_2(4 - x)$$, which statement correctly describes the domain of $$r$$ and explains why it differs from $$f$$? - (A) Domain $$(-\infty, 4)$$ because the argument $$4 - x$$ must be positive, requiring $$4 - x > 0$$, which means $$x < 4$$. - (B) Domain $$(0, \infty)$$ because logarithmic functions always require positive arguments, and the transformation does not change this requirement. - (C) Domain $$(4, \infty)$$ because the $$4 - x$$ inside the logarithm shifts the domain right by 4 units from the original domain $$(0, \infty)$$. - (D) Domain $$(-\infty, 4)$$ because $$4 - x > 0$$; vertical asymptote at $$x = 4$$ because the asymptote shifts from $$x = 0$$ to $$x = 4$$. **Answer: (A)** The transformation $$\log_2(4 - x)$$ involves both a horizontal reflection (from the negative coefficient on $$x$$) and a horizontal shift. The domain restriction comes from requiring the argument to be positive: $$4 - x > 0$$ yields $$x < 4$$, giving domain $$(-\infty, 4)$$. ### Question 2 The function $$f(x) = \log_2(x)$$ has domain $$(0, \infty)$$ and vertical asymptote at $$x = 0$$. If $$t(x) = 2f(\frac{x}{4} - 1) - 3$$, what is the domain of $$t$$ and where is its vertical asymptote? - (A) Domain: $$(4, \infty)$$; Vertical asymptote: $$x = 4$$ - (B) Domain: $$(0, \infty)$$; Vertical asymptote: $$x = 0$$ - (C) Domain: $$(1, \infty)$$; Vertical asymptote: $$x = 1$$ - (D) Domain: $$(4, \infty)$$; Vertical asymptote: $$x = 1$$ **Answer: (A)** The transformation $$f(\frac{x}{4} - 1)$$ requires $$\frac{x}{4} - 1 > 0$$, so $$\frac{x}{4} > 1$$, giving $$x > 4$$. The domain is $$(4, \infty)$$. The vertical asymptote occurs where the argument equals zero: $$\frac{x}{4} - 1 = 0$$, so $$x = 4$$. Vertical and horizontal dilations do not affect the domain or asymptote location. ### Question 3 Let $$f(x) = 3\sin(x)$$ and $$g(x) = x + \frac{\pi}{4}$$. Which expression represents $$(f \circ g)(x)$$? - (A) $$3\sin\left(x + \frac{\pi}{4}\right)$$ - (B) $$3\sin(x) + \frac{\pi}{4}$$ - (C) $$3\sin(x) + 3\frac{\pi}{4}$$ - (D) $$\sin\left(3x + \frac{\pi}{4}\right)$$ **Answer: (A)** Composition $$(f \circ g)(x) = f(g(x)) = f\left(x + \frac{\pi}{4}\right) = 3\sin\left(x + \frac{\pi}{4}\right)$$. The argument of sine is replaced by the entire expression $$g(x)$$, resulting in a horizontal translation of the sine function. ### Question 4 A population model uses $$P(t) = 1000 \cdot 2^t$$. A biologist discovers that a second population follows the same exponential pattern but starts at 2500 individuals and grows at half the rate. Which transformed function models the second population? - (A) $$Q(t) = 2500 \cdot 2^{t/2}$$ - (B) $$Q(t) = 2500 \cdot 2^{2t}$$ - (C) $$Q(t) = 1000 \cdot 2^t + 1500$$ - (D) $$Q(t) = 2.5 \cdot 2^t$$ **Answer: (A)** The initial population of 2500 replaces the coefficient 1000. Growing at half the rate means replacing $$t$$ with $$t/2$$, which represents a horizontal dilation by factor 2, slowing the exponential growth. ### Question 5 A population of bacteria is modeled by $$N(t) = 1000 \cdot 2^t$$, where t is days. A researcher applies a treatment that eliminates half the bacteria immediately and then slows growth to one-third the original rate. Which function models the treated population? - (A) $$N(t) = 500 \cdot 2^{t/3}$$ - (B) $$N(t) = 1000 \cdot 2^{t - 1}$$ - (C) $$N(t) = 500 \cdot 2^t$$ - (D) $$N(t) = 1000 \cdot 2^{0.33t}$$ **Answer: (A)** Eliminating half the bacteria multiplies the initial population by 0.5 (additive effect on the constant). Slowing growth to one-third the rate requires replacing t with t/3, a horizontal dilation with $$b = 1/3$$. Combined: $$N(t) = 0.5 \cdot 1000 \cdot 2^{t/3} = 500 \cdot 2^{t/3}$$, demonstrating LO 1.12.A.5. ### Question 6 A financial analyst models investment growth using $$V(t) = 1000 \cdot 1.05^t$$, where $$t$$ is years and $$V(t)$$ is value in dollars. To model a different investment with a 2-year delayed start, the analyst uses $$W(t) = 1000 \cdot 1.05^{t-2}$$. Which assumption about this transformation is essential for the model to accurately represent the delayed investment? - (A) The 5% annual growth rate remains constant; only the start time shifts. - (B) The initial investment is reduced by 2 dollars from the principal. - (C) The growth rate is reduced to 3% annually instead of 5%. - (D) The investment value is doubled at all times due to the delay. **Answer: (A)** The transformation $$1.05^{t-2}$$ delays the exponential curve by 2 years while preserving the base 1.05, which represents the 5% annual return. This assumes market conditions and investment terms remain stable during the delay; if the growth rate itself changed, the base would need adjustment. ### Question 7 A financial analyst models the value of an investment as $$V(t) = 1500(1.06)^{t-2} + 400$$, where $$V$$ is value in dollars and $$t$$ is years since 2020. This model is derived from the parent function $$f(t) = (1.06)^t$$ through additive and multiplicative transformations. Which assumption about the investment's behavior is critical for this model to remain valid beyond the observed data? - (A) Assumes the growth rate remains constant at 6% annually and that external conditions (market, interest rates) do not change; if these change, extrapolations will be unreliable. - (B) Assumes the growth rate of 6% annually is an average that may fluctuate year-to-year, but these fluctuations average out over time; if growth becomes consistently higher or lower, the model remains valid because averaging still occurs. - (C) Assumes the growth rate remains constant at 6% annually and that external conditions do not change; if these conditions do change, the model parameters (1500 and 400) must be recalibrated but the exponential form remains valid for prediction. - (D) Assumes the growth rate remains constant at 6% annually and that external conditions do not change; this assumption is critical for the model to fit the observed historical data accurately from 2020 to 2024. **Answer: (A)** The exponential form with base 1.06 reflects an assumed constant annual growth rate. Because the model has no mechanism to detect or adjust for changing conditions, its accuracy beyond the observed data depends entirely on the growth rate and market conditions staying the same. ### Question 8 A bacterial culture doubles every 3 hours. The table records the population at selected times: Hour 0: 500 cells, Hour 3: 1,000 cells, Hour 6: 2,000 cells, Hour 9: 4,000 cells. If P(t) represents the population after t hours, which function models this data? - (A) $$P(t) = 500 \cdot 2^{t/3}$$ - (B) $$P(t) = 500 + 500t$$ - (C) $$P(t) = 500 \cdot 2^t$$ - (D) $$P(t) = 500 \cdot 3^{t/3}$$ **Answer: (A)** The population doubles every 3 hours, so the common ratio over 3-hour intervals is 2. The exponent t/3 scales the time to match the doubling period. Verification: $$P(0) = 500 \cdot 2^0 = 500$$, $$P(3) = 500 \cdot 2^1 = 1000$$, $$P(6) = 500 \cdot 2^2 = 2000$$. ### Question 9 A radioactive isotope's mass is recorded at the end of each year. Which statement correctly describes the relationship between consecutive terms of this sequence of masses? - (A) Each term is 0.75 times the previous term, forming a geometric sequence with ratio 0.75. - (B) Each term decreases by 20 grams, forming an arithmetic sequence with common difference −20. - (C) Each term is multiplied by 1.25, indicating the mass increases by a factor of 1.25. - (D) The ratio between consecutive terms varies, so the sequence is neither arithmetic nor geometric. **Answer: (A)** Dividing consecutive terms confirms the constant ratio: 60/80 = 0.75, 45/60 = 0.75, 33.75/45 = 0.75. This constant proportional change defines a geometric sequence with common ratio r = 0.75. ### Question 10 A video goes viral with views following a geometric pattern. The table records daily views: Day 1: 500 views, Day 2: 1,500 views, Day 3: 4,500 views, Day 4: 13,500 views. If g_n denotes the views on day n, which formula correctly expresses g_n? - (A) $$g_n = 500 \cdot 3^{n-1}$$ - (B) $$g_n = 500 \cdot 3^n$$ - (C) $$g_n = 500 + 1000(n-1)$$ - (D) $$g_n = 1000n - 500$$ **Answer: (A)** The common ratio is 1500/500 = 3. Using the formula $$g_n = g_1 \cdot 3^{n-1}$$ with g_1 = 500 gives: $$g_1 = 500 \cdot 3^0 = 500$$, $$g_2 = 500 \cdot 3^1 = 1500$$, $$g_3 = 500 \cdot 3^2 = 4500$$, matching all table entries. ## Review the Unit - [2.1 Change in Arithmetic and Geometric Sequences](/ap-pre-calc/unit-2/change-arithmetic-geometric-sequences/study-guide/TjmiwbtDpN420iuL) - [2.2 Change in Linear and Exponential Functions](/ap-pre-calc/unit-2/change-linear-exponential-functions/study-guide/lfvcJiKaWrIXcYHo) - [2.3 Exponential Functions](/ap-pre-calc/unit-2/exponential-functions/study-guide/5hZXVBTYwi72vxCy) - [2.4 Exponential Function Manipulation](/ap-pre-calc/unit-2/exponential-function-manipulation/study-guide/wgkA05QIw8C2351F) - [2.5 Exponential Function Context and Data Modeling](/ap-pre-calc/unit-2/exponential-function-context-data-modeling/study-guide/6gd54uXCIKjI00Sn) - [2.6 Competing Function Model Validation](/ap-pre-calc/unit-2/competing-function-model-validation/study-guide/VeTW7I04PfukXfeT) - [2.7 Composition of Functions](/ap-pre-calc/unit-2/composition-functions/study-guide/glFlt2HgsCSjvjSL) - [2.8 Inverse Functions](/ap-pre-calc/unit-2/inverse-functions/study-guide/JkTPSAR9TH5LfSXP) - [2.9 Logarithmic Expressions](/ap-pre-calc/unit-2/logarithmic-expressions/study-guide/OcOVToNeiQNxP11V) - [2.10 Inverses of Exponential Functions](/ap-pre-calc/unit-2/inverses-exponential-functions/study-guide/7mdx6zi19alJ4hK3) - [2.11 Logarithmic Functions](/ap-pre-calc/unit-2/logarithmic-functions/study-guide/U0eLmF48zLQJSMJA) - [2.12 Logarithmic Function Manipulation](/ap-pre-calc/unit-2/logarithmic-function-manipulation/study-guide/tBicQgqFwNoesetP) - [2.13 Exponential and Logarithmic Equations and Inequalities](/ap-pre-calc/unit-2/exponential-logarithmic-equations-inequalities/study-guide/Mor8iJ1w4VWPX8Wi) - [2.14 Logarithmic Function Context and Data Modeling](/ap-pre-calc/unit-2/logarithmic-function-context-data-modeling/study-guide/viEiJjFy5ba6uEyw) - [2.15 Semi-log Plots](/ap-pre-calc/unit-2/semi-log-plots/study-guide/sGsPkQ8aoU7UKBna) ## Other Unit Tests - [Unit 1: Polynomial and Rational Functions](/ap-pre-calc/unit-test/1) - [Unit 3: Trigonometric and Polar Functions](/ap-pre-calc/unit-test/3) ## About This Document Canonical Fiveable pages are available as Markdown at the same path plus `.md`. - [llms.txt](https://fiveable.me/llms.txt): index of Fiveable's sections and URL patterns - [llms-full.txt](https://fiveable.me/llms-full.txt): complete subject and unit listing - [MCP server](https://fiveable.me/mcp): call Fiveable as tools instead of fetching pages (`https://fiveable.me/api/mcp`) - [MCP server for AP teachers](https://fiveable.me/mcp/teachers): a teacher's classes, assignments and AP-rubric grading (`https://fiveable.me/api/mcp/teacher`)