Composition of Functions - AP Precalculus

A sound sensor models intensity by $g(x)=10^x$ and converts to decibels with $f(I)=10\log_{10}(I)$ . Based on the scenario, what is $f(g(3))$ ?
Answer: $30$ This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential functions and logarithmic decibel conversion. Composition requires applying one function to the output of another, where g(x) = $10^x$ models intensity and f(I) = 10log₁₀(I) converts to decibels. In this scenario, we need to find f(g(3)), which means first calculating g(3) = 10³ = 1000, then applying f to get the decibel level. Choice A is correct because f(g(3)) = f(1000) = 10log₁₀(1000) = 10·3 = 30 decibels. Choice D is incorrect because it gives g(3) = 1000 instead of f(g(3)), confusing the intensity value with the decibel measurement. To help students: Draw diagrams showing the flow from input through each function, practice with the decibel formula, and emphasize that log₁₀( $10^n$ ) = n is a key logarithm property.
A sample decays by $g(t)=100\left(\tfrac12\right)^t$ and analysis uses $f(x)=\log_{10}(x)$ . Using the function defined, what is $f(g(2))$ ?
Answer: $\log_{10}(25)$ This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential decay and logarithmic analysis. Composition requires evaluating g(t) first to find the remaining amount, then applying f(x) = log₁₀(x) to analyze the scale. In this scenario, a sample decays by g(t) = 100(½)^t and we need to find f(g(2)), which means first calculating g(2) then applying the logarithm. Choice A is correct because g(2) = 100(½)² = 100(¼) = 25, so f(g(2)) = log₁₀(25). Choice B is incorrect because it gives log₁₀(50), which would be g(1) not g(2), confusing the decay rate. To help students: Emphasize evaluating functions step-by-step, practice with powers of ½ like (½)² = ¼, and verify answers by checking that log₁₀(25) ≈ 1.398.
A culture grows as $g(t)=500\cdot 2^{t/3}$ and analysis uses $f(x)=\ln(x)$ . Using the function defined, what is $f(g(6))$ ?
Answer: $\ln(2000)$ This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and natural logarithm functions. Composition f(g(x)) means we evaluate g(x) first, then apply f to that result, requiring careful attention to function order and properties. In this scenario, a culture grows as g(t) = 500· $2^{t/3}$ and analysis uses f(x) = ln(x), so we need to find f(g(6)). Choice A is correct because g(6) = 500· $2^{6/3}$ = 500·2² = 500·4 = 2000, so f(g(6)) = ln(2000). Choice B is incorrect because it represents ln(1000), which might come from miscalculating the exponential growth. To help students: Emphasize evaluating the inner function completely before applying the outer function, practice with specific numerical values, and reinforce properties of exponential functions like $2^{6/3}$ = 2².
A greenhouse warms as $g(t)=20+15(1.08)^t$ and uses $f(x)=\ln(x-20)$ to isolate growth. Using the function defined, what is $f(g(t))$ ?
Answer: $\ln\!\bbig(15(1.08)^t\u00big)$ This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential growth with a vertical shift and logarithmic transformation. Composition f(g(t)) requires substituting the entire expression g(t) into f(x), being careful with domain restrictions and algebraic manipulation. In this scenario, greenhouse temperature is g(t) = 20 + 15 $(1.08)^t$ and f(x) = ln(x - 20) isolates the growth component by removing the baseline temperature. Choice A is correct because f(g(t)) = ln(g(t) - 20) = ln((20 + 15 $(1.08)^t$ ) - 20) = ln(15 $(1.08)^t$ ). Choice B is incorrect because it includes the constant 20 inside the logarithm, failing to subtract it as required by f(x) = ln(x - 20). To help students: Emphasize careful substitution and simplification, highlight how f(x) = ln(x - 20) specifically removes the baseline value, and practice with functions that have domain restrictions.
Radioactive mass is $N(t)=200e^{-0.4t}$ and $g(x)=\ln(x/200)$ . Based on the scenario, solve for $t$ in $(g\circ N)(t)=-2$ .
Answer: $t=5$ This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and logarithmic functions. We need to solve (g∘N)(t) = -2, which means g(N(t)) = -2. In this radioactive decay scenario, the logarithmic function extracts time information from the remaining mass. Setting up: ln( $200e^{-0.4t}$ /200) = -2, which simplifies to ln( $e^{-0.4t}$ ) = -2, giving -0.4t = -2, so t = 5. Choice B is correct because solving -0.4t = -2 yields t = 5. Choice A shows 0.8, which might result from arithmetic errors. To help students: use the property ln( $e^x$ ) = x to simplify, check the answer by substituting back (N(5) = $200e^{-2}$ and g( $200e^{-2}$ ) = ln( $e^{-2}$ ) = -2), and practice solving equations involving compositions of exponentials and logarithms.
A radioactive sample has $N(t)=80\left(\tfrac12\right)^{t/4}$ grams and $h(x)=\log_{1/2}(x/80)$ . Based on the scenario, what is $(h\circ N)(12)$ ?
Answer: $3$ This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and logarithmic functions. Composition requires evaluating (h∘N)(12) = h(N(12)), applying h to the output of N. In this scenario, radioactive decay follows an exponential model, and the logarithmic function extracts the time information. First calculate N(12) = 80× $(1/2)^{12/4}$ = 80× $(1/2)^3$ = 80×(1/8) = 10, then apply h: h(10) = log₁/₂(10/80) = log₁/₂(1/8) = log₁/₂( $(1/2)^3$ ) = 3. Choice B is correct because $\log_b$ ( $b^x$ ) = x applies here. Choice A might result from a sign error, as some students confuse logs with base less than 1. To help students: review that log₁/₂(1/8) asks 'what power of 1/2 gives 1/8?', practice with fractional bases, and verify answers by checking $(1/2)^3$ = 1/8.
A population grows as $g(t)=300e^{0.04t}$ and analysis uses $f(x)=\ln(x/300)$ . Using the function defined, what is $f(g(t))$ ?
Answer: $0.04t$ This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential growth with base e and natural logarithm transformations. Composition requires substituting g(t) into f(x) and simplifying using logarithm properties, particularly ln( $e^x$ ) = x. In this scenario, population grows as g(t) = $300e^{0.04t}$ and f(x) = ln(x/300) normalizes relative to initial population, requiring careful algebraic manipulation. Choice B is correct because f(g(t)) = ln( $300e^{0.04t}$ /300) = ln( $e^{0.04t}$ ) = 0.04t, using the fundamental property that ln and e are inverse functions. Choice A is incorrect because it fails to simplify ln( $e^{0.04t}$ ) = 0.04t, keeping the unnecessary ln(300) term. To help students: Emphasize the inverse relationship between ln and e, practice simplifying expressions like ln( $ae^x$ /a) = ln( $e^x$ ) = x, and reinforce that f removes the initial population factor.
In a greenhouse, temperature is $T(t)=20+15(1.1)^t$ and $f(x)=\log_{1.1}\!\left(\frac{x-20}{15}\right)$ . Using the function defined, what is $(f\circ T)(4)$ ?
Answer: $4$ This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and logarithmic functions. Composition means evaluating (f∘T)(4) = f(T(4)), where f is applied to the temperature at time 4. In this greenhouse scenario, temperature grows exponentially from a baseline, and the logarithmic function extracts the time from temperature readings. Calculate T(4) = 20 + 15 $(1.1)^4$ , then apply f: f(20 + 15 $(1.1)^4$ ) = log₁.₁((20 + 15 $(1.1)^4$ - 20)/15) = log₁.₁( $(1.1)^4$ ) = 4. Choice A is correct because the functions are designed to be inverses of each other. Choice B shows $(1.1)^4$ without the logarithm applied. To help students: recognize when functions undo each other, practice identifying inverse relationships, and verify by checking that f recovers the original time input.
A sound meter models intensity by $I(d)=10^{6-0.2d}$ and converts to decibels with $L(x)=10\log_{10}(x)$ . Based on the scenario, what is $(L\circ I)(10)$ ?
Answer: $40$ This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and logarithmic functions. Composition involves applying one function to the result of another, denoted as (L∘I)(10) = L(I(10)). In this scenario, sound intensity decreases exponentially with distance, and the logarithmic function converts intensity to decibels. To find (L∘I)(10), first calculate I(10) = $10^{6-0.2×10}$ = $10^{6-2}$ = $10^4$ , then apply L: L( $10^4$ ) = 10log₁₀( $10^4$ ) = 10×4 = 40. Choice A is correct because it accurately computes the composition, yielding 40 decibels. Choice D might tempt students who stop at I(10) = $10^4$ without applying the second function. To help students: emphasize the order of operations in composition, practice substituting step-by-step, and reinforce that log₁₀( $10^n$ ) = n.
A population grows as $P(t)=500\cdot 3^t$ and $f(x)=\log_3(x)$ . Using the function defined, what transformation occurs in $(f\circ P)(t)$ ?
Answer: Vertical shift up by $\log_3(500)$ This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and logarithmic functions. The composition (f∘P)(t) = f(P(t)) = log₃( $500×3^t$ ) requires understanding logarithmic properties. In this population growth scenario, applying the logarithm to the exponential function reveals the transformation. Using logarithm properties: log₃( $500×3^t$ ) = log₃(500) + log₃( $3^t$ ) = log₃(500) + t. Choice B is correct because this represents a vertical shift up by log₃(500) units from the basic function y = t. Choice A incorrectly suggests a vertical stretch, but logarithms convert multiplication to addition. To help students: review logarithm properties like log(ab) = log(a) + log(b), graph both the original exponential and the composition, and recognize that log₃( $3^t$ ) = t creates a linear function with a vertical shift.
A bacteria culture follows $P(t)=200e^{0.3t}$ and $g(x)=\ln(x/200)$ recovers elapsed time. Based on the scenario, what is $(g\circ P)(5)$ ?
Answer: $1.5$ This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and logarithmic functions. Composition means applying g to the output of P, denoted (g∘P)(5) = g(P(5)). In this scenario, bacteria grow exponentially, and the logarithmic function recovers the elapsed time from the population count. Calculate P(5) = $200e^{0.3×5}$ = $200e^{1.5}$ , then apply g: g( $200e^1$ .5) = ln( $200e^1$ .5/200) = ln( $e^1$ .5) = 1.5. Choice A is correct because it uses the fundamental property that ln( $e^x$ ) = x. Choice C shows $e^{1.5}$ , which is P(5)/200, not the final composition result. To help students: emphasize that ln and e are inverse functions, practice recognizing when expressions simplify, and work through the composition step-by-step to avoid shortcuts.
An investment grows as $A(t)=500\cdot 2^{0.5t}$ dollars, and $f(x)=\log_2(x/500)$ analyzes growth. Using the function defined, what is $(f\circ A)(6)$ ?
Answer: $3$ This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and logarithmic functions. Composition requires applying f to the result of A, written as (f∘A)(6) = f(A(6)). In this scenario, an investment grows exponentially, and the logarithmic function analyzes the growth factor relative to the initial amount. First calculate A(6) = 500× $2^{0.5×6}$ = 500× $2^3$ = 500×8 = 4000, then apply f: f(4000) = log₂(4000/500) = log₂(8) = log₂( $2^3$ ) = 3. Choice A is correct because it properly evaluates the composition. Choice B might result from confusing the input value 6 with the output. To help students: draw function diagrams showing the flow from input to output, practice identifying which function to apply first, and use the property that $\log_b$ ( $b^x$ ) = x.

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