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.

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.

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.

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.

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.

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.

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.

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.

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.

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.