Inverses of Exponential Functions
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AP Precalculus › Inverses of Exponential Functions
A cooling model is $f(t)=80\left(\dfrac34\right)^t$, where $t$ is minutes. Using the inverse, solve for $t$ when $f(t)=45$.
$t=-\log_{3/4}!\left(\dfrac{45}{80}\right)$
$t=\log_{3/4}!\left(\dfrac{45}{80}\right)$
$t=80\log_{3/4}(45)$
$t=\log_{80}!\left(45\cdot \dfrac34\right)$
Explanation
This question tests understanding of inverses of exponential functions in AP Precalculus, focusing on converting exponential forms to their inverses. Finding the inverse of an exponential function involves expressing the dependent variable in terms of the independent variable, often using logarithms. In this problem, the function f(t) = $80(3/4)^t$ represents exponential decay, and students must solve for t when f(t) = 45. Choice A is correct because starting with 45 = $80(3/4)^t$, dividing both sides by 80 gives 45/80 = $(3/4)^t$, which converts to t = log₃/₄(45/80). Choice D is incorrect because it adds a negative sign, possibly confusing the fractional base with negative exponents. To help students: Emphasize that fractional bases like 3/4 represent decay without needing negative signs. Practice distinguishing between the base being a fraction and the exponent being negative.
A population model is $f(t)=250\cdot 4^t$, where $t$ is years. Using the inverse, solve for $t$ when $f(t)=16000$.
$t=\log_4!\left(\dfrac{16000}{250}\right)$
$t=\log_{250}(16000\cdot 4)$
$t=-\log_4!\left(\dfrac{16000}{250}\right)$
$t=250\log_4(16000)$
Explanation
This question tests understanding of inverses of exponential functions in AP Precalculus, focusing on converting exponential forms to their inverses. Finding the inverse of an exponential function involves expressing the dependent variable in terms of the independent variable, often using logarithms. In this problem, the function f(t) = $250·4^t$ represents exponential growth, and students must solve for t when f(t) = 16000. Choice A is correct because starting with 16000 = $250·4^t$, dividing both sides by 250 gives 64 = $4^t$, which converts to t = log₄(64) = log₄(16000/250). Choice D is incorrect because it adds an unnecessary negative sign, possibly confusing this with a decay function. To help students: Emphasize checking whether the base is greater than 1 (growth) or between 0 and 1 (decay). Practice recognizing that positive results are expected when solving growth models for positive outputs.
A cooling process is $f(t)=90\left(\dfrac23\right)^t$, where $t$ is minutes. What is the inverse of the function $f(t)$?
$f^{-1}(x)=90\log_{2/3}(x)$
$f^{-1}(x)=\log_{2/3}!\left(\dfrac{x}{90}\right)$
$f^{-1}(x)=\left(\dfrac{x}{90}\right)^{3/2}$
$f^{-1}(x)=\log_{90}(\tfrac23 x)$
Explanation
This question tests understanding of inverses of exponential functions in AP Precalculus, focusing on converting exponential forms to their inverses. Finding the inverse of an exponential function involves expressing the dependent variable in terms of the independent variable, often using logarithms. In this problem, the function f(t) = $90(2/3)^t$ represents exponential decay, and students must find f^(-1)(x). Choice A is correct because swapping variables gives x = $90(2/3)^t$, then dividing by 90 gives x/90 = $(2/3)^t$, which inverts to t = log₂/₃(x/90), so f^(-1)(x) = log₂/₃(x/90). Choice D is incorrect because it uses a fractional exponent (3/2) rather than a logarithm, confusing reciprocals with logarithms. To help students: Emphasize that the inverse of $b^x$ is log_b(x), not x^(1/b). Practice distinguishing between different types of inverse operations.
A city’s population is approximated by $f(t)=900\cdot 2^t$, where $t$ is years. A demographer uses the inverse to determine when the population reaches a target value. Here, $f(t)$ is the population after $t$ years and 900 is the initial population. The base 2 indicates doubling per year in this model. What is the inverse of $f(t)$?
$f^{-1}(x)=\log_2!\left(\dfrac{x}{900}\right)$
$f^{-1}(x)=2^{-x/900}$
$f^{-1}(x)=\log_{900}(2x)$
$f^{-1}(x)=900,\log_2(x)$
Explanation
This question tests understanding of inverses of exponential functions in AP Precalculus, focusing on converting exponential forms to their inverses. Finding the inverse of an exponential function involves expressing the independent variable in terms of the dependent variable, using logarithms to extract the exponent. In this problem, the function f(t) = $900·2^t$ models population growth, and students must find f^(-1)(x) to determine time t given population x. Choice A is correct because starting with x = $900·2^t$, dividing by 900 gives x/900 = $2^t$, then applying log base 2 yields t = log₂(x/900). Choice B is incorrect because it multiplies by 900 outside the logarithm instead of dividing inside, reversing the proper operation needed to isolate the exponential term. To help students: Reinforce the connection between exponential and logarithmic forms. Practice checking answers by composing the function with its proposed inverse.
A population of deer is modeled by $f(t)=120\cdot 4^t$, where $t$ is measured in years. A wildlife manager needs the inverse to find the year when the herd reaches a given size. Here, $f(t)$ is the number of deer after $t$ years and 120 is the initial herd size. The base 4 indicates the herd quadruples each year in this model. What is the inverse of the function $f(t)$?
$f^{-1}(x)=\log_4!\left(\dfrac{x}{120}\right)$
$f^{-1}(x)=\log_{120}(4x)$
$f^{-1}(x)=\log_4(x)-120$
$f^{-1}(x)=4^{x}/120$
Explanation
This question tests understanding of inverses of exponential functions in AP Precalculus, focusing on converting exponential forms to their inverses. Finding the inverse of an exponential function involves expressing the independent variable in terms of the dependent variable, using logarithms to extract the exponent. In this problem, the function f(t) = $120·4^t$ models deer population growth, and students must find f^(-1)(x) to determine time t given population x. Choice A is correct because starting with x = $120·4^t$, dividing by 120 gives x/120 = $4^t$, then applying log base 4 yields t = log₄(x/120). Choice C is incorrect because it subtracts 120 instead of dividing inside the logarithm, confusing additive and multiplicative relationships in exponential functions. To help students: Reinforce that the coefficient is multiplied with the exponential term, requiring division to isolate. Practice the systematic approach: isolate the exponential expression, then apply the logarithm.
A town’s population is $f(t)=1000\cdot 1.5^t$, where $t$ is years. Using the inverse, solve for $t$ when $f(t)=3375$.
$t=\log_{1000}(3375\cdot 1.5)$
$t=\left(\dfrac{3375}{1000}\right)^{1/1.5}$
$t=1000\log_{1.5}(3375)$
$t=\log_{1.5}!\left(\dfrac{3375}{1000}\right)$
Explanation
This question tests understanding of inverses of exponential functions in AP Precalculus, focusing on converting exponential forms to their inverses. Finding the inverse of an exponential function involves expressing the dependent variable in terms of the independent variable, often using logarithms. In this problem, the function f(t) = $1000·1.5^t$ represents exponential growth, and students must solve for t when f(t) = 3375. Choice A is correct because starting with 3375 = $1000·1.5^t$, dividing both sides by 1000 gives 3.375 = $1.5^t$, which converts to t = log₁.₅(3375/1000). Choice D is incorrect because it attempts to solve using a fractional exponent rather than logarithms, confusing inverse operations. To help students: Emphasize that logarithms are the inverse operation for exponentials, not roots or fractional powers. Practice recognizing when each inverse operation is appropriate.
A town’s population follows $f(t)=500\cdot 2^t$, where $t$ is years. City planners need the inverse to predict when the population reaches a target size. Here, $f(t)$ is the population after $t$ years and 500 is the initial population. Assume the model remains valid over the time interval considered. Using the inverse, solve for $t$ when $f(t)=P$.
$f^{-1}(P)=\log_{500}(2P)$
$f^{-1}(P)=\log_2!\left(\dfrac{P}{500}\right)$
$f^{-1}(P)=\log_2(P)-500$
$f^{-1}(P)=500,\log_2(P)$
Explanation
This question tests understanding of inverses of exponential functions in AP Precalculus, focusing on converting exponential forms to their inverses. Finding the inverse of an exponential function involves expressing the independent variable in terms of the dependent variable, using logarithms to 'undo' the exponential operation. In this problem, the function f(t) = $500·2^t$ represents population growth, and students must find f^(-1)(P) to determine time t given population P. Choice B is correct because starting with P = $500·2^t$, dividing both sides by 500 gives P/500 = $2^t$, then applying log base 2 to both sides yields t = log₂(P/500). Choice A is incorrect because it subtracts 500 instead of dividing, misunderstanding how to isolate the exponential term. To help students: Emphasize the systematic process of isolating the exponential expression before applying logarithms. Practice the algebraic steps: divide by the coefficient, then apply the logarithm with the same base as the exponential.
A startup’s user base grows according to $f(t)=150\cdot 4^t$, where $t$ is weeks. The product team uses the inverse to find when users reach a benchmark. Here, $f(t)$ is the number of users after $t$ weeks and 150 is the initial number. Assume the weekly growth factor remains 4. Using the inverse, solve for $t$ when $f(t)=U$.
$f^{-1}(U)=\log_4(U)+150$
$f^{-1}(U)=\log_4!\left(\dfrac{U}{150}\right)$
$f^{-1}(U)=\dfrac{U}{150}$
$f^{-1}(U)=\log_{150}(4U)$
Explanation
This question tests understanding of inverses of exponential functions in AP Precalculus, focusing on converting exponential forms to their inverses. Finding the inverse of an exponential function involves expressing the independent variable in terms of the dependent variable, using logarithms to solve for the original input variable. In this problem, the function f(t) = $150·4^t$ models user growth, and students must find f^(-1)(U) to determine time t given user count U. Choice A is correct because starting with U = $150·4^t$, dividing by 150 gives U/150 = $4^t$, then applying log base 4 yields t = log₄(U/150). Choice C is incorrect because it adds 150 outside the logarithm instead of dividing inside, misunderstanding how to properly isolate the exponential expression. To help students: Emphasize the systematic approach to finding inverses of exponential functions. Practice recognizing that the logarithm base must match the base of the original exponential function.
A population of fish in a pond follows $f(t)=40\cdot 3^t$, where $t$ is months. A biologist needs the inverse to find how many months it takes to reach a measured population. Here, $f(t)$ is the fish count after $t$ months and 40 is the initial count. Assume no limiting factors affect the growth in this interval. Using the inverse, solve for $t$ when $f(t)=F$.
$f^{-1}(F)=\log_3(F)-40$
$f^{-1}(F)=\log_{40}(3F)$
$f^{-1}(F)=3^{F}/40$
$f^{-1}(F)=\log_3!\left(\dfrac{F}{40}\right)$
Explanation
This question tests understanding of inverses of exponential functions in AP Precalculus, focusing on converting exponential forms to their inverses. Finding the inverse of an exponential function involves expressing the independent variable in terms of the dependent variable, using logarithms to solve for the exponent. In this problem, the function f(t) = $40·3^t$ models fish population growth, and students must find f^(-1)(F) to determine time t given fish count F. Choice A is correct because starting with F = $40·3^t$, dividing by 40 gives F/40 = $3^t$, then applying log base 3 yields t = log₃(F/40). Choice C is incorrect because it subtracts 40 outside the logarithm instead of dividing inside, confusing the role of the initial value in the exponential model. To help students: Emphasize understanding what each part of the exponential function represents. Practice isolating the exponential term systematically before applying logarithms.
A savings account balance is modeled by $f(t)=1000\cdot 1.5^t$, where $t$ is years. An investor uses the inverse to determine how long it takes to reach a desired balance. Here, $f(t)$ is the balance after $t$ years and 1000 is the initial deposit. Assume the growth factor 1.5 stays constant. Using the inverse, solve for $t$ when $f(t)=B$.
$f^{-1}(B)=\log_{1000}(1.5B)$
$f^{-1}(B)=\dfrac{B}{1000}$
$f^{-1}(B)=\log_{1.5}!\left(\dfrac{B}{1000}\right)$
$f^{-1}(B)=\log_{1.5}(B)-1000$
Explanation
This question tests understanding of inverses of exponential functions in AP Precalculus, focusing on converting exponential forms to their inverses. Finding the inverse of an exponential function involves expressing the independent variable in terms of the dependent variable, using logarithms with the appropriate base. In this problem, the function f(t) = $1000·1.5^t$ models account balance growth, and students must find f^(-1)(B) to determine time t given balance B. Choice C is correct because starting with B = $1000·1.5^t$, dividing by 1000 gives B/1000 = $1.5^t$, then applying log base 1.5 yields t = log₁.₅(B/1000). Choice A is incorrect because it subtracts 1000 instead of dividing, misunderstanding the algebraic manipulation needed to isolate the exponential term. To help students: Emphasize that coefficients multiply the exponential term, so division is needed to isolate it. Practice recognizing that the logarithm base must match the exponential base for proper inverse operations.