Exponential Function Manipulation - AP Precalculus

Carbon-14 follows $N(t)=N_0\left(\tfrac12\right)^{t/5730}$ . When $N(t)=0.30N_0$ , find $t$ (years).
Answer: $t=5730\,\dfrac{\ln(0.30)}{\ln(1/2)}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model radioactive decay using the half-life formula N(t) = N₀ $(1/2)^{t/5730}$ , where carbon-14 has a half-life of 5730 years, and need to find when 30% of the original amount remains. Choice A is correct because setting 0.30N₀ = N₀ $(1/2)^{t/5730}$ and dividing by N₀ gives 0.30 = $(1/2)^{t/5730}$ , then taking ln of both sides yields ln(0.30) = (t/5730)·ln(1/2), so t = 5730·ln(0.30)/ln(1/2). Choice C is incorrect because it uses ln(2) instead of ln(1/2) in the denominator, which changes the sign since ln(2) = -ln(1/2), resulting in a negative time value. To help students: Emphasize that half-life problems use base 1/2 for decay, practice manipulating exponents with fractions, and verify that decay times are positive. Watch for: Sign errors with logarithms of fractions, forgetting to multiply by the half-life constant, and confusion about when material has decayed versus when it remains.
A medicine amount is $M(t)=80e^{-0.22t}$ (t in hours). Solve for $t$ when $M(t)=12$ mg.
Answer: $t=\dfrac{\ln(12/80)}{-0.22}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model medicine decay using M(t) = $80e^{-0.22t}$ , where the negative exponent indicates exponential decay, and need to find when the amount reaches 12 mg. Choice A is correct because setting 12 = $80e^{-0.22t}$ and dividing by 80 gives 12/80 = $e^{-0.22t}$ , then taking ln of both sides yields ln(12/80) = -0.22t, so t = ln(12/80)/(-0.22), which properly accounts for the negative coefficient. Choice D is incorrect because it uses positive 0.22 in the denominator, which would give a negative time value since ln(12/80) < 0. To help students: Emphasize careful attention to signs in exponential decay problems, practice working with continuous decay models using base e, and always verify that time values are positive and reasonable. Watch for: Sign errors with negative exponents, confusion about when to use negative coefficients, and computational mistakes with fractions.
A sample decays by 12% yearly: $A(t)=500(0.88)^t$ . When is $A(t)=200$ grams?
Answer: $t=\dfrac{\ln(0.4)}{\ln(0.88)}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model radioactive decay with A(t) = 500 $(0.88)^t$ , requiring manipulation to find when the amount reaches 200 grams. Choice A is correct because it applies logarithms properly: starting with 200 = 500 $(0.88)^t$ , dividing by 500 gives 0.4 = $(0.88)^t$ , then taking ln of both sides yields ln(0.4) = t·ln(0.88), so t = ln(0.4)/ln(0.88). Choice C is incorrect due to using 2.5 (which is 500/200) instead of 0.4 (which is 200/500), a common error when students confuse the direction of the ratio in decay problems. To help students: Emphasize that decay uses bases between 0 and 1, practice identifying whether to divide final by initial or initial by final, and verify that ln(0.4) and ln(0.88) are both negative (making t positive). Watch for: Confusion about decay versus growth, incorrect ratio setup, and sign errors when working with logarithms of numbers less than 1.
A savings account uses $A(t)=1500(1+r)^6$ after 6 years. If $A=2000$ , find $r$ .
Answer: $r=\left(\dfrac{4}{3}\right)^{1/6}-1$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model compound interest with A(t) = 1500 $(1+r)^6$ , requiring manipulation to find the interest rate r when the final amount is $2000. Choice A is correct because it applies the proper algebraic steps: starting with 2000 = 1500 $(1+r)^6$ , dividing by 1500 gives 4/3 = $(1+r)^6$ , taking the sixth root yields $(4/3)^{1/6}$ = 1+r, so r = $(4/3)^{1/6}$ - 1. Choice C is incorrect because it uses logarithms unnecessarily and doesn't isolate r properly, which often occurs when students default to logarithmic methods even when root extraction is more direct. To help students: Emphasize that when solving for the base (not the exponent), taking roots is often more efficient than logarithms, practice recognizing when the unknown is in the base versus the exponent, and verify the final answer makes sense as an interest rate. Watch for: Forgetting to subtract 1 to isolate r, confusion about when to use roots versus logarithms, and errors in fraction simplification.
Carbon-14 decay is $N(t)=N_0\left(\tfrac12\right)^{t/5730}$ . If $N(t)=0.25N_0$ , find $t$ in years.
Answer: $t=\dfrac{5730\ln(0.25)}{\ln(1/2)}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model carbon-14 decay with N(t) = N₀ $(1/2)^{t/5730}$ , requiring manipulation to find when 75% has decayed (leaving 25%). Choice A is correct because it properly handles the half-life formula: starting with 0.25N₀ = N₀ $(1/2)^{t/5730}$ , dividing by N₀ gives 0.25 = $(1/2)^{t/5730}$ , taking ln of both sides yields ln(0.25) = (t/5730)·ln(1/2), so t = 5730·ln(0.25)/ln(1/2). Choice C is incorrect due to omitting the factor 5730, which often occurs when students forget that the exponent contains t/5730, not just t. To help students: Emphasize understanding half-life formulas where the exponent is t divided by the half-life period, and practice problems involving fractional remaining amounts. Watch for: Forgetting to multiply by the half-life period, confusion about what fraction remains versus what has decayed, and sign errors with ln(1/2).
A city's population is $P(t)=120{,}000(1.03)^t$ (years). When will it reach $200{,}000$ people?
Answer: $t=\dfrac{\ln(200000/120000)}{\ln(1.03)}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model population growth with P(t) = 120,000 $(1.03)^t$ , requiring manipulation to find when the population reaches 200,000. Choice B is correct because it applies the change of base formula correctly: starting with 200,000 = 120,000 $(1.03)^t$ , dividing gives 200,000/120,000 = $1.03^t$ , then taking ln of both sides yields ln(200,000/120,000) = t·ln(1.03), so t = ln(200,000/120,000)/ln(1.03). Choice A is incorrect due to confusing the continuous growth rate 0.03 with ln(1.03), which often occurs when students mix up discrete and continuous growth formulas. To help students: Emphasize the difference between discrete growth $(1+r)^t$ and continuous growth $e^{rt}$ , and practice converting between exponential and logarithmic forms. Watch for: Confusion between growth rate and logarithm of growth factor, and errors in simplifying ratios before taking logarithms.
A cup of cocoa cools by $T(t)=20+65(0.85)^t$ (t in minutes). When will $T(t)=40^\circ\text{C}$ ?
Answer: $t=\dfrac{\ln(20/65)}{\ln(0.85)}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model Newton's law of cooling using T(t) = 20 + 65 $(0.85)^t$ , where temperature approaches room temperature (20°C) exponentially, and need to find when the temperature reaches 40°C. Choice A is correct because substituting gives 40 = 20 + 65 $(0.85)^t$ , subtracting 20 yields 20 = 65 $(0.85)^t$ , dividing by 65 gives 20/65 = $(0.85)^t$ , then taking ln of both sides results in ln(20/65) = t·ln(0.85), so t = ln(20/65)/ln(0.85). Choice D is incorrect because it uses ln(40) - ln(20) = ln(2) in the numerator instead of the correct ratio after accounting for the ambient temperature. To help students: Stress the importance of isolating the exponential term by first dealing with constants, practice recognizing cooling/heating models with asymptotic behavior, and always check that the solution makes physical sense. Watch for: Forgetting to subtract the ambient temperature, errors in fraction simplification, and confusion about which values represent initial versus final states.
A city's population is $P(t)=90{,}000e^{0.018t}$ (years). When will it reach $150{,}000$ people?
Answer: $t=\dfrac{\ln(150000/90000)}{0.018}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model continuous population growth with P(t) = 90, $000e^{0.018t}$ , requiring manipulation to find when population reaches 150,000. Choice B is correct because with continuous growth, the solution simplifies nicely: starting with 150,000 = 90, $000e^{0.018t}$ , dividing gives 150,000/90,000 = $e^{0.018t}$ , taking ln of both sides yields ln(150,000/90,000) = 0.018t, so t = ln(150,000/90,000)/0.018. Choice A is incorrect due to using ln(0.018) instead of 0.018 in the denominator, which often occurs when students apply logarithms unnecessarily to the growth rate. To help students: Emphasize that with $e^{rt}$ , taking ln directly gives rt without needing to take ln of r, and practice recognizing when logarithms are and aren't needed. Watch for: Over-application of logarithms, confusion between discrete and continuous models, and arithmetic errors in simplifying ratios.
A town's population follows $P(t)=52{,}000(1.03)^t$ (t in years). When will $P(t)=70{,}000$ ?
Answer: $t=\dfrac{\ln(70{,}000/52{,}000)}{\ln(1.03)}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model population growth using P(t) = 52,000 $(1.03)^t$ and need to find when the population reaches 70,000. Choice A is correct because it applies logarithms properly: starting with 70,000 = 52,000 $(1.03)^t$ , dividing both sides by 52,000 gives 70,000/52,000 = $(1.03)^t$ , then taking the natural logarithm of both sides yields ln(70,000/52,000) = t·ln(1.03), and solving for t gives t = ln(70,000/52,000)/ln(1.03). Choice B is incorrect due to inverting the fraction, which would give a negative time value since the population is growing. To help students: Emphasize the importance of maintaining the correct order when setting up ratios, practice isolating exponential terms before applying logarithms, and always check that the answer makes sense in context. Watch for: Inverted fractions, forgetting to apply logarithms to both sides, and arithmetic errors when simplifying.
A hot drink cools as $T(t)=20+70(0.92)^t$ . When will $T(t)=40^\circ\text{C}$ ?
Answer: $t=\dfrac{\ln(2/7)}{\ln(0.92)}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model Newton's law of cooling with T(t) = 20 + 70 $(0.92)^t$ , requiring manipulation to find when the temperature reaches 40°C. Choice A is correct because it applies logarithms after properly isolating the exponential term: starting with 40 = 20 + 70 $(0.92)^t$ , subtracting 20 gives 20 = 70 $(0.92)^t$ , dividing by 70 gives 2/7 = $(0.92)^t$ , taking ln of both sides yields ln(2/7) = t·ln(0.92), so t = ln(2/7)/ln(0.92). Choice B is incorrect because it uses ln(7/2) instead of ln(2/7), a common error when students invert the fraction or confuse which quantity is being divided. To help students: Emphasize the importance of isolating the exponential term by first dealing with added constants, practice setting up the ratio correctly (final - ambient)/(initial - ambient), and verify that both logarithms are negative for cooling problems. Watch for: Forgetting to subtract the ambient temperature, fraction inversion errors, and confusion about the physical meaning of the model.
A medication amount is $M(t)=80e^{-0.35t}$ (t in hours). When is $M(t)=20$ mg?
Answer: $t=\dfrac{\ln(1/4)}{-0.35}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model medication decay with M(t) = $80e^{-0.35t}$ , requiring manipulation to find when the amount reaches 20 mg. Choice A is correct because it applies logarithms properly to the decay model: starting with 20 = $80e^{-0.35t}$ , dividing by 80 gives 1/4 = $e^{-0.35t}$ , taking ln of both sides yields ln(1/4) = -0.35t, so t = ln(1/4)/(-0.35). Choice C is incorrect because it uses ln(4) instead of ln(1/4), a common error when students forget that 20/80 = 1/4, not 4, or when they incorrectly handle the reciprocal. To help students: Emphasize careful fraction simplification before applying logarithms, practice recognizing that ln(1/4) = -ln(4), and verify that dividing a negative by a negative yields a positive time. Watch for: Sign errors with negative exponents, confusion about ln(1/x) = -ln(x), and mistakes in simplifying the initial fraction.
A savings account uses $A(t)=5000(1.06)^t$ (years). When will the balance reach $10{,}000$ ?
Answer: $t=\dfrac{\ln(2)}{\ln(1.06)}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model compound interest with A(t) = 5000 $(1.06)^t$ , requiring manipulation to find when the balance doubles to 10,000. Choice B is correct because it applies logarithms properly: starting with 10,000 = 5000 $(1.06)^t$ , dividing gives 2 = $1.06^t$ , taking ln of both sides yields ln(2) = t·ln(1.06), so t = ln(2)/ln(1.06). Choice A is incorrect due to using 0.06 instead of ln(1.06) in the denominator, which often occurs when students confuse the interest rate with the logarithm of the growth factor. To help students: Practice distinguishing between the rate r and the growth factor (1+r), and emphasize that logarithms must be applied to the entire growth factor. Watch for: Mixing up discrete and continuous growth formulas, and errors in recognizing when an account doubles.
Money grows continuously: $A(t)=1200e^{0.045t}$ . How long until the balance reaches $2000$ dollars?
Answer: $t=\dfrac{\ln(2000/1200)}{0.045}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model continuous compound interest using A(t) = $1200e^{0.045t}$ and need to find when the balance reaches $2000. Choice A is correct because for continuous growth with base e, we solve 2000 = $1200e^{0.045t}$ by dividing both sides by 1200 to get 2000/1200 = $e^{0.045t}$ , then taking the natural logarithm gives ln(2000/1200) = 0.045t, so t = ln(2000/1200)/0.045. Choice D is incorrect because it mistakenly applies ln to the growth rate 0.045, treating it as if it were the base of the exponential rather than the coefficient in the exponent. To help students: Emphasize the difference between discrete compound interest (base like 1.045) and continuous compound interest (base e with rate in exponent), practice recognizing when to use ln versus when to divide directly, and always verify units match. Watch for: Confusion between continuous and discrete models, incorrect application of logarithms, and computational errors.
A bacteria culture is $B(t)=900\cdot 1.25^t$ (t in hours). Solve for $t$ when $B(t)=5000$ .
Answer: $t=\dfrac{\ln(5000/900)}{\ln(1.25)}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model bacteria growth using B(t) = 900· $1.25^t$ , where the population increases by 25% each hour, and need to find when it reaches 5000. Choice A is correct because setting 5000 = 900· $1.25^t$ and dividing by 900 gives 5000/900 = $1.25^t$ , then taking the natural logarithm yields ln(5000/900) = t·ln(1.25), so t = ln(5000/900)/ln(1.25). Choice D is incorrect because it uses ln(0.25) in the denominator instead of ln(1.25), likely confusing the growth rate (25% increase means multiplying by 1.25, not 0.25). To help students: Clarify that a 25% increase means multiplying by 1.25 (not 0.25), practice converting percentage changes to growth factors, and always check that growth problems yield positive time values. Watch for: Confusion between growth rates and growth factors, sign errors with logarithms, and misunderstanding percentage increases.
Carbon-14 model: $N(t)=N_0\left(\tfrac12\right)^{t/5730}$ . If $N(t)=0.30N_0$ , find $t$ .
Answer: $t=5730\,\dfrac{\ln(0.30)}{\ln(1/2)}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model carbon-14 decay with N(t) = N₀ $(1/2)^{t/5730}$ , requiring manipulation to find when 30% of the original amount remains. Choice A is correct because it applies the formula properly: starting with 0.30N₀ = N₀ $(1/2)^{t/5730}$ , dividing by N₀ gives 0.30 = $(1/2)^{t/5730}$ , taking ln of both sides yields ln(0.30) = (t/5730)·ln(1/2), so t = 5730·ln(0.30)/ln(1/2). Choice C is incorrect because it omits the half-life constant 5730, which often occurs when students forget that the exponent contains t/5730, not just t. To help students: Emphasize the structure of half-life formulas where the exponent is t divided by the half-life period, practice identifying all components of the exponential model, and verify units match (years in this case). Watch for: Forgetting to multiply by the half-life constant, confusion about whether to use ln(1/2) or ln(2) in the denominator, and errors in handling the fraction in the exponent.
A radioactive sample decays by 12% yearly: $A(t)=500(0.88)^t$ . Solve when $A(t)=200$ grams.
Answer: $t=\dfrac{\ln(2/5)}{\ln(0.88)}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model radioactive decay using A(t) = 500 $(0.88)^t$ , where the sample decreases by 12% yearly, and need to find when the amount reaches 200 grams. Choice A is correct because it applies the logarithm properly: starting with 200 = 500 $(0.88)^t$ , dividing by 500 gives 200/500 = 2/5 = $(0.88)^t$ , then taking the natural logarithm yields ln(2/5) = t·ln(0.88), so t = ln(2/5)/ln(0.88). Choice B is incorrect because it inverts the fraction to 5/2, which would give a negative time since ln(5/2) > 0 and ln(0.88) < 0. To help students: Stress that decay problems result in fractions less than 1 (2/5 < 1), practice recognizing decay rates (0.88 represents 88% remaining, or 12% decay), and verify that time values are positive. Watch for: Confusion between growth and decay rates, inverted fractions, and sign errors with logarithms.
A savings account uses $A(t)=5000(1.06)^t$ . When will the balance be $8000$ dollars?
Answer: $t=\dfrac{\ln(8000/5000)}{\ln(1.06)}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model compound interest using A(t) = 5000 $(1.06)^t$ , representing 6% annual growth, and need to find when the balance reaches $8000. Choice A is correct because setting 8000 = 5000 $(1.06)^t$ and dividing both sides by 5000 gives 8000/5000 = 8/5 = $(1.06)^t$ , then taking the natural logarithm yields ln(8000/5000) = t·ln(1.06), so t = ln(8000/5000)/ln(1.06). Choice D is incorrect because it uses log (common logarithm) instead of ln and divides by 1.06 directly rather than its logarithm, showing confusion about the logarithmic solution process. To help students: Practice using natural logarithms consistently in exponential problems, emphasize that any logarithm base works but ln is conventional, and verify solutions make financial sense. Watch for: Mixing logarithm bases, forgetting to apply logarithms to solve for exponents, and computational errors with ratios.
A city's population is $P(t)=180{,}000(1.018)^t$ . How many years until it reaches $250{,}000$ ?
Answer: $t=\dfrac{\ln(250{,}000/180{,}000)}{\ln(1.018)}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model population growth using P(t) = 180,000 $(1.018)^t$ , representing 1.8% annual growth, and need to find when the population reaches 250,000. Choice A is correct because setting 250,000 = 180,000 $(1.018)^t$ and dividing both sides by 180,000 gives 250,000/180,000 = $(1.018)^t$ , then taking the natural logarithm yields ln(250,000/180,000) = t·ln(1.018), so t = ln(250,000/180,000)/ln(1.018). Choice C is incorrect because it divides by 1.018 directly instead of ln(1.018), confusing the base of the exponential with its logarithm. To help students: Emphasize that solving exponential equations requires logarithms to 'bring down' the exponent, practice identifying growth rates from expressions like 1.018 (representing 1.8% growth), and verify answers by substitution. Watch for: Forgetting to apply logarithms, confusion between the base and its logarithm, and arithmetic errors in calculating ratios.
An artifact has 62% of its original carbon-14: $N(t)=N_0\left(\tfrac12\right)^{t/5730}$ . Find $t$ (years).
Answer: $t=5730\,\dfrac{\ln(0.62)}{\ln(1/2)}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we use the carbon-14 decay formula N(t) = N₀ $(1/2)^{t/5730}$ to find the age of an artifact that has 62% of its original carbon-14 remaining. Choice A is correct because setting 0.62N₀ = N₀ $(1/2)^{t/5730}$ and dividing by N₀ gives 0.62 = $(1/2)^{t/5730}$ , then taking ln of both sides yields ln(0.62) = (t/5730)·ln(1/2), so t = 5730·ln(0.62)/ln(1/2). Choice C is incorrect because it uses ln(2) instead of ln(1/2), which differs by a negative sign since ln(2) = -ln(1/2), resulting in a negative age. To help students: Emphasize that radioactive decay uses base 1/2 (not 2) for half-life calculations, practice working with percentages as decimals, and verify that calculated ages are positive. Watch for: Sign errors when working with ln(1/2), forgetting to multiply by the half-life period, and confusion about fractions versus their reciprocals.
A city grows 2.4% yearly: $P(t)=P_0(1.024)^t$ . If $P(10)=610{,}000$ , find $P_0$ .
Answer: $P_0=\dfrac{610{,}000}{(1.024)^{10}}$ This question tests AP Precalculus skills in manipulating exponential functions, including solving for variables using logarithms. Exponential functions model situations where quantities grow or decay at a constant relative rate, and logarithms allow us to solve for variables in the exponent. In this scenario, we model population growth with P(t) = P₀ $(1.024)^t$ , requiring manipulation to find the initial population P₀ when P(10) = 610,000. Choice A is correct because it properly isolates P₀: starting with 610,000 = P₀ $(1.024)^{10}$ , dividing both sides by $(1.024)^{10}$ gives P₀ = 610,000/ $(1.024)^{10}$ . Choice B is incorrect because it multiplies instead of divides, a common error when students confuse solving for the coefficient versus solving for the exponent. To help students: Emphasize the difference between solving for variables in different positions (coefficient vs. base vs. exponent), practice algebraic manipulation to isolate the unknown, and verify by substituting back into the original equation. Watch for: Confusion about when to multiply versus divide, errors in calculating powers of decimals, and misunderstanding the role of P₀ as the initial value.

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