Doubling Time & Half-Life - Differential Equations

1. Isotope A has a half-life of 20 days. Isotope B has a half-life of 30 days. A sample is prepared with an initial ratio of 4 atoms of Isotope A to 1 atom of Isotope B. After how many days will the ratio of atoms of Isotope A to Isotope B be 1-to-1?
   Answer: 120 days Let $N_A(t)$ and $N_B(t)$ be the number of atoms. Let $N_A(0) = 4N_0$ and $N_B(0) = N_0$. The decay constants are $k_A = -\ln(2)/20$ and $k_B = -\ln(2)/30$. The models are: $N_A(t) = 4N_0 e^{k_A t}$ and $N_B(t) = N_0 e^{k_B t}$. We want to find $t$ where $N_A(t) = N_B(t)$: $4N_0 e^{k_A t} = N_0 e^{k_B t}$ $4 = \frac{e^{k_B t}}{e^{k_A t}} = e^{(k_B - k_A)t}$ Taking the natural logarithm of both sides: $\ln(4) = (k_B - k_A)t$ $2\ln(2) = \left(-\frac{\ln(2)}{30} - \left(-\frac{\ln(2)}{20}\right)\right)t$ $2\ln(2) = \left(\frac{1}{20} - \frac{1}{30}\right) \ln(2) t$ $2 = \left(\frac{3-2}{60}\right)t = \frac{1}{60}t$ $t = 120 \text{ days.}$
2. A drug in the bloodstream follows first-order elimination kinetics with a half-life of 6 hours. If a patient takes 400 mg at time $t = 0$ and then 200 mg exactly 6 hours later, what is the total amount of drug in the bloodstream immediately after the second dose?
   Answer: 400 mg After 6 hours (one half-life), the original 400 mg dose decreases to $400 \times \frac{1}{2} = 200$ mg. When the 200 mg second dose is added immediately after, the total becomes $200 + 200 = 400$ mg. Choice A ignores the remaining amount from the first dose. Choice B incorrectly calculates the remaining first dose. Choice D simply adds both doses without accounting for elimination.
3. A colony of bacteria grows exponentially. Under current conditions, its doubling time is $T_d$. A new nutrient is introduced that increases the relative growth rate, $k$, by 50%. What is the new doubling time in terms of $T_d$?
   Answer: $(2/3) T_d$ The doubling time $T_d$ is inversely proportional to the growth rate $k$, given by the formula $T_d = \ln(2)/k$. The original growth rate is $k$. The new growth rate, $k_{new}$, is 50% larger than $k$. This means $k_{new} = k + 0.50k = 1.5k$. The new doubling time, $T_{new}$, is: $T_{new} = \frac{\ln(2)}{k_{new}} = \frac{\ln(2)}{1.5k} = \frac{1}{1.5} \left(\frac{\ln(2)}{k}\right) = \frac{1}{3/2} T_d = \frac{2}{3} T_d$
4. A cell culture starts with $N_0$ cells and its population doubles every 6 hours. After 24 hours, an inhibitor is added that causes the population to decay with a half-life of 3 hours. How many hours after the inhibitor is added will the population return to its original size, $N_0$?
   Answer: 12 hours First, calculate the population after the 24-hour growth phase. The doubling time is 6 hours, so in 24 hours, there are $24/6 = 4$ doubling periods. The population becomes $N(24) = N_0 \times 2^4 = 16N_0$. Next, this population of $16N_0$ begins to decay with a half-life of 3 hours. We want to find the time $t$ it takes to return to $N_0$. The decay model is $N(t) = (16N_0) (1/2)^{t/3}$. $N_0 = (16N_0) \left(\frac{1}{2}\right)^{t/3}$ $\frac{1}{16} = \left(\frac{1}{2}\right)^{t/3}$ Since $1/16 = (1/2)^4$, we have: $\left(\frac{1}{2}\right)^4 = \left(\frac{1}{2}\right)^{t/3}$ Equating the exponents: $4 = t/3$, which gives $t = 12$ hours.
5. The mass of a radioactive substance is measured at two points in time. At $t = 5$ years, there are 80 grams remaining. At $t = 20$ years, there are 10 grams remaining. Assuming the substance decays exponentially, what is its half-life?
   Answer: 5 years The amount of substance $M(t)$ follows the model $M(t) = M_0 e^{kt}$. We have two data points: $M(5) = 80 = M_0 e^{5k}$ $M(20) = 10 = M_0 e^{20k}$ Dividing the second equation by the first gives: $\frac{10}{80} = \frac{M_0 e^{20k}}{M_0 e^{5k}} \implies \frac{1}{8} = e^{15k}$ Alternatively, one can reason that the mass decreased by a factor of $80/10 = 8$ in $20 - 5 = 15$ years. Since $8 = 2^3$, this period of 15 years corresponds to 3 half-lives. Therefore, the half-life $T_h$ is $15 \text{ years} / 3 = 5$ years.
6. The rate of decay of a radioactive substance is proportional to the amount of substance present. When there are 100 grams of the substance, its mass is decreasing at a rate of 2 grams per day. What is the half-life of the substance?
   Answer: $50 \ln(2)$ days When you encounter radioactive decay problems, you're dealing with exponential decay where the rate of change is proportional to the current amount. This translates to the differential equation $\frac{dN}{dt} = -kN$, where $k$ is the decay constant. Given that 100 grams decreases at 2 grams per day, you can find the decay constant: $-2 = -k(100)$, so $k = 0.02 \text{ day}^{-1}$. The general solution to exponential decay is $N(t) = N_0 e^{-kt}$. Half-life occurs when $N(t) = \frac{N_0}{2}$, giving us: $\frac{N_0}{2} = N_0 e^{-kt_{1/2}}$ Dividing by $N_0$ and taking the natural logarithm: $\ln\left(\frac{1}{2}\right) = -kt_{1/2}$ $-\ln(2) = -0.02 \cdot t_{1/2}$ $t_{1/2} = \frac{\ln(2)}{0.02} = \frac{\ln(2)}{1/50} = 50\ln(2)$ Answer A) $50\ln(2)$ days is correct. Answer B) 50 days incorrectly assumes the decay constant equals 0.01 instead of 0.02, or confuses the relationship between half-life and decay rate. Answer C) 25 days likely comes from incorrectly thinking half-life is simply half the initial amount divided by the decay rate (100÷2÷2). Answer D) $50/\ln(2)$ days inverts the correct formula, a common algebraic error when solving $\ln(2) = kt_{1/2}$. Remember: in exponential decay problems, always set up the differential equation first, find the decay constant from given conditions, then use $t_{1/2} = \frac{\ln(2)}{k}$ for half-life calculations.
7. Population A starts with 1000 individuals and has a doubling time of 10 years. Population B starts with 250 individuals and has a doubling time of 5 years. In how many years will the two populations be equal?
   Answer: 20 years When you encounter exponential growth problems involving doubling times, you're dealing with exponential functions where populations grow according to $P(t) = P_0 \cdot 2^{t/d}$, where $P_0$ is the initial population, $t$ is time, and $d$ is the doubling time. For Population A: $P_A(t) = 1000 \cdot 2^{t/10}$ For Population B: $P_B(t) = 250 \cdot 2^{t/5}$ To find when they're equal, set the equations equal: $1000 \cdot 2^{t/10} = 250 \cdot 2^{t/5}$ Dividing both sides by 250: $4 \cdot 2^{t/10} = 2^{t/5}$ Since $4 = 2^2$: $2^2 \cdot 2^{t/10} = 2^{t/5}$ Using exponent rules: $2^{2 + t/10} = 2^{t/5}$ Therefore: $2 + \frac{t}{10} = \frac{t}{5}$ Solving for t: $2 = \frac{t}{5} - \frac{t}{10} = \frac{2t - t}{10} = \frac{t}{10}$ So $t = 20$ years, confirming answer D. Choice A (10 years) is Population A's doubling time, but ignores Population B's faster growth. Choice B (15 years) might come from averaging the doubling times, which doesn't work with exponential functions. Choice C (25 years) could result from incorrectly adding the doubling times. Study tip: In exponential growth problems, always set up the proper exponential equations rather than using linear thinking like averaging rates. The faster-growing population will eventually catch up, even if it starts smaller.
8. The remaining mass of a decaying radioactive isotope is reduced by 20% in 10 hours. What is the half-life of this isotope?
   Answer: $10 \frac{\ln(2)}{\ln(1.25)}$ hours If the mass is reduced by 20%, then 80% of the original mass remains. Let $M(t)$ be the mass at time $t$. We have $M(10) = 0.8 M_0$. The decay model is $M(t) = M_0 e^{kt}$. $0.8 M_0 = M_0 e^{10k} \implies 0.8 = e^{10k} \implies k = \frac{\ln(0.8)}{10}$ The half-life $T_h$ is given by $T_h = -\ln(2)/k$. $T_h = \frac{-\ln(2)}{\ln(0.8)/10} = -10 \frac{\ln(2)}{\ln(0.8)}$ Using the logarithm property $\ln(0.8) = \ln(4/5) = -\ln(5/4) = -\ln(1.25)$, we can rewrite the expression: $T_h = -10 \frac{\ln(2)}{-\ln(1.25)} = 10 \frac{\ln(2)}{\ln(1.25)} \text{ hours.}$
9. Two radioactive isotopes A and B have half-lives of 10 years and 15 years, respectively. Initially, there are equal amounts of both isotopes. After how many years will the amount of isotope A be exactly half the amount of isotope B?
   Answer: 30 years Let initial amount be $N_0$ for both. After time $t$: $N_A(t) = N_0 e^{-\frac{\ln 2}{10}t}$ and $N_B(t) = N_0 e^{-\frac{\ln 2}{15}t}$. We need $N_A(t) = \frac{1}{2}N_B(t)$: $N_0 e^{-\frac{\ln 2}{10}t} = \frac{1}{2}N_0 e^{-\frac{\ln 2}{15}t}$. This gives us $e^{-\frac{\ln 2}{10}t + \frac{\ln 2}{15}t} = \frac{1}{2}$, so $e^{\ln 2 \cdot t(-\frac{1}{10} + \frac{1}{15})} = \frac{1}{2}$. Since $-\frac{1}{10} + \frac{1}{15} = -\frac{1}{30}$, we get $e^{-\frac{\ln 2}{30}t} = \frac{1}{2}$, giving $t = 30$ years. Other choices represent errors in combining the decay rates.
10. A bacterial culture doubles in size every 4 hours. If the culture starts with 500 bacteria and grows according to the differential equation $\frac{dN}{dt} = kN$, what is the value of the growth constant $k$?
    Answer: $k = \frac{\ln 2}{4}$ per hour For exponential growth $N(t) = N_0 e^{kt}$, doubling means $2N_0 = N_0 e^{4k}$, so $2 = e^{4k}$. Taking natural log: $\ln 2 = 4k$, therefore $k = \frac{\ln 2}{4}$. Choice B incorrectly uses the ratio 2/4 without logarithm. Choice C inverts the correct expression. Choice D incorrectly uses $\ln 4$ and multiplies by 2.
11. A bacterial culture's population doubles every 4 hours. Assuming the population grows exponentially according to the model $dP/dt = kP$, approximately how long does it take for the population to triple its initial size?
    Answer: 6.3 hours The doubling time $T_d$ is related to the growth constant $k$ by $T_d = \ln(2)/k$. Given $T_d = 4$ hours, we have $k = \ln(2)/4$. We want to find the time $t$ when the population $P(t)$ is three times the initial population $P_0$. The model is $P(t) = P_0 e^{kt}$. We set $P(t) = 3P_0$: $3P_0 = P_0 e^{kt} \implies 3 = e^{kt} \implies t = \frac{\ln(3)}{k}$ Substituting the value of $k$: $t = \frac{\ln(3)}{\ln(2)/4} = 4 \frac{\ln(3)}{\ln(2)} = 4 \log_2(3)$ Using the approximations $\ln(3) \approx 1.0986$ and $\ln(2) \approx 0.6931$, we get: $t \approx 4 \times \frac{1.0986}{0.6931} \approx 4 \times 1.585 \approx 6.34 \text{ hours.}$
12. A investment account grows according to $\frac{dA}{dt} = rA$ where $r = 0.08$ per year. If the doubling time for this account is approximately 8.66 years, and the account starts with $5,000, what will be the account balance after exactly 3 doubling periods?
    Answer: $40,000 After 3 doubling periods, the amount becomes $A_0 \times 2^3 = 5000 \times 8 = 40,000$ dollars. Each doubling period multiplies the amount by 2, so after 3 periods we multiply by $2^3 = 8$. Choice A incorrectly adds $10,000 for each doubling period. Choice B uses $2^2 = 4$ instead of $2^3 = 8$. Choice C uses an incorrect multiplication factor of 6.
13. A radioactive isotope has a half-life of 8 years. If a sample initially contains 120 grams of the isotope, after how many years will the sample contain exactly 7.5 grams?
    Answer: 32 years Using the exponential decay formula $N(t) = N_0 e^{-\lambda t}$ where $\lambda = \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{8}$. We need to solve $7.5 = 120 e^{-\frac{\ln 2}{8} \cdot t}$. This gives us $\frac{7.5}{120} = \frac{1}{16} = e^{-\frac{\ln 2}{8} \cdot t}$. Taking natural log: $\ln(\frac{1}{16}) = -\frac{\ln 2}{8} \cdot t$. Since $\ln(\frac{1}{16}) = -4\ln 2$, we get $t = 32$ years. Choice A uses 3 half-lives instead of 4. Choice C incorrectly uses 5 half-lives. Choice D uses 6 half-lives.
14. A radioactive material has a half-life of $T$. How much time must pass until only $1/10$ of the original material remains?
    Answer: $T \log_2(10)$ The amount of material $P(t)$ is given by $P(t) = P_0 e^{kt}$. The half-life $T$ is related to $k$ by $T = -\ln(2)/k$, so $k = -\ln(2)/T$. We want to find the time $t$ when $P(t) = (1/10)P_0$. $\frac{1}{10}P_0 = P_0 e^{kt} \implies \frac{1}{10} = e^{kt}$ Taking the natural logarithm of both sides: $\ln(1/10) = kt \implies -\ln(10) = kt$ $t = \frac{-\ln(10)}{k} = \frac{-\ln(10)}{-\ln(2)/T} = T \frac{\ln(10)}{\ln(2)}$ Using the change of base formula for logarithms, $\log_b(x) = \ln(x)/\ln(b)$, this is equivalent to $T \log_2(10)$.
15. A radioactive sample contains two isotopes with half-lives of 4 hours and 12 hours. Initially, 60% of the radioactivity comes from the first isotope and 40% from the second. After how many hours will both isotopes contribute equally to the total radioactivity?
    Answer: 6 hours Let total initial activity be $A_0$. Initially: isotope 1 has $0.6A_0$, isotope 2 has $0.4A_0$. After time $t$: $A_1(t) = 0.6A_0 e^{-\frac{\ln 2}{4}t}$ and $A_2(t) = 0.4A_0 e^{-\frac{\ln 2}{12}t}$. Setting $A_1(t) = A_2(t)$: $0.6 e^{-\frac{\ln 2}{4}t} = 0.4 e^{-\frac{\ln 2}{12}t}$. This gives us $\frac{0.6}{0.4} = e^{\frac{\ln 2}{4}t - \frac{\ln 2}{12}t}$, so $1.5 = e^{\frac{\ln 2}{6}t}$. Therefore $t = \frac{6 \ln(1.5)}{\ln 2} \approx 6$ hours. Other choices represent errors in setting up or solving the exponential equation.
16. A scientist has two samples. Sample X is a bacterial culture that doubles its mass every 3 hours. Sample Y is a radioactive isotope that loses half its mass every 4 hours. At $t=0$, the mass of Sample X is 10 grams and the mass of Sample Y is 160 grams. At what time $t$ are the masses of the two samples equal?
    Answer: $48/7$ hours When you encounter problems involving exponential growth and decay, you're dealing with processes that change at rates proportional to their current values. This creates exponential functions that either grow (like bacteria) or decay (like radioactive isotopes). For Sample X (bacterial growth), since it doubles every 3 hours, its mass follows $M_X(t) = 10 \cdot 2^{t/3}$. For Sample Y (radioactive decay), since it halves every 4 hours, its mass follows $M_Y(t) = 160 \cdot (1/2)^{t/4}$. To find when the masses are equal, set them equal: $10 \cdot 2^{t/3} = 160 \cdot (1/2)^{t/4}$ Dividing by 10: $2^{t/3} = 16 \cdot (1/2)^{t/4}$ Since $16 = 2^4$ and $(1/2)^{t/4} = 2^{-t/4}$: $2^{t/3} = 2^4 \cdot 2^{-t/4} = 2^{4-t/4}$ Taking logarithms: $\frac{t}{3} = 4 - \frac{t}{4}$ Multiplying by 12: $4t = 48 - 3t$, so $7t = 48$, giving $t = \frac{48}{7}$ hours. Answer A is correct. Answer B (16 hours) might come from incorrectly setting up the initial ratio. Answer C (12 hours) could result from computational errors in solving the logarithmic equation. Answer D (48 hours) appears if you forget to divide by 7 in the final step. Remember: exponential growth/decay problems require careful attention to the base and exponent setup. Always verify your exponential functions match the given doubling/halving times before solving.
17. A bacterial population grows exponentially with a doubling time of 20 minutes. Due to limited nutrients, the growth rate decreases linearly over time according to $k(t) = k_0(1 - 0.02t)$ where $t$ is in minutes and $k_0$ is the initial growth rate. At what time will the instantaneous doubling time first reach 40 minutes?
    Answer: 25 minutes This problem combines exponential growth with time-varying parameters, testing your understanding of how growth rates relate to doubling times. The key insight is recognizing that doubling time and growth rate are inversely related. For exponential growth, the relationship between doubling time $T_d$ and growth rate $k$ is $T_d = \frac{\ln(2)}{k}$. Since the initial doubling time is 20 minutes, we can find the initial growth rate: $k_0 = \frac{\ln(2)}{20}$. At any time $t$, the growth rate is $k(t) = k_0(1 - 0.02t) = \frac{\ln(2)}{20}(1 - 0.02t)$. The instantaneous doubling time becomes $T_d(t) = \frac{\ln(2)}{k(t)} = \frac{\ln(2)}{\frac{\ln(2)}{20}(1 - 0.02t)} = \frac{20}{1 - 0.02t}$. To find when the doubling time reaches 40 minutes, we solve: $40 = \frac{20}{1 - 0.02t}$. Cross-multiplying gives $40(1 - 0.02t) = 20$, so $40 - 0.8t = 20$, which yields $0.8t = 20$, therefore $t = 25$ minutes. Answer A (20 minutes) gives the original doubling time, not when it doubles. Answer B (30 minutes) would result from incorrectly setting up the proportion as $\frac{40}{20} = 1.5$ and multiplying by 20. Answer D (35 minutes) might come from arithmetic errors in the algebraic manipulation. When working with time-varying exponential processes, always establish the relationship between the parameter you're given and what you need to find. Here, converting between growth rate and doubling time was crucial for setting up the correct equation.
18. A population follows logistic growth but in its early stages approximates exponential growth with doubling time of 12 days. If the population reaches 10,000 individuals after 36 days of exponential-like growth, what was the initial population size?
    Answer: 1,250 individuals Using $N(t) = N_0 e^{kt}$ where $k = \frac{\ln 2}{12}$ per day. After 36 days: $10,000 = N_0 e^{\frac{\ln 2}{12} \cdot 36} = N_0 e^{3 \ln 2} = N_0 \cdot 2^3 = 8N_0$. Therefore $N_0 = \frac{10,000}{8} = 1,250$. Choice A uses $2^4 = 16$ instead of $2^3 = 8$. Choice C uses $2^2 = 4$. Choice D uses $2^1 = 2$.
19. The quantity $Q(t)$ of a substance follows the law of exponential decay, $dQ/dt = -kQ$ with $k > 0$. Let $T_h$ be the half-life. Let $T_{1/3}$ be the time it takes for the substance to decay to one-third of its initial amount. What is the ratio $T_{1/3} / T_h$?
    Answer: $\log_2(3)$ For exponential decay $Q(t) = Q_0 e^{-kt}$, the half-life $T_h$ is the time when $Q(T_h) = (1/2)Q_0$. $(1/2)Q_0 = Q_0 e^{-k T_h} \implies \ln(1/2) = -k T_h \implies -\ln(2) = -k T_h \implies T_h = \frac{\ln(2)}{k}$ Similarly, $T_{1/3}$ is the time when $Q(T_{1/3}) = (1/3)Q_0$. $(1/3)Q_0 = Q_0 e^{-k T_{1/3}} \implies \ln(1/3) = -k T_{1/3} \implies -\ln(3) = -k T_{1/3} \implies T_{1/3} = \frac{\ln(3)}{k}$ The ratio is: $\frac{T_{1/3}}{T_h} = \frac{\ln(3)/k}{\ln(2)/k} = \frac{\ln(3)}{\ln(2)}$ Using the change of base formula for logarithms, $\log_b(x) = \ln(x)/\ln(b)$, this ratio is equal to $\log_2(3)$.
20. An initial investment $P_0$ is made into an account that earns interest compounded continuously. The investment triples in value in 18 years. How many years does it take for the investment to double in value?
    Answer: $18 \log_3(2)$ years The value of the investment is $V(t) = P_0 e^{rt}$. We are given that the investment triples in 18 years, so $V(18) = 3P_0$. $3P_0 = P_0 e^{r \cdot 18} \implies 3 = e^{18r} \implies \ln(3) = 18r \implies r = \frac{\ln(3)}{18}$ We want to find the doubling time, $T_d$, which is the time when $V(T_d) = 2P_0$. $2P_0 = P_0 e^{r T_d} \implies 2 = e^{r T_d} \implies \ln(2) = r T_d$ $T_d = \frac{\ln(2)}{r}$ Substituting the expression for $r$: $T_d = \frac{\ln(2)}{\ln(3)/18} = 18 \frac{\ln(2)}{\ln(3)}$ Using the change of base formula for logarithms, this is $18 \log_3(2)$ years.

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