Differential Equations Quiz: Doubling Time And Half Life

What this quiz covers

This quiz focuses on Doubling Time And Half Life, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 6.0 hours
2. 6.3 hours (correct answer)
3. 8.0 hours
4. 5.8 hours

Explanation: 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: $t \approx 4 \times \frac{1.0986}{0.6931} \approx 4 \times 1.585 \approx 6.34 \text{ hours.}$

Question 2

1. $3/2$
2. $2/3$
3. $\log_3(2)$
4. $\log_2(3)$ (correct answer)

Explanation: 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 $\log_2(3)$.

Question 3

1. 24 days
2. 60 days
3. 120 days (correct answer)
4. 150 days

Explanation: 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: $\ln(4) = (k_B - k_A)t$ $2\ln(2) = \left(-\frac{\ln(2)}{30} - \left(-\frac{\ln(2)}{20}\right)\right)t$ $2\ln(2) = \left(\frac{1}{20} - \frac{1}{30}\right) \ln(2) t$ $2 = \left(\frac{3-2}{60}\right)t = \frac{1}{60}t$ $t = 120 \text{ days.}$

Question 4

1. $T \log_{10}(2)$
2. $T \log_2(10)$ (correct answer)
3. $3.5 T$
4. $5 T$

Explanation: 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 $T \log_2(10)$.

Question 5

1. 4 hours
2. 9 hours
3. 12 hours (correct answer)
4. 24 hours

Explanation: 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.

Question 6

1. $50 \ln(2)$ days (correct answer)
2. 50 days
3. 25 days
4. $50 / \ln(2)$ days

Explanation: 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.

Question 7

1. 12 years
2. $18 \log_3(2)$ years (correct answer)
3. $18 \log_2(3)$ years
4. 9 years

Explanation: 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.

Question 8

1. $1.5 T_d$
2. $0.5 T_d$
3. $(2/3) T_d$ (correct answer)
4. $T_d / \ln(1.5)$

Explanation: 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: $T_{new} = \frac{\ln(2)}{k_{new}} = \frac{\ln(2)}{1.5k} = \frac{1}{1.5} \left(\frac{\ln(2)}{k}\right) = \frac{1}{3/2} T_d = \frac{2}{3} T_d$

Question 9

1. 3 years
2. 5 years (correct answer)
3. 7.5 years
4. 15 years

Explanation: 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.

Question 10

1. $10 \frac{\ln(2)}{\ln(1.25)}$ hours (correct answer)
2. $10 \frac{\ln(2)}{\ln(1.2)}$ hours
3. 25 hours
4. 31 hours

Explanation: 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: $T_h = -10 \frac{\ln(2)}{-\ln(1.25)} = 10 \frac{\ln(2)}{\ln(1.25)} \text{ hours.}$

Question 11

1. 10 years
2. 15 years
3. 25 years
4. 20 years (correct answer)

Explanation: 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.

Question 12

1. $48/7$ hours (correct answer)
2. 16 hours
3. 12 hours
4. 48 hours

Explanation: 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.

Question 13

1. 15 years
2. 20 years
3. 25 years
4. 30 years (correct answer)

Explanation: 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.

Question 14

1. $15,000
2. $20,000
3. $30,000
4. $40,000 (correct answer)

Explanation: 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.

Question 15

1. 6 hours (correct answer)
2. 8 hours
3. 10 hours
4. 12 hours

Explanation: 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.

Question 16

1. 24 years
2. 32 years (correct answer)
3. 40 years
4. 48 years

Explanation: 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.

Question 17

1. 625 individuals
2. 1,250 individuals (correct answer)
3. 2,500 individuals
4. 5,000 individuals

Explanation: 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 $2^1 = 2$.

Question 18

1. $k = \frac{\ln 2}{4}$ per hour (correct answer)
2. $k = \frac{2}{4} = 0.5$ per hour
3. $k = \frac{1}{4 \ln 2}$ per hour
4. $k = 2 \ln 4$ per hour

Explanation: 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.

Question 19

1. 200 mg
2. 300 mg
3. 400 mg (correct answer)
4. 600 mg

Explanation: 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.

Question 20

1. 20 minutes
2. 30 minutes
3. 25 minutes (correct answer)
4. 35 minutes

Explanation: 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.