Exponential growth and decay models — half-life, doubling time: Exercises

Everything here uses only tools built in the parent note the parent topic. The two facts we lean on constantly:

To "bring an exponent down to the ground floor" we take the natural logarithm $ln$ , which is the question " $e$ to the what power gives this number?" — see Laws of logarithms for the rule $ln (a^{b}) = b ln a$ we use again and again.

Level 1 — Recognition

Goal: read a situation and pick the right form and the right numbers. No algebra tricks yet.

Exercise 1.1

A colony of $200$ cells grows with rate constant $k = 0.3$ per hour. Write $N (t)$ and find the population after $5$ hours.

Recall Solution

WHAT form? We are handed $k$ directly, so use the continuous form. $N (t) = 200 e^{0.3 t}$ WHY: $N_{0} = 200$ is the start, $k = 0.3 > 0$ means growth. Plug $t = 5$ : $N (5) = 200 e^{0.3 \times 5} = 200 e^{1.5} = 200 (4.4817) \approx 896 cells.$

Exercise 1.2

Iodine-131 has a half-life of $8$ days. Starting from $40$ mg, write $N (t)$ and state how much remains after $24$ days without a calculator.

Recall Solution

WHAT form? We are handed a half-life, so use the halving form. $N (t) = 40 (\frac{1}{2})^{t /8}$ WHY no calculator? $24 = 3 \times 8$ is exactly three half-lives. $N (24) = 40 \cdot (\frac{1}{2})^{3} = 40 \cdot \frac{1}{8} = 5 mg.$ Each half-life halves what remains: $40 \to 20 \to 10 \to 5$ .

Level 2 — Application

Goal: rearrange the formula to solve for an unknown time, rate, or amount using logs.

Exercise 2.1

A bacterial population obeys $N (t) = 1000 e^{0.25 t}$ (hours). How long until it reaches $6000$ ?

Recall Solution

Set up the equation we want to solve: $6000 = 1000 e^{0.25 t} \Rightarrow 6 = e^{0.25 t} .$ WHY take $ln$ ? The unknown $t$ is stuck up in the exponent. $ln$ is the tool that pulls an exponent down, because $ln (e^{x}) = x$ . $ln 6 = 0.25 t \Rightarrow t = \frac{l n 6}{0.25} = \frac{1.79176}{0.25} \approx 7.17 h.$

Exercise 2.2

A radioactive source has half-life $12$ years. What is its rate constant $k$ ?

Recall Solution

Connect the two forms. From the parent note, $T_{1/2} = \frac{ln 2}{∣ k ∣}$ . Rearrange for $k$ , and remember decay means $k < 0$ : $k = - \frac{l n 2}{T _{1/2}} = - \frac{0.69315}{12} \approx - 0.05776 /year.$ WHY the minus: the amount is falling, so the rate constant must be negative or the model would grow.

Exercise 2.3

A drug concentration drops from $80$ mg/L to $50$ mg/L in $3$ hours. Find $k$ .

Recall Solution

Model: $C (t) = 80 e^{k t}$ . Use the data point $C (3) = 50$ : $50 = 80 e^{3 k} \Rightarrow e^{3 k} = \frac{50}{80} = 0.625.$ Take $ln$ of both sides: $3 k = ln 0.625 = - 0.47000 \Rightarrow k = \frac{- 0.47000}{3} \approx - 0.1567 /h.$

Level 3 — Analysis

Goal: extract the model from two data points, or reason about ratios and fractions.

Exercise 3.1

A tree-ring sample contains $65%$ of its original Carbon-14. C-14 half-life is $5730$ years. How old is the sample?

Recall Solution

Model: $N (t) = N_{0} 2^{- t /5730}$ , and the fraction remaining is $\frac{N}{N _{0}} = 0.65$ . $0.65 = 2^{- t /5730} .$ Take $ln$ to free the exponent, using $ln (2^{x}) = x ln 2$ : $ln 0.65 = - \frac{t}{5730} ln 2 \Rightarrow t = - 5730 \cdot \frac{l n 0.65}{l n 2} .$ $t = - 5730 \cdot \frac{- 0.43078}{0.69315} \approx 3561 years.$ WHY the answer is positive: $ln 0.65 < 0$ and the leading minus flips it — the two negatives cancel, giving a sensible positive age. Linked idea: Radioactive decay (Physics).

Exercise 3.2

A population is measured at $N = 1200$ at $t = 2$ h and $N = 1800$ at $t = 5$ h. Find $k$ and $N_{0}$ .

Recall Solution

Two unknowns, two equations. Write both readings from $N = N_{0} e^{k t}$ : $1200 = N_{0} e^{2 k}, 1800 = N_{0} e^{5 k} .$ WHY divide the equations? Dividing kills the unknown $N_{0}$ , isolating $k$ : $\frac{1800}{1200} = \frac{N _{0} e ^{5 k}}{N _{0} e ^{2 k}} = e^{3 k} \Rightarrow 1.5 = e^{3 k} .$ $3 k = ln 1.5 = 0.40546 \Rightarrow k = 0.13515 /h.$ Back-substitute into the first equation to get $N_{0}$ : $N_{0} = \frac{1200}{e ^{2 k}} = \frac{1200}{e ^{0.27031}} = \frac{1200}{1.31037} \approx 915.8.$

Level 4 — Synthesis

Goal: combine the model with the ODE it comes from, or chain several ideas together.

Exercise 4.1

Starting only from $\frac{d N}{d t} = k N$ with $N (0) = N_{0}$ , derive the doubling time $T_{d}$ and then evaluate it for $k = 0.05$ /year.

Recall Solution

Solve the ODE (the machinery lives in Solving first-order separable ODEs): separate $\frac{1}{N} d N = k d t$ , integrate to $ln N = k t + C$ , exponentiate to $N = A e^{k t}$ , and set $t = 0$ to get $A = N_{0}$ . Hence $N (t) = N_{0} e^{k t}$ . Doubling condition: $N (T_{d}) = 2 N_{0}$ : $2 N_{0} = N_{0} e^{k T_{d}} \Rightarrow 2 = e^{k T_{d}} \Rightarrow k T_{d} = ln 2 \Rightarrow T_{d} = \frac{l n 2}{k} .$ Evaluate: $T_{d} = \frac{0.69315}{0.05} \approx 13.86 years.$

Exercise 4.2

Two isotopes start with equal amounts. Isotope A has half-life $4$ days; isotope B has half-life $10$ days. After how many days is A's remaining fraction exactly half of B's remaining fraction?

Recall Solution

Write both fractions (take $N_{0} = 1$ for each, since they start equal): $f_{A} (t) = 2^{- t /4}, f_{B} (t) = 2^{- t /10} .$ Condition: $f_{A} = \frac{1}{2} f_{B}$ : $2^{- t /4} = \frac{1}{2} \cdot 2^{- t /10} = 2^{- t /10 - 1} .$ WHY compare exponents? Same base $2$ on both sides means the exponents must be equal: $- \frac{t}{4} = - \frac{t}{10} - 1.$ Solve — multiply through by $20$ : $- 5 t = - 2 t - 20 \Rightarrow - 3 t = - 20 \Rightarrow t = \frac{20}{3} \approx 6.67 days.$

Level 5 — Mastery

Goal: build the model from a messy real scenario and interpret the result.

Exercise 5.1

A patient receives a drug. Concentration is $C_{0}$ right after the dose and the drug has half-life $6$ h. The dose must be repeated once the concentration drops to $25%$ of $C_{0}$ . Doctors also want to know the continuous rate constant $k$ . Find the redose time and $k$ .

Recall Solution

Redose time — we need $C = 0.25 C_{0}$ , i.e. two quarterings... actually $0.25 = (\frac{1}{2})^{2}$ , exactly two half-lives: $t = 2 \times 6 = 12 h.$ Formally: $0.25 = 2^{- t /6} \Rightarrow lo g_{2} 0.25 = - t /6 \Rightarrow - 2 = - t /6 \Rightarrow t = 12$ . Rate constant from the half-life relation (decay, so $k < 0$ ): $k = - \frac{l n 2}{6} = - \frac{0.69315}{6} \approx - 0.11552 /h.$

Exercise 5.2

A savings scheme grows continuously. It triples in $9$ years. (a) Find $k$ . (b) Find the doubling time. (c) What multiple of the original is reached in $30$ years?

Recall Solution

(a) Triple in 9 years means $N (9) = 3 N_{0}$ : $3 = e^{9 k} \Rightarrow 9 k = ln 3 \Rightarrow k = \frac{l n 3}{9} = \frac{1.09861}{9} \approx 0.12207 /year.$ (b) Doubling time — the "same ratio, same time" idea, now with factor $2$ : $T_{d} = \frac{l n 2}{k} = \frac{0.69315}{0.12207} \approx 5.68 years.$ (c) Multiple after 30 years: $\frac{N ( 30 )}{N _{0}} = e^{30 k} = e^{30 \times 0.12207} = e^{3.66205} \approx 38.9.$ So the money grows to roughly $39 \times$ the original. This is the continuous cousin of Compound interest.

Exercise 5.3

Two populations: X starts at $100$ and doubles every $3$ h; Y starts at $6400$ and halves every $2$ h. When are they equal?

Recall Solution

Write both models: $X (t) = 100 \cdot 2^{t /3}, Y (t) = 6400 \cdot 2^{- t /2} .$ Set equal: $100 \cdot 2^{t /3} = 6400 \cdot 2^{- t /2} .$ Gather the powers of 2 — divide both sides by $100$ and by $2^{- t /2}$ : $2^{t /3} \cdot 2^{t /2} = 64 \Rightarrow 2^{t /3 + t /2} = 2^{6} .$ WHY $64 = 2^{6}$ ? Rewriting the constant as a power of the same base lets us equate exponents. $\frac{t}{3} + \frac{t}{2} = 6 \Rightarrow \frac{2 t + 3 t}{6} = 6 \Rightarrow \frac{5 t}{6} = 6 \Rightarrow t = \frac{36}{5} = 7.2 h.$

Recall Quick self-check ledger (answers only)

1.1 $\approx 896$ · 1.2 $5$ mg · 2.1 $\approx 7.17$ h · 2.2 $\approx - 0.05776$ /yr · 2.3 $\approx - 0.1567$ /h · 3.1 $\approx 3561$ yr · 3.2 $k \approx 0.1352$ , $N_{0} \approx 916$ · 4.1 $\approx 13.86$ yr · 4.2 $\approx 6.67$ d · 5.1 $12$ h, $k \approx - 0.1155$ /h · 5.2 $k \approx 0.1221$ , $T_{d} \approx 5.68$ yr, $\approx 38.9 \times$ · 5.3 $7.2$ h.

Connections

Exponential growth and decay models — half-life, doubling time (index 3.2.5) — the parent this drills.
Solving first-order separable ODEs — used in 4.1.
Laws of logarithms — the log rules behind every L2–L5 solve.
Natural logarithm and e — why $ln$ undoes $e^{k t}$ .
Compound interest — the finance reading of 5.2.
Radioactive decay (Physics) — the physics reading of 3.1.