In 1949, Willard Libby developed a technique that revolutionised archaeology: by measuring the carbon-14 remaining in ancient organic material, he could determine its age with remarkable precision. The mathematics behind this technique is exponential decay — a process that also models radioactive half-life, drug elimination, and depreciation.
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A student makes two errors:
Both statements are wrong. Identify the error in each, explain what the correct answer is, and describe the fundamental misunderstanding that leads to these mistakes.
Core Content
The fundamental technique for solving exponential equations is to take the logarithm of both sides. This uses the key property of logarithms:
$$\log_a(a^x) = x$$
This says that logarithms "undo" exponentials. If $a^x = b$, then taking $\log_a$ of both sides gives:
$$x = \log_a(b) = \frac{\ln b}{\ln a}$$
The change of base formula ($\log_a b = \frac{\ln b}{\ln a}$) lets us calculate any log using the natural log key on a calculator.
$$5^x = 100$$
$$\ln(5^x) = \ln(100)$$
$$x \cdot \ln 5 = \ln 100$$
$$x = \frac{\ln 100}{\ln 5} \approx \frac{4.605}{1.609} \approx 2.861$$
The general model for continuous exponential change is:
$$A(t) = A_0 \cdot e^{kt}$$
where $A_0$ is the initial amount, $k$ is the rate constant, and $t$ is time.
The half-life is the time for the amount to halve. Setting $A(t) = \frac{1}{2}A_0$:
$$\frac{1}{2}A_0 = A_0 \cdot e^{kt_{1/2}}$$
$$\frac{1}{2} = e^{kt_{1/2}}$$
$$\ln(\frac{1}{2}) = kt_{1/2}$$
$$t_{1/2} = \frac{\ln(1/2)}{k} = \frac{-\ln 2}{k} = \frac{\ln 2}{|k|}$$
Notice that half-life depends only on $k$, not on $A_0$. This is why carbon-14 dating works: every 5,730 years, half the carbon-14 decays, regardless of how much you started with.
A sample of wood has 25% of its original carbon-14 remaining. The half-life of carbon-14 is 5,730 years.
The age of the sample.
A = A_0 · e^(kt)
At t = 5730: A = 0.5·A_0
0.5 = e^(5730k)
k = ln(0.5)/5730 ≈ -0.000121
Now 0.25 = e^(kt)
t = ln(0.25)/k = ln(0.25)/(-0.000121)
t ≈ 11,460 years
Approximately 11,460 years (two half-lives)
A substance decays from 80g to 20g in 30 days. Find the half-life.
Answer:
20 = 80·e^(30k) ⇒ e^(30k) = 0.25 ⇒ 30k = ln(0.25) ⇒ k = -0.0462. Half-life = ln(2)/0.0462 ≈ 15 days.
$a^x = b \Rightarrow x = \frac{\ln b}{\ln a}$
$A = A_0 e^{kt}$ ($k > 0$)
$A = A_0 e^{kt}$ ($k < 0$)
$t_{1/2} = \frac{\ln 2}{|k|}$
A population of rabbits grows from 200 to 800 in 6 months.
The errors were: (1) $2^3 = 8$, not 6 — exponentiation means repeated multiplication, not multiplication of base and exponent. (2) $\log_2(8) = 3$, not 4 — logarithms answer "what power?", not "what times?" The fundamental misunderstanding is treating exponentiation and logarithms as multiplication rather than understanding their true definitions.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Solve the following equations: (a) $2^x = 32$ (exact) (b) $3^{x+1} = 81$ (exact) (c) $5^x = 100$ (to 3 decimal places). 3 MARKS
9. A population of rabbits grows from 200 to 800 in 6 months. (a) Find $k$ assuming $P = P_0 e^{kt}$. (b) Predict the population after 1 year. (c) Find the doubling time. 3 MARKS
10. A student argues that since carbon-14 dating assumes exponential decay, it cannot be accurate because "nothing in nature is perfectly exponential." Evaluate this claim, explaining both the strengths and limitations of the exponential decay model in radiometric dating. 3 MARKS
1. $2^x = 32 = 2^5 \Rightarrow x = 5$.
2. $3^{x+1} = 81 = 3^4 \Rightarrow x+1 = 4 \Rightarrow x = 3$.
3. $x = \frac{\ln 100}{\ln 5} \approx \frac{4.605}{1.609} \approx 2.861$.
1. $800 = 200e^{6k} \Rightarrow e^{6k} = 4 \Rightarrow 6k = \ln 4 \Rightarrow k = \frac{\ln 4}{6} \approx 0.231$ per month.
2. $P(12) = 200e^{0.231 \times 12} = 200e^{2.773} \approx 200 \times 16 = 3,200$ rabbits.
3. $t_{\text{double}} = \frac{\ln 2}{0.231} \approx 3$ months.
Q8 (3 marks): (a) $x = 5$ [1]. (b) $x + 1 = 4 \Rightarrow x = 3$ [1]. (c) $x = \frac{\ln 100}{\ln 5} \approx 2.861$ [1].
Q9 (3 marks): (a) $k = \frac{\ln 4}{6} \approx 0.231$ [1]. (b) $P(12) = 200e^{12k} = 200 \times 16 = 3,200$ [1]. (c) $t_{\text{double}} = \frac{\ln 2}{k} = \frac{6\ln 2}{\ln 4} = 3$ months [1].
Q10 (3 marks): The student's claim has some validity — real processes involve noise, environmental variation, and measurement error [1]. However, radioactive decay is one of the most precisely exponential processes in nature because it arises from quantum mechanical probabilities that are constant for a given isotope. The limitations are: contamination, sample size, and the assumption of closed systems. Despite these, carbon-14 dating achieves accuracies within 1–2% and has been independently validated against tree-ring chronologies and historical records [1]. The exponential model is not perfect, but it is sufficiently accurate for its intended applications — which is the hallmark of a good scientific model [1].
Climb platforms using your knowledge of exponential equations, half-life, and growth models. Pool: lesson 4.
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