Banks advertise interest compounded monthly, weekly, or even daily — but what if they compounded every single instant? The answer is not infinity; it is a single, remarkable number that appears in population growth, radioactive decay, and every branch of physics and engineering. That number is $e$.
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Case entry — Compound interest of $100\%$ compounded continuously gives $e \approx 2.718$. Why isn't it exactly 2 or 3?
As the number of compounding periods increases, the result creeps upward — but does it keep growing forever, or does it settle on a limit? Make a prediction before reading on.
Type your initial prediction below — you will revisit this at the end of the lesson.
Write your initial prediction in your book. You will revisit it at the end of the lesson.
Core Content
The number $e$ is one of the most important constants in mathematics. It arises naturally when we ask: what happens if we compound interest more and more frequently?
Suppose you invest $\$1$ at $100\%$ annual interest. If compounded once per year:
$$A = 1 \cdot (1 + 1)^1 = 2$$
If compounded twice per year ($50\%$ each period):
$$A = 1 \cdot \left(1 + \frac{1}{2}\right)^2 = (1.5)^2 = 2.25$$
If compounded $n$ times per year:
$$A = \left(1 + \frac{1}{n}\right)^n$$
As $n \to \infty$, this expression approaches a limit. That limit is $e$:
$$e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.71828...$$
The value creeps upward — $2$, $2.25$, $2.370$, $2.488$... — but the increments get smaller and smaller, settling on $e \approx 2.718$. It is not 2, not 3, but an irrational number with infinite non-repeating decimals, just like $\pi$.
The function $y = e^x$ has all the same features as $y = a^x$ for any base $a > 1$:
But $y = e^x$ has one extraordinary property that no other exponential shares: the gradient of the curve at any point equals the value of the function at that point. In particular, at $x = 0$:
$$\text{Gradient of } y = e^x \text{ at } x = 0 \text{ is exactly } 1$$
For comparison:
$e$ is the unique base where the gradient at the $y$-intercept is exactly 1. This makes $e^x$ the simplest exponential to differentiate and integrate — which is why it dominates calculus, physics, and engineering.
Show three exponential curves on one set of axes with $x$ from -2 to 2 and $y$ from 0 to 8. Curve A (blue): $y = 2^x$ passing through (0, 1) with shallow slope. Curve B (green): $y = e^x \approx 2.718^x$ passing through (0, 1) and (1, 2.718) with slope 1 at x=0. Curve C (orange): $y = 3^x$ passing through (0, 1) and (1, 3) with steeper slope. Label each curve. Mark the common $y$-intercept $(0, 1)$ with a dot and the horizontal asymptote $y = 0$ as a dashed line. Show that $e^x$ lies between $2^x$ and $3^x$ for $x > 0$.
When growth (or decay) happens continuously — every instant, not at discrete intervals — the model is:
$$A = P \cdot e^{rt}$$
Where:
This formula appears in finance (continuous compounding), biology (bacterial growth), physics (radioactive decay, capacitor charging), and chemistry (reaction rates). Any process where change is proportional to current quantity follows this pattern.
For decay, $r$ is negative. For example, radioactive decay might follow $A = A_0 \cdot e^{-0.00012t}$ where the negative exponent causes the quantity to decrease over time.
$\$5,000$ is invested at $4\%$ per annum compounded continuously.
The value of the investment after 10 years (to the nearest dollar).
A = P · e^(rt)
P = 5000, r = 0.04, t = 10
A = 5000 · e^(0.04 × 10)
A = 5000 · e^0.4
e^0.4 ≈ 1.49182...
A ≈ 5000 × 1.49182
A ≈ 7459.12
$\$7,459$ (to the nearest dollar).
$\$10,000$ is invested at $6\%$ per annum compounded continuously for 5 years. Find the final value to the nearest dollar.
Answer:
$A = 10000 \cdot e^{0.06 \times 5} = 10000 \cdot e^{0.3}$
$e^{0.3} \approx 1.34986$
$A \approx 10000 \times 1.34986 = 13,498.6$
So $\$13,499$ to the nearest dollar.
$e = \lim_{n \to \infty} (1 + \frac{1}{n})^n$
$\approx 2.71828...$
$A = P \cdot e^{rt}$
$P$ = initial, $r$ = rate, $t$ = time
$(0, 1)$ since $e^0 = 1$
$(1, e) \approx (1, 2.718)$
Gradient at $x = 0$ equals 1
Gradient = value everywhere
Use your calculator and the continuous growth formula where needed.
A student claims that because $e \approx 2.718$, the graph of $y = e^x$ is almost identical to $y = 2.7^x$ and there is no real reason to use $e$.
Look back at what you wrote in the Think First section. Compounding $100\%$ interest more and more frequently produces values that increase but approach a ceiling: $2$, $2.25$, $2.370$, $2.488$, $2.594$... settling on $e \approx 2.718$. The limit is finite because the extra compounding periods add diminishing returns.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. (a) Calculate $e^2$ to 3 decimal places. (b) Solve $e^x = 10$, giving $x$ to 3 decimal places. (c) Compare $e^2$ with $2^e$. Which is larger? 3 MARKS
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Answer in your workbook.
9. A culture of 500 bacteria grows continuously at $12\%$ per hour. (a) Write the model. (b) Find the population after 8 hours. (c) How long until the population reaches 2,000? 3 MARKS
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Answer in your workbook.
10. Explain why $e$ is considered the "natural" base for exponential functions in calculus, with reference to the gradient of $y = e^x$ at $x = 0$ compared to $y = 2^x$ and $y = 3^x$ at $x = 0$. 3 MARKS
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Answer in your workbook.
1. $e^1 \approx 2.718$; $e^2 \approx 7.389$; $e^{-1} \approx 0.368$.
2. $A = 8000 \cdot e^{0.05 \times 7} = 8000 \cdot e^{0.35} \approx 8000 \times 1.41907 \approx 11,353$.
3. $P(6) = 1000 \cdot e^{0.15 \times 6} = 1000 \cdot e^{0.9} \approx 1000 \times 2.4596 \approx 2,460$ bacteria.
1. $e^2 \approx 7.389$ and $2.7^2 = 7.29$. Close for $x = 2$, but the gap widens for larger $x$ because the bases differ by $0.018$.
2. In calculus, the derivative of $a^x$ is $\ln(a) \cdot a^x$. For $a = e$, this becomes $\ln(e) \cdot e^x = e^x$ — the derivative equals the original function. For $a = 2.7$, the derivative would be $\ln(2.7) \cdot 2.7^x \approx 0.993 \cdot 2.7^x$, introducing an annoying constant factor that complicates every calculation.
3. (i) Continuous compound interest: $A = Pe^{rt}$ is the standard formula in finance. (ii) Population dynamics: the differential equation $\frac{dP}{dt} = kP$ has solution $P = P_0 e^{kt}$, which is the natural form arising from the mathematics.
Q8 (3 marks): (a) $e^2 \approx 7.389$ [1]. (b) $e^x = 10 \Rightarrow x = \ln 10 \approx 2.303$ [1]. (c) $e^2 \approx 7.389$ and $2^e = 2^{2.718} \approx 6.580$. Since $7.389 > 6.580$, $e^2$ is larger [1].
Q9 (3 marks): (a) $P(t) = 500 \cdot e^{0.12t}$ [1]. (b) $P(8) = 500 \cdot e^{0.12 \times 8} = 500 \cdot e^{0.96} \approx 500 \times 2.6117 \approx 1306$ bacteria [1]. (c) $500 \cdot e^{0.12t} = 2000 \Rightarrow e^{0.12t} = 4 \Rightarrow 0.12t = \ln 4 \Rightarrow t = \frac{\ln 4}{0.12} \approx \frac{1.3863}{0.12} \approx 11.55$ hours, so approximately 11.6 hours [1].
Q10 (3 marks): For $y = a^x$, the gradient at $x = 0$ equals $\ln a$ [1]. For $y = 2^x$, this is $\ln 2 \approx 0.693$; for $y = 3^x$, it is $\ln 3 \approx 1.099$; for $y = e^x$, it is $\ln e = 1$ exactly [1]. Because the gradient of $e^x$ equals its own value everywhere (not just at $x = 0$), $e^x$ is the unique exponential that simplifies differentiation and integration, making it the natural choice for calculus and all continuous growth models [1].
Climb platforms using your knowledge of $e$, continuous growth, natural exponentials, and limits. Pool: lesson 3.
Tick when you've finished all activities and checked your answers.