You have journeyed from the bacteria that doubles every 20 minutes to the archaeologist dating 50,000-year-old bones. From the decibel scale that spans a trillion-fold range of sound to the pH scale that keeps your blood alive. From the number $e$ that is its own derivative to the logarithm that undoes every exponential. This lesson weaves all fifteen threads into a single tapestry — the mathematics of change, growth, and decay that underpins modern science.
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Core Content
This module has four interconnected themes:
$y = a^x$ models growth and decay. The number $e \approx 2.718$ is special because $\frac{d}{dx}(e^x) = e^x$. Continuous compounding $A = Pe^{rt}$ is the natural form.
$y = \log_a x$ is the inverse of $y = a^x$. Log laws turn multiplication into addition, division into subtraction, and powers into multiplication. Change of base lets us evaluate any logarithm.
$\frac{d}{dx}(e^x) = e^x$, $\frac{d}{dx}(e^{kx}) = ke^{kx}$, $\frac{d}{dx}(a^x) = a^x \ln a$. The chain rule is essential for composite functions.
$\frac{d}{dx}(\ln x) = \frac{1}{x}$, $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$. The chain rule gives $\frac{d}{dx}(\ln u) = \frac{u'}{u}$.
Exponential growth/decay models, half-life, doubling time, and optimisation with exponentials bring everything together.
| Mistake | Why It Is Wrong | Correct Approach |
|---|---|---|
| $\frac{d}{dx}(a^x) = xa^{x-1}$ | This is the power rule for $x^n$, not the exponential rule | $\frac{d}{dx}(a^x) = a^x \ln a$ |
| $\log_a(M + N) = \log_a M + \log_a N$ | No log law for addition inside the log | Only $\log_a(MN) = \log_a M + \log_a N$ |
| Forgetting domain checks | Log equations can have extraneous solutions | Always check all log arguments are positive |
| $\ln(a + b) = \ln a + \ln b$ | This "distributes" incorrectly | $\ln(ab) = \ln a + \ln b$ only |
| Missing chain rule factor | Composite functions need $u'$ | $\frac{d}{dx}(e^{u}) = e^u \cdot u'$ |
Solve $3^{2x} = 5^{x+1}$, giving $x$ in exact form.
The exact value of $x$.
Take ln of both sides:
ln(3^{2x}) = ln(5^{x+1})
2x ln(3) = (x+1) ln(5)
2x ln(3) = x ln(5) + ln(5)
2x ln(3) - x ln(5) = ln(5)
x(2 ln(3) - ln(5)) = ln(5)
x = ln(5) / (2 ln(3) - ln(5))
= ln(5) / ln(9/5)
$x = \frac{\ln 5}{\ln(9/5)}$ or $\frac{\ln 5}{2\ln 3 - \ln 5}$.
Differentiate $f(x) = \ln(x^2 + 1) \cdot e^{3x}$.
Answer:
$f'(x) = \frac{2x}{x^2+1} \cdot e^{3x} + \ln(x^2+1) \cdot 3e^{3x} = e^{3x}\left(\frac{2x}{x^2+1} + 3\ln(x^2+1)\right)$.
The module follows a natural progression:
Every formula in this module is connected. The derivative of $e^x$ being $e^x$ is why $e^{kt}$ appears in growth models. The derivative of $\ln x$ being $\frac{1}{x}$ is why logarithms appear in utility theory and information entropy. The change of base formula connects all bases to the natural base $e$.
$\frac{d}{dx}(e^x) = e^x$
$\frac{d}{dx}(a^x) = a^x \ln a$
$\frac{d}{dx}(\ln x) = \frac{1}{x}$
$\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$
$\log_a(MN) = \log_a M + \log_a N$
$\log_a(M^n) = n\log_a M$
$A = A_0 e^{kt}$
$T_{1/2} = \frac{\ln 2}{|k|}$
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Solve $2^{x+1} = 5^{x-1}$, giving your answer in exact form. 3 MARKS
9. Differentiate $f(x) = \frac{e^{2x}}{\ln x}$. Show all working. 3 MARKS
10. A lake is being restocked with fish. The population is modelled by $P = \frac{10000}{1 + 9e^{-0.4t}}$ where $t$ is in months. This is a logistic model. Find the initial population and the population after 10 months. Show that $\frac{dP}{dt} = 0.4P(1 - \frac{P}{10000})$ and explain what this tells us about the growth rate at different population levels. Discuss why this model is more realistic than simple exponential growth for fish populations. 3 MARKS
1. $(x+1)\ln 2 = (x-1)\ln 5 \Rightarrow x = \frac{\ln 5 + \ln 2}{\ln 5 - \ln 2} = \frac{\ln 10}{\ln(5/2)}$.
2. $f'(x) = 2x\ln(3x) + x^2 \cdot \frac{3}{3x} = 2x\ln(3x) + x = x(2\ln(3x) + 1)$.
3. $k = \frac{\ln 4}{6} = \frac{\ln 2}{3} \approx 0.231$. After 10 days: $P = 1000e^{0.231 \times 10} \approx 1000 \times 10.08 \approx 10{,}080$.
4. $f'(x) = 2xe^{-0.5x} - 0.5x^2e^{-0.5x} = xe^{-0.5x}(2 - 0.5x) = 0 \Rightarrow x = 4$. $f(4) = 16e^{-2} = 16/e^2 \approx 2.17$.
1. $e$ is the only base where $\frac{d}{dx}(a^x) = a^x$ (since $\ln e = 1$). This makes $e$ the most natural base for calculus.
2. If $y = e^x$, then $\ln y = x$. Differentiating: $\frac{1}{y}\frac{dy}{dx} = 1$, so $\frac{dy}{dx} = y = e^x$.
3. (i) Population growth: $A = A_0e^{kt}$. (ii) Drug dosing: optimisation with $C(t) = Ate^{-kt}$. (iii) Carbon dating: half-life calculations.
Q8 (3 marks): $(x+1)\ln 2 = (x-1)\ln 5$ [0.5]. $x\ln 2 + \ln 2 = x\ln 5 - \ln 5$ [0.5]. $x(\ln 2 - \ln 5) = -\ln 5 - \ln 2$ [0.5]. $x = \frac{-\ln 10}{\ln(2/5)} = \frac{\ln 10}{\ln(5/2)}$ [1.5].
Q9 (3 marks): $u = e^{2x}$, $v = \ln x$ [0.5]. $u' = 2e^{2x}$, $v' = 1/x$ [0.5]. $f'(x) = \frac{2e^{2x}\ln x - e^{2x} \cdot \frac{1}{x}}{(\ln x)^2}$ [1] $= \frac{e^{2x}(2\ln x - \frac{1}{x})}{(\ln x)^2}$ [0.5] $= \frac{e^{2x}(2x\ln x - 1)}{x(\ln x)^2}$ [0.5].
Q10 (3 marks): At $t = 0$: $P = \frac{10000}{1 + 9} = 1000$ fish [0.25]. At $t = 10$: $P = \frac{10000}{1 + 9e^{-4}} \approx \frac{10000}{1 + 9(0.0183)} \approx \frac{10000}{1.165} \approx 8584$ fish [0.5]. $\frac{dP}{dt} = 10000 \cdot \frac{-1}{(1+9e^{-0.4t})^2} \cdot 9e^{-0.4t} \cdot (-0.4)$ [0.5] $= \frac{36000e^{-0.4t}}{(1+9e^{-0.4t})^2}$ [0.25]. Rewrite: $\frac{dP}{dt} = 0.4 \cdot \frac{10000}{1+9e^{-0.4t}} \cdot \frac{9e^{-0.4t}}{1+9e^{-0.4t}}$ [0.5] $= 0.4P(1 - \frac{P}{10000})$ [0.5]. This tells us growth rate is proportional to both current population $P$ and remaining capacity $(1 - P/10000)$ [0.25]. When $P$ is small, growth is nearly exponential. As $P$ approaches 10000, growth slows to zero [0.25]. This is more realistic than simple exponential growth because real populations face limited food, space, and resources — the carrying capacity of 10000 represents these limits [0.5].
Climb platforms using mixed problems from across the whole module. Pool: lesson 15.
Tick when you've finished all activities and checked your answers.