Jacob Bernoulli and the Compound Interest Puzzle
In 1683, the Swiss mathematician Jacob Bernoulli was studying a seemingly simple question about money: if you invest \(\$1\) at \(100\%\) interest for one year, and you compound the interest more and more frequently, how much money do you end up with?
He expected the amount to grow without bound. What he found instead was a mysterious ceiling — a number that would later be called e.
It is the base of the natural exponential function $y = e^x$.
It asks: "To what power must e be raised to give x?"
The functions $y = e^x$ and $y = \ln x$ are inverses: one "undoes" the other.
• $\ln(ab) = \ln a + \ln b$
• $\ln\!\left(\dfrac{a}{b}\right) = \ln a - \ln b$
• $\ln(a^n) = n\,\ln a$
$\ln(2 \times 3) = \ln(6) \approx 1.792, \quad \ln 2 + \ln 3 \approx 1.792 \;\; \checkmark$
$\ln\!\left(\dfrac{2}{3}\right) = \ln(0.67) \approx -0.405, \quad \ln 2 - \ln 3 \approx -0.405 \;\; \checkmark$
$\ln(2^2) = \ln(4) \approx 1.386, \quad 2\,\ln 2 \approx 1.386 \;\; \checkmark$
• When $k > 0$: the graph rises from left to right (growth).
• When $k < 0$: the graph falls from left to right (decay).
• The $y$-intercept is always $(0, 1)$, and $y = 0$ is the horizontal asymptote.
Interactive: drag $k$ between negative and positive values to switch between growth and decay.
$y = e^{0.5x}$
• When $k > 0$: $\ln(kx) = \ln k + \ln x$, so the graph is $y = \ln x$ shifted vertically by $\ln k$ (domain $x > 0$).
• When $k < 0$: the domain is $x < 0$, so the graph reflects across the $y$-axis onto the left side.
• The $x$-intercept is $(1/k, 0)$; $x = 0$ is always the vertical asymptote, and $k = 0$ is undefined.
Interactive: drag $k$ from $-2$ to $2$. For $k > 0$ the solid curve $y = \ln(kx)$ hugs the positive $x$-axis; for $k < 0$ it flips to the negative side. The amber dot marks $(1/k, 0)$.
$\ln(kx) = \ln k + \ln x$
Instructions (6 min): For each statement, decide whether it is Correct or contains an Error, then tap the matching button. You'll get instant feedback — green when your judgment is right, red when it's wrong.
Instructions (5 min): Match each function card with its correct feature card. Click one card from each column to pair them up.
Function
Features
Solve the equation $\ln(3 - x) = 4 + \ln x$.
$$\ln(3 - x) - \ln x = 4$$
Using $\ln A - \ln B = \ln(A/B)$:
$$\ln\!\left(\dfrac{3 - x}{x}\right) = 4$$
$$\dfrac{3 - x}{x} = e^4$$
$$3 - x = x\,e^4$$
$$3 = x + x\,e^4 = x(1 + e^4)$$
$$x = \dfrac{3}{1 + e^4} \approx \dfrac{3}{55.598} \approx 0.0540 \;\text{(3 s.f.)}$$
$x > 0$ and $3 - x > 0$ ✓. Substitute back: $\ln(3 - 0.0540) \approx \ln 2.946 \approx 1.080$ and $4 + \ln 0.0540 \approx 4 - 2.920 \approx 1.080$. ✓
Make $x$ the subject of $\ln x - \ln x_0 = kt$.
$$\ln x - \ln x_0 = \ln\!\left(\dfrac{x}{x_0}\right) = kt$$
$$\dfrac{x}{x_0} = e^{kt}$$
$$x = x_0\,e^{kt}$$
Solve the equation $\ln(x + 5) = 5 + \ln x$.
$$\ln(x + 5) - \ln x = 5$$
$$\ln\!\left(\dfrac{x + 5}{x}\right) = 5$$
$$\dfrac{x + 5}{x} = e^5$$
$$1 + \dfrac{5}{x} = e^5$$
$$\dfrac{5}{x} = e^5 - 1$$
$$x = \dfrac{5}{e^5 - 1} \approx \dfrac{5}{147.413} \approx 0.0339 \;\text{(3 s.f.)}$$
$x > 0$ and $x + 5 > 0$ ✓. Substitute: $\ln(0.0339 + 5) \approx \ln 5.034 \approx 1.616$ and $5 + \ln 0.0339 \approx 5 - 3.384 \approx 1.616$. ✓
Solve the equation $4(\log_5 x)^2 - \log_5(x^3) = 10$.
Since $\log_5(x^3) = 3\log_5 x$, the equation becomes:
$$4(\log_5 x)^2 - 3\log_5 x = 10$$
$$4u^2 - 3u - 10 = 0$$
$$(4u + 5)(u - 2) = 0$$
$$u = -\dfrac{5}{4} \quad \text{or} \quad u = 2$$
Since $u = \log_5 x$, we have $x = 5^u$:
$$u = 2: \quad x = 5^2 = 25$$
$$u = -\dfrac{5}{4}: \quad x = 5^{-5/4} = \dfrac{1}{5^{5/4}} \approx 0.134 \;\text{(3 s.f.)}$$
$x = 25$: $\log_5 25 = 2$ and $\log_5(25^3) = 3 \times 2 = 6$, so $4(2)^2 - 6 = 16 - 6 = 10$ ✓. $x = 5^{-5/4}$: $\log_5 x = -\tfrac{5}{4}$, so $4\left(-\tfrac{5}{4}\right)^2 - 3\left(-\tfrac{5}{4}\right) = \tfrac{25}{4} + \tfrac{15}{4} = 10$ ✓.
Test your understanding of all four lesson objectives. Each question targets one objective — try each before moving on!
What is the value of $\ln(e)$?
By definition, $\ln(e) = \log_e(e) = 1$, because $e^1 = e$. The logarithm of the base itself is always $1$.
Simplify $e^{\ln 5}$.
Since $e^x$ and $\ln x$ are inverse functions, $e^{\ln A} = A$ for any $A > 0$. So $e^{\ln 5} = 5$.
Which expression is equal to $\ln 8 - \ln 2$?
Using $\ln A - \ln B = \ln(A/B)$: $\ln 8 - \ln 2 = \ln\!\left(\dfrac{8}{2}\right) = \ln 4$.
Which statement about $y = e^{kx}$ is correct when $k < 0$?
For $y = e^{kx}$, when $k < 0$ the exponent becomes more negative as $x$ increases, so $y \to 0$ — the graph decays (falls). The $y$-intercept is always $e^0 = 1$, and $y = 0$ is the horizontal asymptote.
Make $t$ the subject of $s = s_0\,e^{-kt}$.
$\dfrac{s}{s_0} = e^{-kt}$
$\ln\!\left(\dfrac{s}{s_0}\right) = -kt$
$t = -\dfrac{1}{k}\,\ln\!\left(\dfrac{s}{s_0}\right) = \dfrac{1}{k}\,\ln\!\left(\dfrac{s_0}{s}\right)$
Make $p$ the subject of $\ln p = -0.02t$.
Exponentiate both sides with base $e$:
$p = e^{-0.02t}$
Solve the equation $\ln(2 - x) = 2 + \ln x$.
$\ln(2 - x) - \ln x = 2 \Rightarrow \ln\!\left(\dfrac{2 - x}{x}\right) = 2$
$\dfrac{2 - x}{x} = e^{2}$
$2 - x = x\,e^{2}$
$2 = x(1 + e^{2}) \Rightarrow x = \dfrac{2}{1 + e^{2}} \approx \dfrac{2}{8.389} \approx 0.238 \;\text{(3 s.f.)}$
$x > 0$ and $2 - x > 0$ ✓. Substitute back: $\ln(2 - 0.2384) \approx \ln 1.7616 \approx 0.5661$ and $2 + \ln 0.2384 \approx 2 - 1.4339 \approx 0.5661$ ✓.
It is given that $\ln(y + 5) - \ln y = 2\ln x$. Express $y$ in terms of $x$, in a form not involving logarithms.
$\ln\!\left(\dfrac{y + 5}{y}\right) = \ln(x^{2})$
$\dfrac{y + 5}{y} = x^{2}$
$1 + \dfrac{5}{y} = x^{2}$
$\dfrac{5}{y} = x^{2} - 1 \Rightarrow y = \dfrac{5}{x^{2} - 1}$
Solve the equation $4(\log_2 x)^2 - 2\log_2(x^2) = 3$.
Since $\log_2(x^2) = 2\log_2 x$:
$4(\log_2 x)^2 - 4\log_2 x = 3$
$4u^2 - 4u - 3 = 0$
$(2u - 3)(2u + 1) = 0$
$u = \dfrac{3}{2} \quad \text{or} \quad u = -\dfrac{1}{2}$
Since $u = \log_2 x$, we have $x = 2^u$:
$u = \dfrac{3}{2}: \quad x = 2^{3/2} = 2\sqrt{2}$
$u = -\dfrac{1}{2}: \quad x = 2^{-1/2} = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$
$x = 2\sqrt{2}$: $\log_2 x = \tfrac{3}{2}$ and $\log_2(x^2) = 3$, so $4\left(\tfrac{3}{2}\right)^2 - 2(3) = 9 - 6 = 3$ ✓. $x = \dfrac{\sqrt{2}}{2}$: $\log_2 x = -\tfrac{1}{2}$ and $\log_2(x^2) = -1$, so $4\left(-\tfrac{1}{2}\right)^2 - 2(-1) = 1 + 2 = 3$ ✓.
Solve, correct to 3 significant figures, the equation $e^{x} + e^{2x} = e^{3x}$.
Since $e^{x} > 0$:
$1 + e^{x} = e^{2x}$
$u^{2} - u - 1 = 0$
$u = \dfrac{1 \pm \sqrt{5}}{2}$
Since $u = e^{x} > 0$, take the positive root:
$u = \dfrac{1 + \sqrt{5}}{2}$
$x = \ln\!\left(\dfrac{1 + \sqrt{5}}{2}\right) \approx 0.481 \;\text{(3 s.f.)}$
Exit Ticket
Check the learning objectives you feel confident about: