Papers 2 and 3. Meet \(e \approx 2.718\), the graphs of \(y = e^{kx}\) for growth and decay, and \(\ln x\), the logarithm to base \(e\). Because \(e^x\) and \(\ln x\) are inverse functions, each one undoes the other in equations.
\(\ln x = \log_e x\), so \(y = e^x \iff x = \ln y\). They undo each other: \(\ln(e^x) = x\) for all \(x\), and \(e^{\ln x} = x\) for \(x > 0\). \(y = e^{kx}\) always passes through \((0, 1)\), stays above the \(x\)-axis, grows for \(k > 0\) and decays towards \(y = 0\) for \(k < 0\).
Fig. 1: y = e^x, y = e^(2x) (dashed, steeper) and y = e^(-x) (the reflection of y = e^x in the y-axis). All three pass through (0, 1) and approach the x-axis without reaching it
Section 01
The graph of \(y = e^{kx}\): growth and decay.
In the lab, set a positive percentage rate and watch the curve climb faster and faster; then make the rate negative and watch it decay towards zero. Every curve there has the form \(y = A m^t\), and since \(m = e^{\ln m}\) it can be written \(y = Ae^{kt}\) with \(k = \ln m\): growth when \(k > 0\), decay when \(k < 0\).
The number \(e\): \(e = 2.718\,28\dots\), irrational. \(y = e^x\) is the exponential function whose gradient at every point equals its \(y\)-value.
Shape of \(y = e^{kx}\): passes through \((0, 1)\); \(y > 0\) for all \(x\); the \(x\)-axis (\(y = 0\)) is an asymptote.
\(k > 0\): increasing; a bigger \(k\) gives a steeper curve. \(k < 0\): decreasing, the reflection of \(y = e^{|k|x}\) in the \(y\)-axis.
Worked example Sketching a transformed exponential
Sketch \(y = 4 - 3e^{-x}\), giving the exact coordinates of any points where it meets the axes and the equation of the asymptote.
\(x = 0\): \(y = 4 - 3 = 1\), so \((0, 1)\).
As \(x \to \infty\), \(e^{-x} \to 0\), so \(y \to 4\): the asymptote is \(y = 4\), approached from below.
\(y = 0\): \(3e^{-x} = 4\), so \(e^{-x} = \dfrac{4}{3}\) and \(-x = \ln \dfrac{4}{3}\).
So \(x = -\ln \dfrac{4}{3} = \ln \dfrac{3}{4} \approx -0.288\). As \(x \to -\infty\), \(y \to -\infty\).
Answer: Increasing curve through \(\left(\ln \dfrac{3}{4}, 0\right)\) and \((0, 1)\), levelling off below the asymptote \(y = 4\).
Worked example An exponential decay model
The mass, \(m\) grams, of a substance after \(t\) days is \(m = 100e^{-0.03t}\). Find (a) the mass after 10 days, (b) the time taken for the mass to halve.
(a) \(m = 100e^{-0.3} = 74.08\dots\)
(b) Halving: \(100e^{-0.03t} = 50\), so \(e^{-0.03t} = 0.5\).
Take ln: \(-0.03t = \ln 0.5 = -\ln 2\).
\(t = \dfrac{\ln 2}{0.03} = 23.10\dots\)
Answer: (a) \(74.1\) g (b) \(23.1\) days
Examiner trap
\(e^{-x}\) is never negative: \(y = e^{-x}\) is a decreasing curve ABOVE the \(x\)-axis, not the curve \(y = e^x\) flipped below it. The minus sign in the power reflects in the \(y\)-axis; a minus sign in front, \(-e^x\), reflects in the \(x\)-axis.
Section 02
\(\ln x\): the inverse of \(e^x\).
\(\ln x\) means \(\log_e x\). It reverses \(e^x\), just as \(\log_2 x\) reverses \(2^x\), so its graph is the reflection of \(y = e^x\) in the line \(y = x\): it passes through \((1, 0)\), exists only for \(x > 0\), and has the \(y\)-axis as an asymptote. All the laws of logarithms hold for ln.
\(f(x) = e^{2x} - 3\) for all real \(x\). Find \(f^{-1}(x)\) and state its domain.
Let \(y = e^{2x} - 3\). Then \(e^{2x} = y + 3\).
Take ln: \(2x = \ln(y + 3)\), so \(x = \dfrac{1}{2}\ln(y + 3)\).
The range of \(f\) is \(f(x) > -3\) (since \(e^{2x} > 0\)), which becomes the domain of \(f^{-1}\).
Answer: \(f^{-1}(x) = \dfrac{1}{2}\ln(x + 3)\), for \(x > -3\)
Examiner trap
\(\ln(e^x + 1)\) is NOT \(x + 1\), and \(\ln(2e^x)\) is \(\ln 2 + x\), not \(2x\). ln only undoes \(e\) when the whole inside of the log is a single power of \(e\).
Stage 1 · Learn
Check your understanding
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What is \(\ln(e^3)\)?
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What is \(e^{\ln 7}\)?
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Which describes the graph of \(y = e^{-2x}\)?
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For which values of \(x\) is \(\ln x\) defined?
Section 03
Solving equations with \(e\) and ln.
Isolate the exponential (or the log) first, then apply the inverse function to both sides. Equations that look like quadratics in \(e^x\) are solved by substituting \(u = e^x\); remember that \(e^x\) can never be zero or negative.
Worked example Exponential to ln
Solve \(e^{2x - 1} = 7\), giving \(x\) exactly and to 3 significant figures.
Let \(u = e^x\), so \(e^{2x} = (e^x)^2 = u^2\): \(u^2 - 5u + 6 = 0\).
\((u - 2)(u - 3) = 0\), so \(e^x = 2\) or \(e^x = 3\).
Take ln: \(x = \ln 2\) or \(x = \ln 3\).
Answer: \(x = \ln 2\) or \(x = \ln 3\)
Your turn
Solve \(e^x - 6e^{-x} = 1\).
Multiply by \(e^x\): \(e^{2x} - e^x - 6 = 0\), so \((e^x - 3)(e^x + 2) = 0\). \(e^x = -2\) is impossible, so \(e^x = 3\) and \(x = \ln 3\).
Examiner trap
Take ln only after isolating the exponential: from \(3e^{x} + 1 = 10\), first get \(e^x = 3\), then \(x = \ln 3\). Writing \(\ln 3 + x + \ln 1 = \ln 10\) is not a valid step, because there is no law for the log of a sum.
Stage 2 · Exam
Exam-style questions
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Solve \(e^{2x - 1} = 7\), giving your answer to 3 significant figures.
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Solve \(e^{2x} - 5e^x + 6 = 0\).
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\(f(x) = e^{2x} - 3\) for all real \(x\). Find \(f^{-1}(x)\).
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A drug's concentration is \(C = 50e^{-0.04t}\) units after \(t\) hours. Find the time for \(C\) to fall to 20 units.
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e and ln, inverted.
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