The exponential curve \(y = e^{x}\) shoots upwards but never touches the \(x\)-axis; the logarithm \(y = \ln x\) creeps upwards but never touches the \(y\)-axis. They are inverses, so each is the reflection of the other in \(y = x\). Learn their shapes, their asymptotes and their key points, and every graph of the form \(y = ke^{nx} + a\) or \(y = k\ln(ax + b)\) follows from them.
\(\log_{a} x = y\) means exactly the same as \(a^{y} = x\); \(\ln x\) is \(\log_{e} x\), where \(e \approx 2.718\). \(y = e^{x}\) has domain all real \(x\), range \(y > 0\), asymptote \(y = 0\) and passes through \((0, 1)\). \(y = \ln x\) has domain \(x > 0\), range all real \(y\), asymptote \(x = 0\) and passes through \((1, 0)\). They are inverse functions: \(\ln(e^{x}) = x\) and \(e^{\ln x} = x\), so each graph is the reflection of the other in \(y = x\). For \(y = ke^{nx} + a\): asymptote \(y = a\), \(y\)-intercept \(k + a\). For \(y = k\ln(ax + b)\): asymptote \(x = -\frac{b}{a}\), \(x\)-intercept where \(ax + b = 1\).
Fig. 1: y = eˣ (blue) and y = ln x (orange) are reflections of each other in the line y = x (dashed). y = eˣ has asymptote y = 0 and meets the y-axis at (0, 1); y = ln x has asymptote x = 0 and meets the x-axis at (1, 0)
Section 01
Exponentials and logarithms are inverses.
A logarithm answers the question 'what power?'. \(\log_{2} 8 = 3\) because \(2^{3} = 8\). In general, for a base \(a > 0\), \(a \ne 1\), \(\log_{a} x = y\) means exactly the same as \(a^{y} = x\). The most important base is the number \(e \approx 2.718\), and \(\log_{e} x\) is written \(\ln x\) (the natural logarithm). So \(y = \ln x\) means \(e^{y} = x\), and the functions \(f(x) = e^{x}\) and \(g(x) = \ln x\) undo each other.
Key values: \(e^{0} = 1\) so \(\ln 1 = 0\); \(e^{1} = e\) so \(\ln e = 1\). For any base, \(\log_{a} 1 = 0\) and \(\log_{a} a = 1\).
Inverses: \(\ln(e^{x}) = x\) for every real \(x\), and \(e^{\ln x} = x\) for every \(x > 0\). In the same way \(\log_{a}(a^{x}) = x\) and \(a^{\log_{a} x} = x\).
\(y = e^{x}\): domain all real \(x\), range \(y > 0\). It passes through \((0, 1)\), increases faster and faster, and as \(x \to -\infty\) it gets closer and closer to the \(x\)-axis without meeting it: the asymptote is \(y = 0\).
\(y = e^{-x}\) is the reflection of \(y = e^{x}\) in the \(y\)-axis: still through \((0, 1)\), still asymptote \(y = 0\), but decreasing (decay instead of growth).
\(y = \ln x\): domain \(x > 0\) (you cannot take the log of zero or a negative number), range all real \(y\). It passes through \((1, 0)\), increases more and more slowly, and plunges down towards the \(y\)-axis as \(x \to 0\): the asymptote is \(x = 0\).
Reflection in \(y = x\): inverse functions swap \(x\) and \(y\), so the point \((0, 1)\) on \(y = e^{x}\) becomes \((1, 0)\) on \(y = \ln x\), the asymptote \(y = 0\) becomes \(x = 0\), and domain and range swap (Fig. 1).
Worked example Switching between the two forms
Find the exact value of (a) \(\log_{2} 32\), (b) \(\log_{3} \frac{1}{9}\), (c) \(\ln e^{5}\), (d) \(e^{\ln 7}\). Then (e) write \(\ln x = 3\) and \(e^{y} = 10\) in the other form.
(a) \(\log_{2} 32 = y\) means \(2^{y} = 32 = 2^{5}\), so \(y = 5\).
(b) \(3^{y} = \frac{1}{9} = 3^{-2}\), so \(\log_{3} \frac{1}{9} = -2\). A log can be negative; only the number inside must be positive.
(c) \(\ln\) undoes \(e\) to the power: \(\ln e^{5} = 5\).
(d) \(e\) to the power undoes \(\ln\): \(e^{\ln 7} = 7\).
(e) \(\ln x = 3\) means \(x = e^{3}\); \(e^{y} = 10\) means \(y = \ln 10\). These are exact answers: on Paper 1 leave them like this.
\(f(x) = e^{x}\) for \(x \in \mathbb{R}\) and \(g(x) = \ln x\) for \(x > 0\). Simplify \(gf(x)\) and \(fg(x)\), stating where each is defined, and state the range of each of \(f\) and \(g\).
\(gf(x) = g(e^{x}) = \ln(e^{x}) = x\). Since \(e^{x} > 0\) for every \(x\), this works for all real \(x\).
\(fg(x) = f(\ln x) = e^{\ln x} = x\), but only for \(x > 0\), because \(\ln x\) needs \(x > 0\).
Range of \(f\): \(f(x) > 0\). Range of \(g\): all real numbers.
So \(g = f^{-1}\): the domain of each is the range of the other, exactly as for any function and its inverse in topic 1.
Answer: \(gf(x) = x\) for all real \(x\); \(fg(x) = x\) for \(x > 0\); range of \(f\) is \(f(x) > 0\), range of \(g\) is all real numbers
Your turn
Find the exact value of \(\log_{5} 125\), \(\log_{4} 2\) and \(\ln \frac{1}{e^{2}}\).
\(\ln x\) is only defined for \(x > 0\): \(\ln 0\) and \(\ln(-3)\) do not exist, so a log graph never reaches its asymptote and never appears on the wrong side of it. But the VALUE of a log can be negative or zero: \(\ln 1 = 0\) and \(\ln \frac{1}{e} = -1\). And \(e^{x}\) is never zero or negative, so \(y = e^{x}\) stays strictly above the \(x\)-axis, however far left you go.
Section 02
Exponential graphs: asymptote first.
Every graph \(y = ke^{nx} + a\) (with \(k\), \(n\), \(a\) integers, \(k \ne 0\), \(n \ne 0\)) is the curve \(y = e^{x}\) stretched and shifted. The term \(ke^{nx}\) gets closer and closer to \(0\) at one end, so the curve levels out towards the horizontal line \(y = a\). Draw that asymptote dashed first, then decide whether the curve sits above it or below it.
Fig. 2: y = 4 - 2eˣ. The asymptote is y = 4; since k = -2 is negative, the curve lies below it, meets the y-axis at (0, 2) and crosses the x-axis at (ln 2, 0)
Asymptote: \(y = a\). It is approached as \(x \to -\infty\) when \(n > 0\), and as \(x \to +\infty\) when \(n < 0\).
Above or below: \(ke^{nx}\) has the sign of \(k\). If \(k > 0\) the curve is above the asymptote and the range is \(y > a\); if \(k < 0\) it is below and the range is \(y < a\).
\(y\)-intercept: \(x = 0\) gives \(y = k + a\), since \(e^{0} = 1\).
\(x\)-intercept: solve \(ke^{nx} + a = 0\), that is \(e^{nx} = -\frac{a}{k}\). This has a solution only if \(-\frac{a}{k} > 0\), and then \(x = \frac{1}{n}\ln\left(-\frac{a}{k}\right)\). If \(-\frac{a}{k} \le 0\), the curve never reaches \(y = 0\) and there is no \(x\)-intercept.
Worked example A curve below its asymptote
Sketch \(y = 4 - 2e^{x}\), stating the equation of the asymptote and the exact coordinates of the points where the curve meets the axes. State the range.
In the form \(ke^{nx} + a\): \(k = -2\), \(n = 1\), \(a = 4\). As \(x \to -\infty\), \(e^{x} \to 0\), so \(y \to 4\): the asymptote is \(y = 4\).
\(k < 0\), so the curve is below the asymptote and falls more and more steeply as \(x\) increases.
\(y\)-intercept: \(4 - 2(1) = 2\), so \((0, 2)\).
\(x\)-intercept: \(4 - 2e^{x} = 0\) gives \(e^{x} = 2\), so \(x = \ln 2\): the point \((\ln 2, 0)\).
Range: \(y < 4\) (Fig. 2).
Answer: Asymptote \(y = 4\); meets the axes at \((0, 2)\) and \((\ln 2, 0)\); range \(y < 4\)
Worked example Finding the constants from the graph
The curve \(y = ke^{2x} + a\) has asymptote \(y = 2\) and meets the \(y\)-axis at \((0, 5)\). Find \(k\) and \(a\), and find the exact value of \(y\) when \(x = \ln 2\).
The asymptote of \(y = ke^{2x} + a\) is \(y = a\), so \(a = 2\).
\(y\)-intercept \(k + a = 5\), so \(k = 3\) and \(y = 3e^{2x} + 2\).
Sketch \(y = 3e^{-2x} + 1\): state the asymptote, the \(y\)-intercept, the range and whether the curve meets the \(x\)-axis.
Asymptote \(y = 1\) (approached as \(x \to +\infty\), since \(n = -2 < 0\)); \(y\)-intercept \((0, 4)\); \(k = 3 > 0\), so the curve is a decreasing curve above its asymptote and the range is \(y > 1\). It never meets the \(x\)-axis, because \(y > 1\) everywhere.
Examiner trap
The \(y\)-intercept of \(y = ke^{nx} + a\) is \(k + a\), not \(k\) and not \(a\): \(e^{0} = 1\), so the \(k\) survives. And the asymptote is \(y = a\), not \(y = 0\): a vertical shift moves the asymptote too. Always draw the asymptote dashed and label it with its equation.
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What is the exact value of \(\log_{2} \frac{1}{8}\)?
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What is the equation of the asymptote of \(y = 3e^{2x} - 5\)?
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What is the domain of \(y = \ln(2x - 6)\)?
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Which graph is the reflection of \(y = e^{x}\) in the line \(y = x\)?
Section 03
Log graphs: a wall where the bracket is zero.
For \(y = k\ln(ax + b)\) (with \(k\), \(a\), \(b\) integers), the logarithm needs \(ax + b > 0\). The curve exists on one side of the line where \(ax + b = 0\), and plunges along that line without ever touching it: \(x = -\frac{b}{a}\) is a vertical asymptote. Because \(\ln 1 = 0\), the curve crosses the \(x\)-axis where the bracket equals \(1\), not where it equals \(0\).
Fig. 3: y = 2 ln(x + 3). The bracket must be positive, so the domain is x > -3 and the line x = -3 is the asymptote. The curve crosses the x-axis where x + 3 = 1, at (-2, 0), and meets the y-axis at (0, 2 ln 3)
Domain and asymptote: solve \(ax + b > 0\). The asymptote is \(x = -\frac{b}{a}\). If \(a < 0\), as in \(\ln(4 - x)\), the domain is to the LEFT of the asymptote.
\(x\)-intercept: \(ax + b = 1\), so \(x = \frac{1 - b}{a}\).
\(y\)-intercept: \(y = k\ln b\), which exists only if \(b > 0\) (so that \(x = 0\) is in the domain). Leave it exact, such as \(2\ln 3\).
Shape: the range is all real \(y\). If \(k > 0\) and \(a > 0\) the curve rises slowly to the right; a negative \(k\) reflects it in the \(x\)-axis, a negative \(a\) reflects it left to right.
Link to exponentials: the inverse of \(y = e^{x} + c\) is \(y = \ln(x - c)\). Reflecting in \(y = x\) turns a horizontal asymptote into a vertical one.
Worked example Domain, asymptote and exact intercepts
Sketch \(y = 2\ln(x + 3)\), stating the domain, the equation of the asymptote and the exact coordinates of the points where the curve meets the axes.
The bracket must be positive: \(x + 3 > 0\), so the domain is \(x > -3\) and the asymptote is \(x = -3\).
\(x\)-intercept: \(2\ln(x + 3) = 0\) means \(\ln(x + 3) = 0\), so \(x + 3 = 1\) and \(x = -2\): the point \((-2, 0)\).
\(y\)-intercept: \(x = 0\) gives \(y = 2\ln 3\): the point \((0, 2\ln 3)\).
Shape: \(k = 2 > 0\), so the curve comes up from \(-\infty\) beside the asymptote, crosses at \((-2, 0)\) and keeps rising, more and more slowly (Fig. 3).
Answer: Domain \(x > -3\); asymptote \(x = -3\); meets the axes at \((-2, 0)\) and \((0, 2\ln 3)\)
Worked example An exponential and its inverse on one diagram
\(f(x) = e^{x} + 3\) for \(x \in \mathbb{R}\). Find \(f^{-1}(x)\) and state its domain. On one diagram, sketch \(y = f(x)\) and \(y = f^{-1}(x)\), giving their asymptotes and the points where they meet the axes.
For \(y = -2\ln(x - 1)\), state the domain, the asymptote, any points where the curve meets the axes, and whether it rises or falls.
\(x - 1 > 0\): domain \(x > 1\), asymptote \(x = 1\). \(x - 1 = 1\) gives \((2, 0)\); there is no \(y\)-intercept because \(x = 0\) is not in the domain. \(k = -2 < 0\), so the curve falls: it comes down from \(+\infty\) beside the asymptote and keeps falling.
Examiner trap
The \(x\)-intercept of a log graph comes from bracket \(= 1\), NOT bracket \(= 0\): bracket \(= 0\) gives the asymptote. For \(y = 2\ln(x + 3)\), \(x = -3\) is the asymptote and \(x = -2\) is the intercept. In the next two lessons the laws of logarithms and solving equations such as \(e^{x} = 2\) will make these intercepts quicker to find.
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For the curve \(y = 5 - 3e^{2x}\), which gives the point where it meets the \(y\)-axis and the equation of its asymptote?
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Where does the curve \(y = 3\ln(2x - 5)\) meet the coordinate axes?
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The curve \(y = ke^{x} + a\) passes through \((0, 1)\) and has asymptote \(y = -3\). Find \(k\) and \(a\).
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\(f(x) = e^{2x} + 1\) for \(x \in \mathbb{R}\). Find \(f^{-1}(x)\) and its domain.
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