P3.5 Exponential and logarithmic functions

What is the number e, and what does \(y = e^x\) look like?

Every exponential curve \(y = a^x\) with \(a > 1\) passes through \((0, 1)\), stays above the \(x\)-axis and gets steeper as \(x\) increases. A less obvious fact is that its gradient is always proportional to its height. For \(a = 2\) the gradient is about \(0.69 \times\) height; for \(a = 3\) it is about \(1.10 \times\) height. Somewhere between 2 and 3 there is a base for which the gradient equals the height. That base is \(e = 2.71828\ldots\), an irrational number like \(\pi\).

So \(\frac{d}{dx}(e^x) = e^x\), and more generally \(\frac{d}{dx}(e^{kx}) = ke^{kx}\). This is why e is the natural base for growth and decay: a quantity whose rate of change is proportional to its size follows a curve \(Ae^{kt}\). The curve \(y = e^{-x}\) is the reflection of \(y = e^x\) in the \(y\)-axis; it also passes through \((0, 1)\) but decreases towards the \(x\)-axis.

e also appears as a limit: \(\left(1 + \frac{1}{n}\right)^n\) gets closer to e as \(n\) grows. It is the value of £1 after a year at 100% interest compounded \(n\) times.

How do you sketch \(y = e^{ax+b} + c\)?

Treat the curve as a transformation of \(y = e^x\) (P3.2): \(a\) stretches it horizontally (and reflects it in the \(y\)-axis if \(a < 0\)), \(b\) shifts it sideways, and \(c\) shifts it up or down. Three questions then give the whole sketch.

• Asymptote: it moves with the vertical shift, so it is the line \(y = c\). Draw it dashed and label it.
• Intercepts: put \(x = 0\) for \(e^b + c\). For the \(x\)-intercept put \(y = 0\) and solve \(e^{ax+b} = -c\); it exists only if \(c < 0\), because the curve must start below the axis to cross it.
• Range: \(e^{ax+b} > 0\), so \(y > c\). If the exponential is multiplied by a negative number, as in \(y = 5 - 2e^{-x}\), the curve lies below its asymptote and the range is \(y < 5\).

If \(a > 0\) the curve rises to the right and approaches \(y = c\) on the left; if \(a < 0\) it is the other way round.

What is \(\ln x\), and how is its graph related to \(e^x\)?

The natural logarithm \(\ln x\) is \(\log_e x\): the power you raise e to in order to get \(x\). So \(y = \ln x \iff x = e^y\). This makes ln and the exponential inverse functions, each undoing the other: \(\ln(e^x) = x\) for every real \(x\), and \(e^{\ln x} = x\) for \(x > 0\). All the P2 laws of logarithms hold for ln: \(\ln ab = \ln a + \ln b\), \(\ln\frac{a}{b} = \ln a - \ln b\), \(\ln a^k = k\ln a\), \(\ln 1 = 0\) and \(\ln e = 1\).

Because they are inverses, the graph of \(y = \ln x\) is the reflection of \(y = e^x\) in the line \(y = x\): every point \((p, q)\) on one matches \((q, p)\) on the other. So \(\ln x\) is defined only for \(x > 0\), passes through \((1, 0)\), and has the \(y\)-axis as a vertical asymptote. It increases for ever, but very slowly.

For \(y = \ln(ax + b) + c\) with \(a > 0\), the same three questions as before give the sketch. You need \(ax + b > 0\), so the domain is \(x > -\frac{b}{a}\) and the vertical asymptote is \(x = -\frac{b}{a}\). The \(y\)-intercept (only if 0 is in the domain) is \(\ln b + c\). The \(x\)-intercept comes from \(\ln(ax + b) = -c \Rightarrow ax + b = e^{-c}\).

How do you solve \(e^{ax+b} = p\) and \(\ln(ax + b) = q\)?

The method has one golden rule: isolate first. Get the exponential (or the logarithm) on its own on one side, with a coefficient of 1. Then:

• for an exponential equation, take ln of both sides: \(e^{ax+b} = p \Rightarrow ax + b = \ln p\). This needs \(p > 0\); if \(p \le 0\) there are no solutions, because e to any power is positive;
• for a logarithmic equation, raise e to the power of both sides: \(\ln(ax + b) = q \Rightarrow ax + b = e^q\). Any real \(q\) is allowed.

Exact answers keep ln and e: \(x = \frac{\ln 7 + 2}{3}\), not 1.315. If an equation contains both \(e^{2x}\) and \(e^x\) (or \(e^x\) and \(e^{-x}\)), it is a quadratic in disguise: let \(u = e^x\), so \(e^{2x} = u^2\) and \(e^{-x} = \frac1u\), and reject any root \(u \le 0\).

How can a log graph reveal \(y = ax^n\)?

Data from an experiment may follow a curve whose equation is unknown. A straight line is much easier to recognise and measure than a curve, so we change the variables until the graph is straight. If \(y = ax^n\), take logs (any base, usually 10 or e) of both sides and use the laws of logs: \[\log y = \log a + n\log x\] Compare with \(Y = c + mX\): plotting \(\log y\) against \(\log x\) gives a straight line with gradient \(n\) and intercept \(\log a\) on the vertical axis. So \(n\) is the gradient, and \(a = 10^{\text{intercept}}\) (base 10) or \(e^{\text{intercept}}\) (base e). If the points do not lie close to a straight line, the relationship is not of the form \(ax^n\).

How can a log graph reveal \(y = kb^x\)?

For an exponential relationship \(y = kb^x\) the unknown is in the power, so taking logs gives \[\log y = \log k + x\log b\] Now plot \(\log y\) against \(x\) itself, not \(\log x\). The line has gradient \(\log b\) and intercept \(\log k\), so \(b = 10^{\text{gradient}}\) (or \(e^{\text{gradient}}\)) and \(k = 10^{\text{intercept}}\) (or \(e^{\text{intercept}}\)). The quickest way to tell the two models apart: \(ax^n\) is straight on a log–log graph; \(kb^x\) is straight when only the vertical axis is logged. The constant \(b\) is the growth factor per unit of \(x\): \(b = 1.35\) means a 35% increase for each unit increase in \(x\), and \(0 < b < 1\) means decay.

How do exponential growth and decay models behave?

A model \(y = Ae^{kt}\) describes growth if \(k > 0\) and decay if \(k < 0\). Its rate of change is \(\frac{dy}{dt} = kAe^{kt} = ky\), proportional to the current amount. Many real quantities settle towards a limit rather than to 0, giving \(y = L + Be^{-kt}\) with \(k > 0\); a hot drink cooling to room temperature \(L\) is the classic example.

• Initial value: put \(t = 0\), using \(e^0 = 1\). For \(L + Be^{-kt}\) it is \(L + B\).
• Long-term behaviour: as \(t \to \infty\), \(e^{-kt} \to 0\), so \(y \to L\). Say it in words, in context: "the temperature tends to 18 °C".
• When does \(y\) reach a value? Set \(y\) equal to it, isolate the exponential and take ln.
• Unknown constants: use the information given (often the value at \(t = 0\) and at one other time) to set up equations; the one involving the exponential is solved with ln.

How do you differentiate \(a^x\), and where do models break down?

Any exponential can be rewritten with base e. Since \(e^{\ln a} = a\), we have \(a^x = \left(e^{\ln a}\right)^x = e^{x\ln a}\). Differentiating \(e^{kx}\) with \(k = \ln a\) gives the result below. For example, the gradient of \(y = 2^x\) at \(x = 3\) is \(8\ln 2 \approx 5.545\).

A model \(y = Ab^t\) ("increases by 60% a month" means \(b = 1.6\)) can always be rewritten as \(Ae^{(\ln b)t}\), so everything in the previous section applies. Models are judged by their limitations:

• Unlimited growth. \(Ae^{kt}\) with \(k > 0\) grows without bound; real populations, users or sales are limited by space, food or the size of the market.
• Never reaching zero. A decay model predicts a positive amount for ever; a real quantity eventually becomes negligible, or has a floor such as a scrap value.
• Continuous values. The model gives non-integer numbers of people or bacteria.
• Constant conditions. The rate constant \(k\) is assumed fixed; a change in conditions breaks the model.

Refinements follow from the limitation: add a constant so the model levels off at a realistic limit (\(y = L - Be^{-kt}\) for a quantity growing towards a maximum \(L\); \(V = S + Be^{-kt}\) for a value decaying to a floor \(S\)), or re-fit the constants to more recent data.

Exam tips for P3.5

Common misconceptions

How this topic connects

P2 exponentials and P3.1 functions

The laws of logarithms, solving \(a^x = b\) and change of base all carry over to ln, which is simply \(\log_e\). \(\ln x\) and \(e^x\) are the standard example of a function and its inverse, with domain and range swapped.

P3.6 Differentiation and P3.7 Integration

The chain, product and quotient rules differentiate expressions such as \(x^2e^{3x}\) and \(\ln(x^2 + 1)\), and \(\frac{d}{dx}(\ln x) = \frac1x\) follows from the inverse relationship. In P3.7, \(\int e^{kx}\,dx = \frac1k e^{kx}\) and \(\int\frac1x\,dx = \ln|x|\); in P4, solving \(\frac{dy}{dt} = ky\) produces exactly the models on this page.

Statistics and Physics

The Poisson distribution uses \(e^{-\lambda}\). In IAL Physics, radioactive decay, capacitor discharge and log graphs of experimental data (log–log for power laws) use the same algebra as the log-graph and model sections here.

Key formulae

Check your understanding

Specification checklist

• (P3 3.1) I can sketch \(y = e^x\) and \(y = e^{-x}\), and know that \(e^x\) is positive with the \(x\)-axis as asymptote.
• (P3 3.1) I can sketch \(y = e^{ax+b} + c\), giving exact intercepts, the equation of the asymptote and the range.
• (P3 3.2) I can explain \(\ln x\) as the inverse of \(e^x\), use \(e^{\ln x} = x\) and \(\ln e^x = x\), and sketch \(y = \ln(ax + b) + c\) with its vertical asymptote.
• (P3 3.2) I can solve equations such as \(e^{ax+b} = p\) and \(\ln(ax + b) = q\) exactly, including those that become quadratics in \(e^x\).
• (P3 3.3) I can turn \(y = ax^n\) into a straight line by plotting \(\log y\) against \(\log x\), and find \(a\) and \(n\) from the gradient and intercept.
• (P3 3.3) I can turn \(y = kb^x\) into a straight line by plotting \(\log y\) against \(x\), and find \(k\) and \(b\).
• (P3 4.4) I can use exponential growth and decay models: find initial values, long-term behaviour and the time to reach a given value, and find unknown constants.
• (P3 4.4) I can differentiate \(e^{kx}\) and \(a^x\) (using \(a^x = e^{x\ln a}\)) to find rates of change in a model.
• (P3 4.4) I can state limitations of an exponential model in context and suggest a sensible refinement.