The number e, the graph of y = eˣ, and the gradient property of eᵏˣ

Sketching y = eᵃˣ⁺ᵇ + c: intercepts, asymptote and range

ln x as the inverse of eˣ, and graphs of y = ln(ax + b) + c

Exact solutions of eᵃˣ⁺ᵇ = p, ln(ax + b) = q and equations that reduce to them

Log graphs to estimate a and n in y = axⁿ, and k and b in y = kbˣ

Exponential growth and decay models: initial value, long-term behaviour and rates of change

d/dx(aˣ) = aˣln a, limitations of models and sensible refinements

1. The number e ≈2.718 is the base for which the gradient of y = eˣ equals its height at every point, and its inverse, the natural logarithm ln x, lets you solve exponential equations exactly, straighten curved data and analyse growth and decay models.

2. d/dx(eᵏˣ) = keᵏˣ. The curve y = eᵃˣ⁺ᵇ + c has horizontal asymptote y = c, y-intercept eᵇ + c and range y > c.

3. ln x = logₑ x undoes eˣ: e^(ln x) = x for x > 0 and ln(eˣ) = x. Its graph is the reflection of y = eˣ in y = x.

4. To solve, isolate the exponential or the log, then take ln (or exponentiate): eᵃˣ⁺ᵇ = p ⇒x = (ln p - b)/a; ln(ax + b) = q ⇒x = (e^q - b)/a.

5. y = axⁿ ⇒log y = log a + nlog x: plot log y against log x. y = kbˣ ⇒log y = log k + xlog b: plot log y against x.

6. In a model y = A + Be⁻ᵏᵗ, put t = 0 for the initial value; as t →∞, e⁻ᵏᵗ →0 so y →A.

Every exponential curve y = aˣ 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 × height; for a = 3 it is about 1.10 × height. Somewhere between 2 and 3 there is a base for which the gradient equals the height. That base is e = 2.71828…, an irrational number like π.

So d/dx(eˣ) = eˣ, and more generally d/dx(eᵏˣ) = keᵏˣ.

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 Aeᵏᵗ. The curve y = e⁻ˣ is the reflection of y = eˣ in the y-axis; it also passes through (0, 1) but decreases towards the x-axis.

e also appears as a limit: (1 + 1/n)ⁿ gets closer to e as n grows. It is the value of £1 after a year at 100% interest compounded n times.

Treat the curve as a transformation of y = eˣ (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 eᵇ + c. For the x-intercept put y = 0 and solve eᵃˣ⁺ᵇ = -c; it exists only if c < 0, because the curve must start below the axis to cross it.

Range: eᵃˣ⁺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.

WORKED EXAMPLE · QUESTION

The curve C has equation y = e^(½x+1) - 2. Sketch C, giving the exact coordinates of the points where it meets the axes and the equation of its asymptote. State the range of values of y.

SOLUTION

y-intercept: x = 0 gives y = e - 2, so (0, e - 2) (≈ 0.718).

x-intercept: isolate the exponential, then take ln. e^(½x+1) = 2 ⇒½x + 1 = ln 2 ⇒x = 2ln 2 - 2, so (2ln 2 - 2, 0) (≈ −0.614).

As x →-∞, e^(½x+1) →0: asymptote y = -2, range y > -2. The sketch is an increasing curve, flattening onto y = -2 on the left.

The natural logarithm ln x is logₑ x: the power you raise e to in order to get x. So y = ln x ⇔x = eʸ. This makes ln and the exponential inverse functions, each undoing the other: ln(eˣ) = 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 a/b = ln a - ln b, ln aᵏ = kln a, ln 1 = 0 and ln e = 1.

Because they are inverses, the graph of y = ln x is the reflection of y = eˣ 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 > -b/a and the vertical asymptote is x = -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 ⇒ax + b = e⁻ᶜ.

WORKED EXAMPLE · QUESTION

Sketch the curve with equation y = ln(2x + 6) - 1, stating the equation of its asymptote and the exact coordinates of the points where it meets the coordinate axes.

SOLUTION

Domain: 2x + 6 > 0 ⇒x > -3, so the asymptote is x = -3.

y-intercept: x = 0 gives y = ln 6 - 1, so (0, ln 6 - 1) (≈ 0.792).

x-intercept: isolate the log, then exponentiate. ln(2x + 6) = 1 ⇒2x + 6 = e, so ((e - 6)/2, 0) (≈ −1.64).

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: eᵃˣ⁺ᵇ = p ⇒ax + b = ln p. This needs p > 0; if p ≤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 ⇒ax + b = e^q. Any real q is allowed.

Exact answers keep ln and e: x = (ln 7 + 2)/3, not 1.315. If an equation contains both e²ˣ and eˣ (or eˣ and e⁻ˣ), it is a quadratic in disguise: let u = eˣ, so e²ˣ = u² and e⁻ˣ = 1/u, and reject any root u ≤0.

WORKED EXAMPLE · QUESTION

Find the exact solution of each equation. (a) e³ˣ⁻² = 7 (b) ln(4x - 1) = 2 (c) 5 - 2e⁻ˣ = 1

SOLUTION

(a) Take ln: 3x - 2 = ln 7 ⇒ x = (2 + ln 7)/3 (≈ 1.315).

(b) Exponentiate: 4x - 1 = e² ⇒ x = (e² + 1)/4 (≈ 2.097; check 4x - 1 = e² > 0).

(c) Isolate first: 2e⁻ˣ = 4 ⇒e⁻ˣ = 2 ⇒-x = ln 2 ⇒ x = -ln 2 (or ln½).

WORKED EXAMPLE · QUESTION

Solve, giving exact answers: (a) e²ˣ - 7eˣ + 12 = 0; (b) eˣ - 6e⁻ˣ = 1.

SOLUTION

(a) Let u = eˣ: u² - 7u + 12 = 0 ⇒(u - 3)(u - 4) = 0 ⇒eˣ = 3 or eˣ = 4. x = ln 3 or x = ln 4 (= 2ln 2).

(b) Multiply every term by eˣ, which is never zero: e²ˣ - 6 = eˣ ⇒u² - u - 6 = 0 ⇒(u - 3)(u + 2) = 0. eˣ = -2 has no solutions since eˣ > 0; eˣ = 3 gives x = ln 3 only.

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 = axⁿ, take logs (any base, usually 10 or e) of both sides and use the laws of logs:

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 = 10ⁱⁿᵗᵉʳᶜᵉᵖᵗ (base 10) or eⁱⁿᵗᵉʳᶜᵉᵖᵗ (base e). If the points do not lie close to a straight line, the relationship is not of the form axⁿ.

WORKED EXAMPLE · QUESTION

Two variables are thought to satisfy y = axⁿ. Measurements give:

The graph of log₁₀y against log₁₀x is a straight line passing through (0.25, 0.80) and (1.25, 2.40). (a) Find the values of a and n. (b) Estimate y when x = 25.

SOLUTION

(a) log₁₀y = log₁₀a + nlog₁₀x: gradient n, intercept log₁₀a. n = (2.40 - 0.80)/(1.25 - 0.25) = 1.6. Intercept: 0.80 = c + 1.6 ×0.25 ⇒c = 0.40, so a = 10^(0.40) = 2.51 (3 s.f.).

(b) y = 10^(0.40) ×25^(1.6) ≈ 433. Keep a unrounded here. x = 25 is outside the measured range, so this is an extrapolation and less reliable.

For an exponential relationship y = kbˣ the unknown is in the power, so taking logs gives

Now plot log y against x itself, not log x. The line has gradient log b and intercept log k, so b = 10ᵍʳᵃᵈⁱᵉⁿᵗ (or eᵍʳᵃᵈⁱᵉⁿᵗ) and k = 10ⁱⁿᵗᵉʳᶜᵉᵖᵗ (or eⁱⁿᵗᵉʳᶜᵉᵖᵗ). The quickest way to tell the two models apart: axⁿ is straight on a log–log graph; kbˣ 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.

WORKED EXAMPLE · QUESTION

The number of bacteria, N, in a culture t hours after it is set up is recorded:

It is suggested that N = kbᵗ. A line of best fit on a graph of ln N against t passes through (1, 5.31) and (7, 7.11). (a) Find k and b, to 3 significant figures. (b) Interpret b. (c) Use the line to estimate N when t = 10.

SOLUTION

(a) ln N = ln k + tln b: gradient ln b, intercept ln k. Gradient = (7.11 - 5.31)/(7 - 1) = 0.30; intercept = 5.31 - 0.30 = 5.01. b = e^(0.30) = 1.35, k = e^(5.01) = 150.

(b) The number of bacteria is multiplied by about 1.35 each hour, an increase of about 35% per hour.

(c) ln N = 5.01 + 0.30 ×10 = 8.01 ⇒N = e^(8.01) ≈ 3010. k is the model's value of N at t = 0 (about 150; the recorded value was 154).

A model y = Aeᵏᵗ describes growth if k > 0 and decay if k < 0. Its rate of change is dy/dt = kAeᵏᵗ = 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 e⁰ = 1. For L + Be⁻ᵏᵗ it is L + B.

Long-term behaviour: as t →∞, e⁻ᵏᵗ →0, so y →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.

WORKED EXAMPLE · QUESTION

The temperature, T °C, of a cup of tea t minutes after it is made is modelled by T = 18 + 72e^(-0.04t), t ≥0. (a) State the initial temperature of the tea. (b) State the temperature the tea approaches in the long term, and what this represents. (c) Find the time taken for the tea to cool to 40 °C. (d) Find the rate at which the temperature is falling when t = 10.

SOLUTION

(a) t = 0: T = 18 + 72 = 90 °C.

(b) e^(-0.04t) →0, so T → 18 °C, the temperature of the room.

(c) 72e^(-0.04t) = 22 ⇒e^(-0.04t) = (11/36) ⇒-0.04t = ln(11/36) ⇒t = 25ln(36/11) = 29.6 minutes. ln(11/36) is negative, so dividing by −0.04 gives a positive time; at t = 30 the model gives 39.7 °C, just below 40, which agrees.

(d) dT/dt = 72 ×(-0.04)e^(-0.04t) = -2.88e^(-0.04t). At t = 10: -2.88e^(-0.4) = -1.93.

The temperature is falling at 1.93 °C per minute.

Any exponential can be rewritten with base e. Since e^(ln a) = a, we have aˣ = (e^(ln a))ˣ = e^(xln a). Differentiating eᵏˣ with k = ln a gives the result below. For example, the gradient of y = 2ˣ at x = 3 is 8ln 2 ≈5.545.

A model y = Abᵗ ("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. Aeᵏᵗ 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.

WORKED EXAMPLE · QUESTION

After a new app is launched, the number of active users, N, is modelled by N = 500 ×1.6ᵗ, where t is the time in months after launch. (a) State the number of active users at launch. (b) Show that ln N = ln 500 + tln 1.6, and describe the graph of ln N against t. (c) Find the value of t at which the model first predicts 20 000 active users. (d) Using aˣ = e^(xln a), show that dN/dt = 500 ×1.6ᵗln 1.6, and find the rate of increase of users when t = 6. (e) Give one reason why this model is unlikely to be reliable for large t, and suggest how it could be refined.

SOLUTION

(a) 1.6⁰ = 1, so 500 users.

(b) ln N = ln(500 ×1.6ᵗ) = ln 500 + ln 1.6ᵗ = ln 500 + tln 1.6: a straight line with gradient ln 1.6 ≈0.470 and intercept ln 500 ≈6.21 on the ln N axis.

(c) 500 ×1.6ᵗ = 20 000 ⇒1.6ᵗ = 40 ⇒tln 1.6 = ln 40 ⇒t = ln 40/ln 1.6 = 7.85 months.

(d) N = 500e^(tln 1.6) ⇒dN/dt = 500ln 1.6 ×e^(tln 1.6) = 500 ×1.6ᵗln 1.6, as required. At t = 6: 500 ×16.78 ×0.4700 ≈ 3940 users per month (keep 1.6⁶ = 16.78… unrounded until the end).

(e) The model grows without limit: by t = 12 it predicts about 141 000 users, and the number of possible users is finite. A refinement is a model that levels off at a maximum, for example N = M - Be⁻ᵏᵗ.

P2 exponentials and P3.1 functions

The laws of logarithms, solving aˣ = b and change of base all carry over to ln, which is simply logₑ. ln x and eˣ 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 x²e³ˣ and ln(x² + 1), and d/dx(ln x) = 1/x follows from the inverse relationship. In P3.7, ∫eᵏˣ dx = 1/k eᵏˣ and ∫1/x dx = ln|x|; in P4, solving dy/dt = ky produces exactly the models on this page.

Statistics and Physics

The Poisson distribution uses e^(-λ). 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.

1. Solve e²ˣ⁺¹ = 5, giving an exact answer.

2. Solve ln(3x - 2) = -1, giving an exact answer.

3. State the asymptote and range of y = 4 - 3e⁻²ˣ.

4. A graph of ln y against ln x is a straight line. What form of relationship does this suggest, and what do the gradient and intercept give?

5. Find the gradient of y = 3ˣ at x = 2, exactly.

6. A car's value is modelled by V = 2000 + 18 000e^(-0.25t) (£, t in years). State its value when new and the value it approaches in the long term.

1. 2x + 1 = ln 5, so x = (ln 5 - 1)/2.

2. 3x - 2 = e⁻¹, so x = (e⁻¹ + 2)/3 = (2e + 1)/3e. A negative right-hand side is fine for a log equation.

3. As x →∞, e⁻²ˣ →0, so the asymptote is y = 4. Since 3e⁻²ˣ > 0, y < 4: the curve lies below its asymptote.

4. y = axⁿ: ln y = ln a + nln x, so the gradient is n and the intercept is ln a, giving a = eⁱⁿᵗᵉʳᶜᵉᵖᵗ.

5. dy/dx = 3ˣln 3, so at x = 2 the gradient is 9ln 3 (≈ 9.89).

6. At t = 0: V = 2000 + 18 000 = £20 000. As t →∞, e^(-0.25t) →0, so V →£2000: the model has a floor, like a scrap value, rather than decaying to zero.