1The idea in one paragraph

An exponential function f(x) = aˣ, with a > 0 and a ≠ 1, multiplies its output by the same factor a every time x goes up by 1. If a > 1 it grows ever faster; if 0 < a < 1 it decays towards zero. Every such graph passes through (0, 1) and has the x-axis as a horizontal asymptote. The most important base is the number e ≈ 2.718, giving eˣ. A logarithmic function f(x) = logₐ x undoes aˣ: it answers "what power of a gives x?". So the logarithm is the inverse of the exponential, and its graph is the exponential's graph reflected in y = x: it passes through (1, 0), has the y-axis as a vertical asymptote, and only exists for x > 0. The natural logarithm ln x is the inverse of eˣ. Two identities from the formula booklet tie them together: logₐ aˣ = x = a^(logₐ x), and aˣ = e^(x ln a).

2Exponential functions y = aˣ

Take y = 2ˣ and make a table:

Each step right doubles the output; each step left halves it. Halving forever never reaches zero, so on the left the graph gets closer and closer to the x-axis without touching it. That makes y = 0 a horizontal asymptote. Figure 1(a) shows y = 2ˣ and y = 3ˣ.

What every exponential y = aˣ has in common:

Why a > 0 and a ≠ 1? With a = 1, 1ˣ = 1 for every x: a flat line, not a growth curve. With a negative base, many inputs make no sense: (−4)^(1/2) = √(−4) is not a real number. So the definition only allows positive bases other than 1.

Decay is growth backwards. Figure 1(b) shows y = (½)ˣ. Since (½)ˣ = 2⁻ˣ, its graph is the graph of 2ˣ reflected in the y-axis. Every decreasing exponential is an increasing one with x replaced by −x.

A bigger base is steeper to the right and flatter to the left. For x > 0, 3ˣ > 2ˣ; for x < 0, 3ˣ < 2ˣ. The two curves cross only at (0, 1).

3The number e and the function eˣ

The number e = 2.71828… is a constant like π. It appears when growth is compounded more and more often. Invest €1 at 100% interest a year. Compounded once, you have €2. Compounded n times a year, you have (1 + 1/n)ⁿ euros, and as n grows this settles towards e: with n = 1000 it is 2.7169, with a million it is 2.71828. So e is what €1 becomes after a year of continuous growth at 100%, which links this page to the compound interest of 1.4.

The exponential function f(x) = eˣ has all the features of section 2: through (0, 1), asymptote y = 0, range y > 0, and between 2ˣ and 3ˣ since 2 < e < 3. Its special property appears in calculus (5.6): the gradient of y = eˣ at any point equals its y-value. That is why eˣ, and not 2ˣ or 10ˣ, is the base used in almost every model. Your calculator has an eˣ key, and a question that gives e^(−0.2t) expects you to use it.

Transformed exponentials. Models usually look like y = A·e^(kx) + c. Each constant has a job:

• c moves the asymptote from y = 0 to y = c, and the range becomes y > c (if A > 0).
• A is a vertical stretch; the y-intercept is A + c.
• k sets the rate: k > 0 grows, k < 0 decays.

Figure 2 shows y = 2e⁻ˣ + 3. The asymptote is y = 3, the y-intercept is (0, 2 + 3) = (0, 5), and the range is y > 3. A cup of coffee cooling to room temperature follows exactly this shape, with c as the room temperature.

4Logarithmic functions y = logₐ x

The logarithmic function f(x) = logₐ x asks "to what power must I raise a to get x?". So log₂ 8 = 3, because 2³ = 8, and log₂ ½ = −1, because 2⁻¹ = ½. The natural logarithm ln x is log base e.

By its definition, the logarithm undoes the exponential: if y = aˣ then x = logₐ y. So logₐ x is the inverse function of aˣ, and ln x is the inverse of eˣ. The graph of an inverse is the reflection in y = x (2.2), so every feature of the exponential reappears with x and y swapped. Figure 3 shows y = eˣ and y = ln x.

logₐ x and aˣ are inverse functions. Their graphs are reflections of each other in y = x, and logₐ x only exists for x > 0.

The restriction x > 0 is the one that costs marks. There is no power of a that gives 0 or a negative number, so ln 0 and ln(−3) do not exist. Your calculator returns an error, and the graph has no points to the left of the y-axis.

The logarithm grows very slowly. ln x passes 1 at x = e ≈ 2.72, passes 2 at x = e² ≈ 7.39, and only reaches 10 at x ≈ 22 026. It keeps growing without limit, but it has no horizontal asymptote: that is a common wrong guess.

Different bases. Figure 4 shows log₂ x, ln x and log₁₀ x. All pass through (1, 0), since a⁰ = 1 for every base, and all have x = 0 as a vertical asymptote. Each passes through (a, 1), which tells you which curve is which: log₂ x reaches 1 at x = 2, ln x at x = e, log₁₀ x at x = 10. For a base between 0 and 1 the log graph decreases instead, but such bases are rare in exams.

5The identities that link them

Because the two functions are inverses, composing them in either order gives the identity function (2.5). The formula booklet states it:

These simplify expressions at a glance: e^(ln 7) = 7, ln(e^(3x)) = 3x, 5^(log₅ 9) = 9, log₂(2^(x+1)) = x + 1. Combined with the laws of logarithms from 1.7:

The second booklet identity turns any exponential into base e:

aˣ = e^(x ln a)

Here is why. By the identity above, a = e^(ln a). Raise both sides to the power x: aˣ = (e^(ln a))ˣ = e^(x ln a). So 2ˣ = e^(0.693x) and 1.06ᵗ = e^(0.0583t), since ln 1.06 = 0.0583 to 3 s.f. This is how a model written with a growth factor (a population rising 6% a year, 1.06ᵗ) is converted into the form e^(kt) used in the sciences and in calculus, and back again. It also shows every exponential is eˣ with a horizontal stretch, which is why they all share one shape.

6Transformed logarithms, and finding inverses

A logarithm can only take a positive input, so the domain comes from setting the inside of the log greater than zero, and the vertical asymptote is where the inside equals zero.

Take f(x) = ln(2x − 6).

Now the inverse, by swapping and rearranging (2.5). The step that undoes ln is "take e to the power of both sides".

Figure 5 shows both. The vertical asymptote x = 3 of f becomes the horizontal asymptote y = 3 of f⁻¹; the x-intercept (3.5, 0) becomes the y-intercept (0, 3.5). The domain of f⁻¹ is the range of f, which is all of ℝ, and the range of f⁻¹ is the domain of f, y > 3.

The other direction works the same way. For g(x) = 3e^(x−1) + 2: swap to x = 3e^(y−1) + 2, so e^(y−1) = (x − 2)/3, so y − 1 = ln((x − 2)/3), giving g⁻¹(x) = 1 + ln((x − 2)/3), for x > 2. The asymptote y = 2 of g has become the vertical asymptote x = 2 of g⁻¹.

7Exponential models: growth and decay

An exponential model y = A·e^(kt) (or A·bᵗ) describes any quantity that changes by the same percentage in each equal time step. With whole-number t, its values form a geometric sequence (1.3), with common ratio b = e^k.

A drug is given by injection. The concentration in the blood t hours later is modelled by C(t) = 12e^(−0.2t) mg per litre. Figure 6 shows the curve.

The half-life is the time for the level to halve. Solve e^(−0.2t) = ½: t = ln 2 / 0.2 = 3.47 hours. The striking thing about exponential decay is that this does not depend on the starting level: from 12 to 6 takes 3.47 hours, and from 6 to 3 takes another 3.47 hours, which is why 3 mg/L is reached at 2 × 3.47 = 6.93 hours. In the same way, a quantity growing as e^(kt) has a doubling time of ln 2 / k.

On Paper 2 you can solve C(t) = 3 by intersecting y = 12e^(−0.2x) with y = 3 on the GDC, which gives the same 6.93. On Paper 1 you would be asked for the exact answer, 5 ln 4.

A word on "exponential". In everyday speech "exponential growth" often just means "fast". Mathematically it means growth by a constant percentage per unit time, and it can look slow at first: a quantity growing 3% a year takes about 23 years to double. What makes it remarkable is that it never stops speeding up, not that it starts fast. The guide invites you to ask whether the popular phrase is a misleading use of a mathematical term; used carelessly, it usually is. A real population cannot grow exponentially for ever either: food and space run out, so an exponential model is only valid over a limited domain.