2.9 Exponential and logarithmic functions

Some quantities change by a fixed amount in each time interval; others change by a fixed percentage. A bacterial colony that grows by 30% every hour, money that earns 5% interest a year and a radioactive sample that loses half of its atoms every 5730 years all change by the same factor in equal times. Functions in which the variable is an exponent, the exponential functions, describe this kind of change, and their inverses, the logarithmic functions, answer the question "how long until…?".

Exponential change starts slowly and then becomes very rapid (or, for decay, falls quickly and then levels off), which is why it is often underestimated. Recognising the shapes of exponential and logarithmic graphs, and the meaning of each parameter in an exponential model, is the basis for work on growth and decay throughout science, economics and later in calculus.

Exponential functions

Definition

In a power such as $x^2$ the base varies and the exponent is fixed. In an exponential function the base is fixed and the exponent varies.

The restrictions on the base keep the exponential function defined and interesting for every real $x$. A negative base fails for many inputs: $(-4)^{1/2} = \sqrt{-4}$ is not a real number. The base $a = 1$ gives $1^x = 1$ for all $x$, which is a constant function, and $a = 0$ gives $0^x$, which is undefined for $x \le 0$.

A table of values for $y = 2^x$ shows the characteristic pattern: each step of 1 in $x$ multiplies $y$ by 2.

Adding a constant to the input multiplies the output by a constant, which is the defining property of exponential change. In a linear function, equal steps in $x$ add the same amount to $y$; in an exponential function, equal steps in $x$ multiply $y$ by the same factor.

The graph of y = a^x when a > 1

The graph of $y = a^x$ for $a > 1$ has these key features.

• $y$-intercept $(0, 1)$, because $a^0 = 1$ for every base. The graph also passes through $(1, a)$.

• Horizontal asymptote $y = 0$. As $x \to -\infty$, $a^x \to 0$: for example, $2^{-10} = \frac{1}{1024}$

• Increasing everywhere, and increasing ever more steeply: as $x \to \infty$, $a^x \to \infty$.

• Domain $x \in \mathbb{R}$ and range $y > 0$. There is no $x$-intercept, because $a^x$ is never 0.

The larger the base, the faster the graph rises for $x > 0$ and the faster it approaches the asymptote for $x < 0$. All the curves cross at $(0, 1)$.

Growth and decay

When $0 < a < 1$, the function $y = a^x$ decreases. For example, $\left(\frac{1}{2}\right)^x$ halves with every step of 1 in $x$. Since $\left(\frac{1}{2}\right)^x = \frac{1}{2^x} = 2^{-x}$, the graph of $y = \left(\frac{1}{2}\right)^x$ is the reflection of $y = 2^x$ in the $y$-axis. It still passes through $(0, 1)$ and still has the asymptote $y = 0$, but now the curve approaches the asymptote as $x \to \infty$.

So the base decides between exponential growth ($a > 1$) and exponential decay ($0 < a < 1$), and a negative sign in the exponent turns growth into decay: $5^{-x} = \left(\frac{1}{5}\right)^x$ is a decay function.

The natural exponential function

One base is used far more than any other: Euler's number $e = 2.718\,28\ldots$ (see 1.5). The function $f(x) = e^x$ is the natural exponential function, and it appears on a graphic display calculator (GDC) as a dedicated key. Since $2 < e < 3$, the graph of $y = e^x$ lies between $y = 2^x$ and $y = 3^x$, with the same intercept $(0, 1)$ and asymptote $y = 0$.

What makes $e$ special is the gradient of its graph. The gradient of $y = e^x$ at any point is equal to the $y$-coordinate of that point, so at $(0, 1)$ the gradient is exactly 1. This property, studied in 5.6, makes $e^x$ the simplest exponential function to differentiate and integrate, and the natural choice for modelling continuous change.

Writing any exponential with base e

Every positive number can be written as a power of $e$: since $\ln a$ is the power of $e$ that gives $a$, $a = e^{\ln a}$. Raising both sides to the power $x$ gives a result that is in the formula booklet:

So every exponential function is the natural exponential function with a horizontal stretch: $a^x = e^{kx}$ with $k = \ln a$. Growth and decay can then be read from the sign of $k$: if $a > 1$ then $\ln a > 0$ (growth), and if $0 < a < 1$ then $\ln a < 0$ (decay). This equivalence is why scientists write almost every exponential model in base $e$.

Transformed exponential functions

Exponential models rarely have the plain form $a^x$. Functions such as $f(x) = ka^x + c$ and $f(x) = ka^{-x} + c$ are transformations of $y = a^x$ (see 2.11), and their key features follow from those of the basic graph.

The constant $c$ moves the asymptote, because $ka^x$ tends to 0 and so $f(x)$ tends to $c$. The sign of $k$ decides whether the curve lies above the asymptote ($k > 0$) or below it ($k < 0$). The $x$-intercept, if there is one, is found by solving $f(x) = 0$ with logarithms.

A graph's features also allow the constants in a model to be found. The asymptote gives $c$ directly, and the intercept or another point gives $k$.

Logarithmic functions

The logarithm as the inverse of the exponential

The function $f(x) = a^x$ ($a > 0$, $a \neq 1$) is one-to-one, since it is always increasing or always decreasing, so it has an inverse function. To find it, write $y = a^x$ and swap $x$ and $y$: $x = a^y$. By the definition of a logarithm, $x = a^y$ means $y = \log_a x$. So

For base $e$, the inverse of $e^x$ is the natural logarithm $\ln x$.

Because the two functions are inverses, applying one after the other returns the starting value. This gives the identities in the formula booklet:

The first holds for every real $x$; the second only for $x > 0$, because $\log_a x$ is defined only for positive $x$. With base $e$: $\ln e^x = x$ for all $x$, and $e^{\ln x} = x$ for $x > 0$.

Key features of the graph of y = log_a x

Reflecting $y = a^x$ in the line $y = x$ swaps the roles of $x$ and $y$, so every feature of the exponential graph has a partner on the logarithmic function's graph.

The graph of $y = \log_a x$ for $a > 1$ rises without limit, but very slowly. For example, $\log_{10} x$ reaches only 6 when $x = 1\,000\,000$. As $x \to 0^+$, $\log_a x \to -\infty$, so the $y$-axis is a vertical asymptote.

Different bases give graphs with the same intercept $(1, 0)$ and asymptote $x = 0$. A larger base gives a flatter curve for $x > 1$: $\log_{10} x < \ln x < \log_2 x$ for every $x > 1$. When $0 < a < 1$, the function $\log_a x$ is decreasing, as the reflection of a decaying exponential, but such bases are rarely used.

Transformed logarithmic functions

Functions such as $y = \ln(x - 2) + 1$ and $y = \log_3(x + 9)$ are transformations of the basic logarithmic graph. The most important feature to find first is the domain: the expression inside the logarithm must be positive. The vertical asymptote is where that expression equals zero.

When the variable has a negative coefficient inside the logarithm, as in $\ln(3 - 2x)$, the domain lies to the left of the asymptote. Here $3 - 2x > 0$ gives $x < 1.5$, and the graph is a decreasing curve with vertical asymptote $x = 1.5$.

Finding inverses of exponential and logarithmic functions

Since exponentials and logarithms undo each other, the inverse of a transformed exponential function is a logarithmic function, and the inverse of a transformed logarithmic function is an exponential one. The method is the one used for every inverse (see 2.5): swap $x$ and $y$, then isolate the exponential or logarithmic part before applying its inverse.

The reverse direction works the same way. For $g(x) = \ln(x + 4) - 1$, $x > -4$: swap to get $x = \ln(y + 4) - 1$, so $x + 1 = \ln(y + 4)$, $e^{x + 1} = y + 4$ and $g^{-1}(x) = e^{x + 1} - 4$, $x \in \mathbb{R}$. The vertical asymptote $x = -4$ of $g$ becomes the horizontal asymptote $y = -4$ of $g^{-1}$.

Exponential models

The model N = N₀e^(kt)

A quantity $N$ that grows or decays exponentially over time $t$ is modelled by

where $N_0$ and $k$ are constants with these meanings.

• $N_0$ is the initial value: when $t = 0$, $N = N_0 e^0 = N_0$.

• $k$ is the growth constant (or rate constant). If $k > 0$ the model shows exponential growth; if $k < 0$, exponential decay.

• In each unit of time, $N$ is multiplied by $e^k$. So the model can also be written $N = N_0 a^t$

The value of $k$ is usually found from one further measurement, by substituting it into the model and solving with natural logarithms.

Doubling time and half-life

A feature of exponential change is that the time taken to multiply by a given factor is always the same, wherever you start. A colony that doubles from 500 to 1000 in some time $T$ also doubles from 1000 to 2000 in the same time $T$. This time can be found from $k$.

For growth, solve $N_0 e^{kT} = 2N_0$: the $N_0$ cancels, $e^{kT} = 2$, and so

For decay ($k < 0$), solve $N_0 e^{kT} = \frac{1}{2}N_0$: $e^{kT} = \frac{1}{2}$, so $kT = \ln \frac{1}{2} = -\ln 2$ and

These formulas are not in the formula booklet, and the derivation takes only a line, so many people prefer to set up $e^{kT} = 2$ or $e^{kT} = \frac{1}{2}$ each time. In terms of the half-life $T$, a decay model can also be written $N = N_0 \left(\frac{1}{2}\right)^{t/T}$, which shows directly that $N$ halves every $T$ units of time.

For the bacteria in the last example, the doubling time is $\frac{\ln 2}{0.291\,82\ldots} \approx 2.38$ hours.

Radioactive decay is the classic example of a constant half-life. Carbon-14 is formed in the atmosphere and absorbed by living things; after an organism dies, the carbon-14 in it decays with a half-life of about 5730 years. Measuring the fraction that remains in a sample gives its age, a method called radiocarbon dating.

Continuous compounding

When interest is compounded $n$ times a year at a nominal annual rate $r$, a principal $P$ grows to $P\left(1 + \frac{r}{n}\right)^{nt}$ after $t$ years (see 1.4). Compounding more often gives slightly more interest, but the amount does not grow without limit. As $n \to \infty$, the amount approaches $Pe^{rt}$, which is the definition of continuous compounding.

The table shows the value after 10 years of USD 1000 invested at a nominal annual rate of 5% with different compounding frequencies.

Moving from yearly to monthly compounding adds about USD 18, but moving from daily to continuous compounding adds only 6 cents: the values approach a limit. Continuous compounding is the model $N = N_0 e^{kt}$ with $N_0 = P$ and $k = r$, so all the methods for growth models apply.

Models of the form y = ke^(−at) + c

When a quantity decays towards a value other than zero, the model has the form $y = ke^{-at} + c$, with horizontal asymptote $y = c$. A familiar case is a hot drink cooling in a room. The temperature does not fall to 0 °C; it approaches the room temperature. Newton's law of cooling gives the model

in which the difference between the drink's temperature and the room temperature decays exponentially. Here $T_0$ is the starting temperature and the asymptote $T = T_{\text{room}}$ is the long-term temperature.

Logarithmic scales

When a quantity varies over many powers of 10, it is often measured on a logarithmic scale, where each step of 1 means multiplication by a fixed factor. The moment magnitude scale for earthquakes is an example. The magnitude $M_w$ is calculated from the seismic moment $M_0$ (in newton metres), a measure of the size of the fault slip, by

This is a logarithmic function of $M_0$, so equal steps in magnitude correspond to equal factors in $M_0$. An increase of 1 in $M_w$ means $\log_{10} M_0$ increases by $\frac{3}{2}$, so $M_0$ is multiplied by $10^{1.5} \approx 31.6$.

Exam-style questions on exponential and logarithmic functions

The first question is a typical non-calculator question on graphs and inverses; the second is a growth and decay model with a GDC.

Self-check

• To do:

  I can sketch $y = a^x$ for $a > 1$ and $0 < a < 1$, and $y = e^x$, with the intercept and asymptote labelled.

• To do:

  I can find the asymptote, intercepts and range of $y = ka^x + c$ and similar transformed exponentials.

• To do:

  I can rewrite $a^x$ as $e^{x \ln a}$ and use the sign of $k$ in $e^{kx}$

• To do:

  I can sketch $y = \log_a x$ and transformed logarithmic graphs, stating the domain and vertical asymptote.

• To do:

  I can explain why $\log_a x$ is the inverse of $a^x$, and find inverses of exponential and logarithmic functions.

• To do:

  I can find $N_0$ and $k$ in $N = N_0 e^{kt}$

• To do:

  I can find a doubling time or half-life, and use a given half-life to find $k$.

• To do:

  I can calculate amounts with continuous compounding using $A = Pe^{rt}$.

1. (a)

   Solve the equation $f(x) = 2$, for $0 \le x \le 2\pi$.

   [3]3 marks

   Markscheme (a)

   $\sin x = \ln 2$ (M1)

   $x = 0.766$ $(0.765846\ldots)$ A1

   $x = 2.38$ $(2.37574\ldots)$ A1

   Note: Accept a graphical solution from a GDC.

2. (b)

   Find $f'(x)$.

   [2]2 marks

   Markscheme (b)

   $f'(x) = \cos x\,e^{\sin x}$

3. (c)

   Hence, find the exact coordinates of the local maximum point and the local minimum point on the graph of $f$, for $0 \le x \le 2\pi$.

   [3]3 marks

   Markscheme (c)

   $\cos x = 0$, since $e^{\sin x} > 0$

4. (d)

   Find the area of the region bounded by the curve $y = f(x)$, the lines $x = 0$ and $x = 2\pi$, and the $x$

5. (e)

   Show that $f(x)\,f(x + \pi) = 1$.

   [2]2 marks

   Markscheme (e)

   $\sin(x + \pi) = -\sin x$

6. (f)

   Hence, solve the equation $f(x) = 2f(x + \pi)$, for $0 \le x \le 2\pi$.

   [3]3 marks

   Markscheme (f)

(b) $0.8^x = e^{x \ln 0.8}$, so $k = \ln 0.8 \approx -0.223$. Since $k < 0$ (equivalently, $0 < 0.8 < 1$), the function shows decay.

(c) Use the laws of exponents in reverse: $e^{0.4x} = \left(e^{0.4}\right)^x$, so $a = e^{0.4} \approx 1.49$ (1.4918…). Each increase of 1 in $x$ multiplies $y$ by about 1.49, an increase of about 49%.

(b) $y$-intercept: $f(0) = 3 - 2e^{0} = 3 - 2 = 1$, giving $(0, 1)$.

$x$-intercept: solve $3 - 2e^{-x} = 0$.

The $x$-intercept is $\left(\ln \frac{2}{3}, 0\right)$, about $(-0.405, 0)$.

(c) Here $k = -2 < 0$, so the curve lies below the asymptote $y = 3$. As $x$ decreases, $e^{-x}$ grows rapidly, so $f(x) \to -\infty$. The graph increases from left to right and levels off towards $y = 3$. The range is $y < 3$.

(b) Solve $3 \times 2^x + 2 = 26$:

$x$-intercept: $\log_3(x + 9) = 0$ means $x + 9 = 3^0 = 1$, so $x = -8$. The point is $(-8, 0)$.

(c) The graph is $y = \log_3 x$ translated 9 units left. It rises steeply from the asymptote, crosses the $x$-axis at $(-8, 0)$ and the $y$-axis at $(0, 2)$, then continues to rise slowly.

(b) Write $y = 2e^{x - 1} + 3$ and swap $x$ and $y$:

So $f^{-1}(x) = 1 + \ln\left(\frac{x - 3}{2}\right)$. Its domain is the range of $f$, $x > 3$, which is also exactly where the argument $\frac{x - 3}{2}$ is positive.

Check: $f(1) = 2e^0 + 3 = 5$, and $f^{-1}(5) = 1 + \ln 1 = 1$.

So $k = \frac{1}{3}\ln 2.4 \approx 0.292$ (0.291 82…).

(b) $N = 500e^{8k} = 500e^{\frac{8}{3}\ln 2.4} \approx 5163$. Store the unrounded value of $k$; using $k = 0.292$ would give 5170. The model predicts about 5160 cells.

(c) Solve $500e^{kt} = 10\,000$:

So $k = -\frac{\ln 2}{5730}$, which is about $-1.21 \times 10^{-4}$.

(b) Solve $N = 0.35N_0$:

The wood is about 8680 years old. As a check, 35% lies between 50% (one half-life, 5730 years) and 25% (two half-lives, 11 460 years), and the answer lies between these ages.

(b) $T = 20 + 70e^{-10k} \approx 55.7$ °C. (Exactly, $e^{-10k} = \left(\frac{5}{7}\right)^2 = \frac{25}{49}$, so $T = 20 + \frac{250}{7} = 55.71\ldots$)

(c) Solve $20 + 70e^{-kt} = 40$: $e^{-kt} = \frac{2}{7}$, so $t = \frac{\ln 3.5}{k} \approx 18.6$ minutes.

(d) As $t \to \infty$, $e^{-kt} \to 0$ and $T \to 20$. The asymptote $T = 20$ means the tea cools towards room temperature but, according to the model, never quite reaches it.

(c) Swap $x$ and $y$ in $y = \ln(2x - 4)$:

So $f^{-1}(x) = \frac{e^x + 4}{2}$, $x \in \mathbb{R}$.

(d) The range of $f^{-1}$ is the domain of $f$: $y > 2$. The horizontal asymptote is $y = 2$, since $\frac{e^x}{2} \to 0$ as $x \to -\infty$.

(e) The graph of $f^{-1}$ has $y$-intercept $f^{-1}(0) = \frac{1 + 4}{2} = 2.5$, so it passes through $(0, 2.5)$, the reflection of $(2.5, 0)$ in $y = x$.

So $k = 0.0864$ to 3 significant figures. In a "show that" question, give the unrounded value before the rounded one.

(b) $A = 400e^{-0.086\,427\ldots \times 30} \approx 29.9$ MBq. Use the stored value of $k$ from part (a), not 0.0864.

(c) Solve $400e^{-kt} = 5$ with the GDC or by logarithms: $e^{-kt} = \frac{1}{80}$, so $t = \frac{\ln 80}{k} \approx 50.7$ days. The activity is still above 5 MBq after 50 days, so the sample can be disposed of after 51 complete days.

(d) The graph starts at $(0, 400)$, halves every 8.02 days and approaches the asymptote $A = 0$.

Note: Award A1 for a correct expression with one error, for example $e^{\cos x}$ in place of $e^{\sin x}$.

local maximum $\left(\frac{\pi}{2}, e\right)$ A1

local minimum $\left(\frac{3\pi}{2}, \frac{1}{e}\right)$ A1

[2]2 marks

$f(x)\,f(x + \pi) = e^{\sin x}\,e^{-\sin x}$ A1

$= e^{0} = 1$ AG

[3]3 marks