Exponential and logarithmic functions · AP Precalculus

Chapter 2 · Exponential and logarithmic functions

growth by multiplying

Some quantities grow by adding the same amount over each step. Others grow by multiplying by the same factor. The second kind starts slowly and then outruns every line.

The blue curve is $y = 2^x$ : each step of 1 to the right doubles the height, from 0.25 at x = −2 to 8 at x = 3. The red curve undoes it. $y = \log_2 x$ asks which power of 2 gives x, so it turns 8 back into 3.

2^3 = 8 \iff \log_2 8 = 3

The two curves are mirror images across the dashed line $y = x$ : the point (x, 2^x) on one is (2^x, x) on the other.

This chapter goes from sequences to exponential functions and their models, then through composition and inverses to logarithms, the equations they solve, and the semi-log plots that turn exponential data into a straight line. It is Unit 2 of AP Precalculus.

2.1 · Arithmetic and geometric sequences

adding, or multiplying

A sequence is a function whose inputs are whole numbers n. In an arithmetic sequence each term adds the same common difference d; in a geometric sequence each term multiplies by the same common ratio r.

a_n = a_0 + dn \qquad\qquad g_n = g_0\, r^n

The blue dots are $a_n = 2 + 3n$ and the red dots $g_n = 2(1.5)^n$ . Both start at 2. The arithmetic one climbs by 3 every step; the geometric one climbs by more each time, and by n = 6 it is ahead.

n	0	1	2	3	4	5	6
a_n	2	5	8	11	14	17	20
g_n	2	3	4.5	6.75	10.13	15.19	22.78

Any known term can serve as the starting point: $a_n = a_k + d(n - k)$ and $g_n = g_k\, r^{\,n-k}$ . Example. If $a_2 = 8$ and $a_5 = 17$ , the difference is $d = \frac{17 - 8}{5 - 2} = 3$ . If $g_1 = 3$ and $g_4 = 10.125$ , then $r^3 = 3.375$ , so r = 1.5.

The same two patterns, for every real input instead of whole steps, are linear and exponential functions.

2.2 · Linear and exponential change

equal steps in, equal differences or equal ratios out

Over input intervals of equal length, a linear function changes by equal differences; an exponential function changes by equal ratios. Linear change is additive, exponential change multiplicative.

Slide the window [x, x + 1] along both curves. On $f(x) = 2 + 3x$ the rise is always 3. On $g(x) = 2(1.5)^x$ the rise grows, but the ratio of the two heights is always 1.5:

\begin{gathered} f(x + 1) - f(x) = 3 \\ \frac{g(x + 1)}{g(x)} = 1.5 \end{gathered}

The forms say the same thing. $f(x) = b + mx$ starts at b and adds m per unit; $g(x) = a r^x$ starts at a and multiplies by r per unit. Arithmetic sequences are linear functions on the whole numbers, and geometric sequences exponential ones.

Over an interval of length 2 the linear function adds 6, twice 3, while the exponential multiplies by $1.5^2 = 2.25$ . Given two points, a linear model is $f(x) = y_i + m(x - x_i)$ and an exponential one $g(x) = y_i\, r^{\,x - x_i}$ .

2.3 · Exponential functions

a·bˣ, and what b decides

An exponential function has the form $f(x) = a b^x$ with initial value a ≠ 0 and base b > 0, b ≠ 1. Its graph passes through (0, a).

The red curve is $y = b^x$ as b sweeps from 2 down through 1 to $\tfrac{1}{2}$ and back. For a > 0:

b > 1: exponential growth, always increasing; $f(x) \to \infty$ as $x \to \infty$ and $f(x) \to 0$ as $x \to -\infty$ .
0 < b < 1: exponential decay, always decreasing; the end behavior reverses.
Domain: all real numbers. Range: y > 0. The line y = 0 is a horizontal asymptote.
Always concave up, with no extrema and no inflection points.

At b = 1 the curve is the flat line y = 1, which is why 1 is ruled out as a base. If a < 0 the graph flips below the axis and its concavity reverses.

Example. $f(x) = 3\left(\tfrac{1}{2}\right)^x$ starts at 3 and halves with each unit: $f(2) = 3 \cdot \tfrac{1}{4} = 0.75$ . As x grows it approaches 0 but never reaches it.

2.4 · Manipulating exponentials

a shift is a stretch

The laws of exponents rewrite an exponential function in equivalent forms, and each form shows something different about the graph.

product	$b^m b^n = b^{m+n}$
power	$(b^m)^n = b^{mn}$
negative exponent	$b^{-n} = \dfrac{1}{b^n}$
rational exponent	$b^{1/k} = \sqrt[k]{b}$

By the product law, $2^{x+k} = 2^k \cdot 2^x$ . So moving $y = 2^x$ left by k (blue) gives exactly the same curve as stretching it vertically by $2^k$ (red). The figure sweeps k from 0 to 2: the point (1, 2) slides left to (1 − k, 2) and rises to (1, $2^{1+k}$ ), and both land on the red curve.

2^{x+3} = 2^3 \cdot 2^x = 8 \cdot 2^x

A horizontal translation of an exponential is always a vertical dilation. The other laws give more: $\left(\tfrac{1}{2}\right)^x = 2^{-x}$ is a reflection of $2^x$ in the y-axis, and $8^{x/3} = (8^{1/3})^x = 2^x$ . Every exponential can be written with any positive base other than 1, including the natural base e ≈ 2.7183.

2.5 · Exponential models

a constant percent change

An exponential model fits a quantity that changes by the same proportion over equal times: a population, an investment, a radioactive sample.

Carbon-14 has a half-life of 5,730 years: every 5,730 years, half of what is left decays. With t in years, the percent left is

C(t) = 100\left(\tfrac{1}{2}\right)^{t/5730}

Per-unit rewriting gives the yearly factor: $\left(\tfrac{1}{2}\right)^{1/5730} \approx 0.999879$ . A sample with 30% left is about 9,953 years old, between one and two half-lives (2.13 shows how to solve for t).

Other forms come from the context:

Percent rate r: $f(t) = a(1 + r)^t$ (growth) or $a(1 - r)^t$ (decay).
Compounding n times a year: $1,000 at 6% monthly for 10 years is $1000\left(1 + \tfrac{0.06}{12}\right)^{120} \approx 1819.40$ ; compounded continuously, $1000e^{0.06(10)} \approx 1822.12$ .
Two points fix a model $y = y_i\, b^{\,x - x_i}$ . A hypothetical town of 1,200 that reaches 1,500 in 4 years has $b = \left(\tfrac{1500}{1200}\right)^{1/4} \approx 1.0574$ : growth of 5.74% a year.
Data: exponential regression on a calculator finds the a and b that best fit a scatterplot.

A model is only as good as its assumptions: it may stop fitting outside the data, or once a real limit (food, space) is reached.

2.6 · Choosing a model: residuals

what the model missed

A residual is the actual value minus the value the model predicts. A model is appropriate when its residual plot shows no pattern.

\text{residual} = y - \hat y

The counts are hypothetical: a bacteria culture counted every hour. Two models are fitted to them, the least-squares line $\hat y = -62.18 + 123.11x$ (blue) and the exponential $\hat y = 50.39(1.596)^x$ (red).

hour	0	1	2	3	4	5	6
count	52	77	131	202	322	532	834
linear residual	114	16	−53	−105	−108	−21	158
exponential residual	2	−3	3	−3	−5	10	1

The line's residuals run positive, negative, then positive again: a U shape, the sign of curvature the line cannot follow. The exponential's residuals scatter above and below 0 with no pattern, so the exponential model is the better choice.

The sign reads directly: a positive residual means the model underestimates there, a negative one that it overestimates. At hour 3 the line predicts 307 against an actual 202, an overestimate by 105.

2.7 · Composition of functions

one function’s output, the next one’s input

The composition $f(g(x))$ , also written $(f \circ g)(x)$ , applies g first and then f to the result.

Here $g(x) = x - 1$ and $f(x) = 2^x$ . Follow the dots down: x on the top line, then g(x), one less, then 2 to that power.

x = 3 \;\xrightarrow{\;g\;}\; 2 \;\xrightarrow{\;f\;}\; 2^{2} = 4

So $f(g(x)) = 2^{x-1}$ . The other order is a different function: $g(f(x)) = 2^x - 1$ , and $g(f(3)) = 7$ , not 4. In general $f(g(x)) \ne g(f(x))$ .

The domain of $f(g(x))$ is the inputs of g whose outputs are in the domain of f.
The identity function $I(x) = x$ changes nothing: $f(I(x)) = I(f(x)) = f(x)$ .
Transformations are compositions: $f(x - 1)$ composes f with an additive shift of the input, $3f(x)$ composes a multiplication by 3 after f.
Decomposing a function finds pieces to compose: $h(x) = (x^2 + 1)^3$ is $f(g(x))$ with $g(x) = x^2 + 1$ and $f(x) = x^3$ .

Compositions can be read from tables or graphs too: find g(x), then look that value up as an input of f.

2.8 · Inverse functions

undoing a function

The inverse $f^{-1}$ of a function reverses it: if $f(a) = b$ then $f^{-1}(b) = a$ , and $f^{-1}(f(x)) = f(f^{-1}(x)) = x$ .

Swapping input and output reflects the graph across $y = x$ : the point (a, b) on f becomes (b, a) on $f^{-1}$ . The domain of f becomes the range of $f^{-1}$ , and the range its domain.

An inverse function exists only if f is one-to-one: each output comes from just one input. The blue line $y = c$ meets the whole parabola $y = x^2$ twice, at $x = \pm\sqrt{c}$ , so $x^2$ has no inverse on all reals. Restricting the domain to x ≥ 0 keeps the half where it is one-to-one, and there the inverse is $f^{-1}(x) = \sqrt{x}$ (red).

Finding an inverse. Write $y = 3x - 4$ , swap x and y, and solve for y:

x = 3y - 4 \implies f^{-1}(x) = \frac{x + 4}{3}

Check: $f(3) = 5$ and $f^{-1}(5) = \frac{5 + 4}{3} = 3$ . The inverse undoes f step by step in reverse order: add 4, then divide by 3.

2.9 · Logarithmic expressions

the exponent, asked for

The logarithm $\log_b c$ is the exponent a that turns the base b into c, for b > 0, b ≠ 1 and c > 0:

\log_b c = a \iff b^a = c

expression	$\log_{5} 125$	$\log_{2} \tfrac{1}{16}$	$\log_{10} 1000$	$\log_{3} 1$
because	$5^{3} = 125$	$2^{-4} = \tfrac{1}{16}$	$10^{3} = 1000$	$3^{0} = 1$
value	3	−4	3	0

Two bases have their own names: the common logarithm $\log x$ has base 10, and the natural logarithm $\ln x$ has base e. Always $\log_b 1 = 0$ and $\log_b b = 1$ .

The figure lays the values $c = 10^u$ on a logarithmic scale (red), under their exponents $u = \log_{10} c$ (blue). Equal lengths on the scale are equal ratios: 1 to 10, 10 to 100 and 100 to 1000 are the same distance. That is how a log scale fits values of very different sizes on one line.

Estimates come from neighboring powers: 50 is between 10 and 100, so $\log_{10} 50$ is between 1 and 2; it is 1.699.

2.10 · Inverses of exponential functions

the logarithm undoes bˣ

$f(x) = b^x$ is one-to-one, so it has an inverse, and that inverse is the logarithm $f^{-1}(x) = \log_b x$ .

Swap every row of a table of $y = 2^x$ and you get a table of $y = \log_2 x$ . The red dots in the figure do exactly that: each (x, 2^x) flies across $y = x$ to (2^x, x).

x	−2	−1	0	1	2	3
2^x	0.25	0.5	1	2	4	8

x	0.25	0.5	1	2	4	8
log₂ x	−2	−1	0	1	2	3

So the two functions swap their features. $b^x$ has domain all reals, range y > 0 and the horizontal asymptote y = 0; $\log_b x$ has domain x > 0, range all reals and the vertical asymptote x = 0. Each undoes the other:

\log_b\left(b^x\right) = x \quad\text{for all } x, \qquad b^{\log_b x} = x \quad\text{for } x > 0

For base e, $e^x$ and $\ln x$ are inverses: $\ln e^{3} = 3$ and $e^{\ln 50} = 50$ .

2.11 · Logarithmic functions

multiply the input, add to the output

A logarithmic function has the form $f(x) = a \log_b x$ with a ≠ 0, b > 0 and b ≠ 1. It turns multiplicative change into additive change.

On $y = \log_2 x$ , each doubling of the input adds 1 to the output: the blue staircase runs from (1, 0) to (2, 1), (4, 2), (8, 3), (16, 4). Over input intervals with the same ratio, the outputs change by the same difference: exactly the reverse of an exponential.

x	1	2	4	8	16
log₂ x	0	1	2	3	4

For a > 0 and b > 1:

Domain x > 0, range all real numbers; the x-intercept is (1, 0).
The y-axis is a vertical asymptote: $\log_b x \to -\infty$ as $x \to 0^+$ .
Always increasing and always concave down; $\log_b x \to \infty$ as $x \to \infty$ , but more slowly than any positive power of x.
No extrema and no inflection points. For 0 < b < 1 it is always decreasing (and concave up).

Its slowness is a feature: the output goes up 1 every time the input doubles, so $\log_2 x$ only reaches 10 at x = 1024.

2.12 · Manipulating logarithms

a stretch is a shift

Each law of exponents has a logarithm twin, because a logarithm is an exponent.

product	$\log_b(xy) = \log_b x + \log_b y$
quotient	$\log_b \dfrac{x}{y} = \log_b x - \log_b y$
power	$\log_b x^n = n \log_b x$
change of base	$\log_b x = \dfrac{\log_a x}{\log_a b}$

By the product law, $\log_2(kx) = \log_2 k + \log_2 x$ . Squeezing $y = \log_2 x$ toward the y-axis by a factor 1/k (blue) gives the same curve as lifting it by $\log_2 k$ (red). As k sweeps from 1 to 4 the point (2, 1) moves in to (2/k, 1) and up to (2, 1 + $\log_2 k$ ), both on the red curve.

So a horizontal dilation of a logarithm is a vertical translation, the twin of 2.4. The power law makes $\log_b x^n$ a vertical dilation by n, and $\log_b \tfrac{1}{x} = -\log_b x$ a reflection in the x-axis.

Examples. $\log_2 40 = \log_2 8 + \log_2 5 \approx 3 + 2.3219 = 5.3219$ ; $\log_2 24 - \log_2 3 = \log_2 8 = 3$ ; and by change of base, $\log_3 20 = \dfrac{\ln 20}{\ln 3} \approx 2.7268$ . Every logarithmic function is a constant multiple of $\ln x$ .

2.13 · Exponential and logarithmic equations and inequalities

undo, then check

To solve an exponential equation, isolate the power and take a logarithm; to solve a logarithmic one, isolate the log and exponentiate. The inverse undoes the function.

The figure solves $3 \cdot 2^x = c$ for every level c at once: the curve crosses the blue line at $x = \log_2 \tfrac{c}{3}$ . At $c = 24$ :

\begin{aligned} 3 \cdot 2^x &= 24 \\ 2^x &= 8 \\ x &= \log_2 8 = 3 \end{aligned}

The inequality $3 \cdot 2^x > 24$ needs only the direction: $2^x$ is increasing, so it holds for x > 3, the red stretch of the axis. A decreasing function (base between 0 and 1) reverses the inequality.

With base e. $5e^{0.2t} = 20$ gives $e^{0.2t} = 4$ , so $t = \frac{\ln 4}{0.2} \approx 6.931$ .

A logarithmic equation. $\log_2 x + \log_2(x - 2) = 3$ combines to $\log_2\big(x(x - 2)\big) = 3$ , so $x^2 - 2x = 8$ and x = 4 or x = −2. But −2 is outside the domain of $\log_2 x$ : it is extraneous. The only solution is x = 4. Always check solutions in the original equation.

The same steps find inverses: $y = 3 \cdot 2^x$ has $f^{-1}(x) = \log_2 \tfrac{x}{3}$ .

2.14 · Logarithmic models

quick gains, then slower

A logarithmic model $y = a + b\ln x$ fits a quantity that grows quickly at first and then ever more slowly, or one where equal ratios of the input give equal differences in the output.

The data are hypothetical: one typist’s speed, in words per minute, after each week of practice. Logarithmic regression gives

\hat y = 20.52 + 14.70\ln x

Each doubling of the weeks adds $14.70 \ln 2 \approx 10.19$ words per minute: from week 1 to 2 as much as from week 4 to 8. Predicted at week 10, $\hat y \approx 54.4$ , an extrapolation beyond the data, to be trusted less.

Logarithmic scales measure many real quantities, because the values range over many powers of 10:

Sound level in decibels: $L = 10 \log_{10} \tfrac{I}{I_0}$ . Doubling the intensity adds $10\log_{10} 2 \approx 3.01$ dB; multiplying it by 10 adds 10 dB.
Acidity: $\text{pH} = -\log_{10}[\text{H}^+]$ . A solution with $[\text{H}^+] = 10^{-3}$ has pH 3; ten times the concentration lowers the pH by 1.

Logarithmic and exponential models are inverses: if y grows exponentially with x, then x grows logarithmically with y.

2.15 · Semi-log plots

exponential data, straightened

A semi-log plot keeps a linear x-axis and puts y on a logarithmic scale. On it, exponential data fall on a straight line.

Take the log of an exponential model $y = a b^x$ :

\log_{10} y = \log_{10} a + x \log_{10} b

That is a line in $\log_{10} y$ , with slope $\log_{10} b$ and vertical intercept $\log_{10} a$ . Any base n > 1 works the same way.

The bacteria counts of 2.6 (hypothetical) curve upward on ordinary paper and line up on semi-log paper. Their best-fit line is $\log_{10}\hat y = 1.7023 + 0.2030x$ , so

\begin{aligned} a &= 10^{1.7023} \approx 50.39 \\ b &= 10^{0.2030} \approx 1.596 \end{aligned}

The model is $\hat y = 50.39(1.596)^x$ : about 60% more bacteria each hour.

A semi-log plot is a test as well as a tool: data that line up are exponential; data that still curve are not, and need another model.