7. Exponents and logarithms¶

Some quantities grow by adding the same amount each step; many more grow by multiplying by the same factor: money earning interest, bacteria dividing, a rumour spreading, a radioactive sample decaying. Multiplying again and again is written with exponents, and running it backwards ("how many times must I multiply?") is the logarithm. In this chapter you will see why \(a^0 = 1\) and \(a^{1/2} = \sqrt{a}\) are forced on us, model growth and decay, prove the laws of logarithms, solve exponential equations, meet the number \(e\) as the limit of compound interest, and see why decibels, pH and earthquake magnitudes are all logarithms.

In this chapter

You will learn:

what \(a^n\) means for whole, negative, fractional and irrational \(n\), and why each definition is the only sensible one
how to model exponential growth and decay, with doubling times and half-lives
what a logarithm is, and proofs of the product, quotient and power laws
how to change the base of a logarithm, and count the digits of huge numbers
how to solve exponential and logarithmic equations, and avoid their traps
where the number \(e\) comes from, with a proof that the compound-interest limit exists
why so many measuring scales (sound, acidity, earthquakes, star brightness) are logarithmic

You need: Chapter 6 (functions and inverses) and Chapter 2 (solving equations and inequalities). Gentler versions: Powers, Zero and negative exponents, Exponent laws, Exponential functions, Logarithms and Exponential and log equations on the Topics shelf.

Levels: 7.1 to 7.3 are beginner to intermediate; 7.4 to 7.6 and 7.8 are intermediate; 7.7 is advanced.

7.1 Exponents: whole, negative, fractional, real¶

Whole-number exponents. For a number \(a\) and a whole number \(n \ge 1\),

\[ a^n = \underbrace{a \cdot a \cdots a}_{n \text{ factors}} . \]

\(a\) is the base and \(n\) the exponent (or power). Three laws follow by counting factors:

\[ a^m \cdot a^n = a^{m+n}, \qquad (a^m)^n = a^{mn}, \qquad (ab)^n = a^n b^n . \]

Why. \(a^m a^n\) is \(m\) factors of \(a\) followed by \(n\) more: \(m + n\) factors. \((a^m)^n\) is \(n\) blocks of \(m\) factors: \(mn\) factors. \((ab)^n\) is \(n\) copies of \(ab\), and since multiplication can be reordered (Chapter 1), that is \(n\) \(a\)'s times \(n\) \(b\)'s. \(\square\)

A fourth law, for \(a \neq 0\) and \(m > n\): \(a^m / a^n = a^{m-n}\), because \(n\) of the factors cancel.

Zero exponent. What should \(a^0\) mean? "Zero factors of \(a\)" sounds like nothing at all. But we want the laws to keep working. If \(a^m \cdot a^0 = a^{m+0} = a^m\), then \(a^0\) must be 1 (divide both sides by \(a^m\), allowed when \(a \neq 0\)). So for \(a \neq 0\) we define

\[ a^0 = 1 . \]

This is not a discovery about nature; it is the only choice that keeps the laws true. (The same reasoning makes an empty product equal to 1, just as an empty sum is 0: multiplying by "nothing" should change nothing. The expression \(0^0\) is usually set to 1 in algebra and combinatorics, for this reason, but left undefined in some parts of calculus.)

Negative exponents. Again, keep the laws: \(a^n \cdot a^{-n} = a^0 = 1\), so \(a^{-n}\) must be the reciprocal of \(a^n\):

\[ a^{-n} = \frac{1}{a^n} \qquad (a \neq 0). \]

So \(2^{-3} = 1/8\) and \(10^{-2} = 0.01\). Now \(a^m / a^n = a^{m-n}\) holds for all integers \(m\), \(n\), with no condition \(m > n\). A pattern shows it is natural: going down the list \(2^3 = 8\), \(2^2 = 4\), \(2^1 = 2\), each step halves, so continuing gives \(2^0 = 1\), \(2^{-1} = 1/2\), \(2^{-2} = 1/4\).

Fractional exponents. What should \(a^{1/2}\) be? If the law \((a^m)^n = a^{mn}\) is to survive, then \((a^{1/2})^2 = a^1 = a\). So \(a^{1/2}\) must be a number whose square is \(a\): a square root. For \(a > 0\) there are two, and we pick the positive one. In general, for \(a > 0\) and a whole number \(n \ge 1\),

\[ a^{1/n} = \sqrt[n]{a} \quad \text{(the positive number whose } n\text{th power is } a\text{)}, \qquad a^{m/n} = \left(a^{1/n}\right)^m . \]

So \(8^{2/3} = (\sqrt[3]{8})^2 = 4\), \(16^{3/4} = 2^3 = 8\), \(27^{-2/3} = 1/9\).

There is a small thing to check: \(m/n\) can be written in many ways (\(2/3 = 4/6\)), and the definition must not depend on which. If \(m/n = p/q\), is \((a^{1/n})^m = (a^{1/q})^p\)? Raise both to the power \(nq\): the left gives \(a^{mq}\) (since \((a^{1/n})^{n} = a\)), the right gives \(a^{pn}\), and \(mq = pn\) because \(m/n = p/q\). Two positive numbers with the same \(nq\)-th power are equal (the function \(x \mapsto x^{nq}\) is one-to-one on positive numbers). So the definition is consistent. \(\square\)

Why the base must be positive. For negative bases, fractional powers misbehave. \(\sqrt[3]{-8} = -2\) makes sense, so one might write \((-8)^{1/3} = -2\). But \(1/3 = 2/6\), and \(\left((-8)^2\right)^{1/6} = 64^{1/6} = 2\). The same exponent gives two answers depending on how it is written. To keep the laws, we only allow fractional and real exponents for positive bases. (In Chapter 9 the complex numbers will explain what is really going on: there are three cube roots of \(-8\), and choosing among them is the problem.)

Irrational exponents. What is \(2^{\sqrt 2}\)? \(\sqrt 2 = 1.41421\dots\) is not a fraction, so none of the definitions apply. But \(\sqrt 2\) is approached by fractions, and the powers approach a limit:

\(x\)	1.4	1.41	1.414	1.4142	\(\sqrt 2\)
\(2^x\)	2.63902	2.65737	2.66475	2.66512	2.66514...

We define \(a^x\), for \(a > 0\) and real \(x\), as the limit of \(a^{r}\) as the fraction \(r\) approaches \(x\). That the limit exists, does not depend on the fractions chosen, and still obeys the three laws is proved in a first course in analysis (it uses the completeness of the real numbers and the fact that \(r \mapsto a^r\) is increasing for \(a > 1\)); see MAT158H1 Unit 11 for one careful route, which instead defines the logarithm first, by an integral, and the exponential as its inverse. For algebra, the upshot is simple: for \(a > 0\), \(a^x\) is defined for every real \(x\), is always positive, and the laws

\[ a^x a^y = a^{x+y}, \qquad (a^x)^y = a^{xy}, \qquad (ab)^x = a^x b^x, \qquad a^{-x} = \frac{1}{a^x} \]

hold for all real \(x\), \(y\).

7.2 Exponential growth and decay¶

An exponential function is \(f(x) = a^x\) with base \(a > 0\), \(a \neq 1\), or more generally \(f(t) = C a^t\). Its defining feature: equal steps in \(t\) multiply \(f\) by equal factors. Indeed \(f(t + 1) = C a^{t+1} = a \cdot f(t)\). Compare a linear function \(g(t) = C + mt\), where equal steps add equal amounts.

For \(a > 1\) the graph rises, slowly at first and then steeply; for \(0 < a < 1\) it falls towards zero without ever reaching it. All of them pass through \((0, 1)\), since \(a^0 = 1\), and all are positive. The graph of \((1/a)^x\) is the mirror image of \(a^x\) in the \(y\)-axis, since \((1/a)^x = a^{-x}\) (a reflection from Chapter 6).

Growth. A population of bacteria doubles every 20 minutes. Starting from \(N_0\) cells, after \(t\) hours (three doublings an hour) there are

\[ N(t) = N_0 \cdot 2^{3t} \]

cells. After 3 hours, \(2^9 = 512\) times as many. After a day, \(2^{72} \approx 4.7 \times 10^{21}\) times as many, which is why real colonies stop growing exponentially: they run out of food. Every exponential model in the real world is valid only for a while.

Percentage growth. Money growing at 7% a year is multiplied by 1.07 each year: after \(t\) years, \(P(t) = P_0 (1.07)^t\). How long until it doubles? We need \(1.07^t = 2\). By trial, \(1.07^{10} \approx 1.967\), just short, so a little over 10 years; in 7.6 we will solve it exactly (\(t \approx 10.24\)). Bankers use the rule of 72: at \(r\) percent, money doubles in about \(72/r\) years (\(72/7 \approx 10.3\)). It is an approximation that works well for rates of a few percent.

Decay and half-life. A radioactive isotope decays so that in each fixed time, the half-life \(T\), half of what remains decays. After time \(t\) the amount left is

\[ N(t) = N_0 \left(\tfrac12\right)^{t/T} = N_0 \, 2^{-t/T} . \]

Carbon-14 has a half-life of about 5730 years. After 10 000 years the fraction left is \(2^{-10000/5730} \approx 0.298\), about 30%. Radiocarbon dating runs this backwards: measure the fraction, solve for \(t\); and solving for an exponent is exactly what logarithms are for.

Exponentials beat powers. Compare \(2^n\) with \(n^{10}\). At \(n = 10\), \(2^{10} = 1024\) while \(10^{10}\) is ten billion; the power is far ahead. But from \(n = 59\) on, \(2^n > n^{10}\) forever. Why must the exponential win eventually? Look at what happens from one step to the next: \(2^n\) is multiplied by 2, while \(n^{10}\) is multiplied by \(\left(\frac{n+1}{n}\right)^{10} = \left(1 + \frac1n\right)^{10}\), which gets closer and closer to 1 as \(n\) grows (for \(n \ge 15\) it is below 2, since \((16/15)^{10} \approx 1.91\)). Once each step multiplies the exponential by more than the power, the ratio \(2^n / n^{10}\) grows by a fixed factor above 1 at every step and must eventually pass any bound. The same argument shows that \(a^n\) beats \(n^k\) for any \(a > 1\) and any \(k\).

7.3 Logarithms: the inverse of exponentials¶

\(2^x = 8\) has the solution \(x = 3\). What about \(2^x = 10\)? Somewhere between 3 and 4, since \(2^3 = 8 < 10 < 16 = 2^4\). We need a name for "the exponent that turns 2 into 10".

For \(a > 0\), \(a \neq 1\), the function \(x \mapsto a^x\) is a bijection from \(\mathbb{R}\) onto \((0, \infty)\): it is one-to-one because it is strictly increasing (for \(a > 1\)) or strictly decreasing (for \(a < 1\)), and it is onto \((0, \infty)\) because it takes every positive value. (The second claim needs the intermediate value theorem of calculus; we take it as given here.) So by Theorem 6.1 of Chapter 6 it has an inverse, the logarithm to base \(a\):

\[ \log_a y = x \quad \text{means} \quad a^x = y . \]

In words: \(\log_a y\) is the power you must raise \(a\) to, to get \(y\). So \(\log_2 8 = 3\), \(\log_{10} 1000 = 3\), \(\log_{10} 0.01 = -2\), \(\log_5 1 = 0\), \(\log_9 3 = 1/2\), and \(\log_2 10 \approx 3.3219\).

Because they are inverse functions,

\[ a^{\log_a y} = y \quad (y > 0), \qquad \log_a(a^x) = x \quad (\text{all } x). \]

Facts that follow at once:

\(\log_a y\) is defined only for \(y > 0\). There is no power of 2 that equals 0 or \(-3\).
\(\log_a 1 = 0\) and \(\log_a a = 1\).
The graph of \(\log_a\) is the mirror image of the graph of \(a^x\) in the line \(y = x\); it passes through \((1, 0)\), climbs slowly for \(a > 1\), and plunges down as \(y\) approaches 0.
For \(a > 1\), \(\log_a\) is increasing: bigger numbers have bigger logs. It grows very slowly: \(\log_{10}\) of a billion is only 9.

Two bases have special names. \(\log_{10}\) is the common logarithm, often written just \(\log\) on calculators; \(\log_e\), with the number \(e \approx 2.718\) of 7.7, is the natural logarithm, written \(\ln\). In much of pure mathematics "\(\log\)" means \(\ln\); in computer science it often means \(\log_2\). When it matters, say which.

A logarithm counts multiplications. \(\log_{10} y\) is roughly "how many digits \(y\) has"; \(\log_2 y\) is "how many times can you halve \(y\) before reaching 1". A binary search through a sorted list of a million names takes about \(\log_2 10^6 \approx 20\) halvings. This is why logarithms show up wherever things are repeatedly split or multiplied.

7.4 The laws of logarithms, proved¶

Theorem 7.1. For \(a > 0\), \(a \neq 1\), positive \(x\), \(y\) and any real \(r\):

\(\log_a (xy) = \log_a x + \log_a y\) (product law),
\(\log_a (x/y) = \log_a x - \log_a y\) (quotient law),
\(\log_a (x^r) = r \log_a x\) (power law).

Proof. Let \(u = \log_a x\) and \(v = \log_a y\), so that \(x = a^u\) and \(y = a^v\).

\(xy = a^u a^v = a^{u+v}\), so the power that gives \(xy\) is \(u + v\): \(\log_a(xy) = u + v\).
\(x / y = a^u / a^v = a^{u - v}\), so \(\log_a(x/y) = u - v\).
\(x^r = (a^u)^r = a^{ur}\), so \(\log_a(x^r) = ru\). \(\square\)

Each law is an exponent law read backwards. The product law is the reason logarithms were invented: it turns multiplication into addition. Before electronic calculators, to multiply two eight-digit numbers you looked up their logarithms in a printed table, added them, and looked up the number with that logarithm. A slide rule does the same with two rulers marked in logarithmic distances: sliding one along the other adds lengths, so it multiplies numbers.

Example 1. Expand \(\log_3 \dfrac{9x^4}{\sqrt{y}}\). Product and quotient: \(\log_3 9 + \log_3 x^4 - \log_3 y^{1/2}\). Power law and \(\log_3 9 = 2\): \(2 + 4\log_3 x - \tfrac12 \log_3 y\).

Example 2. Combine \(2\ln x - \ln(x + 1) + \tfrac13 \ln 8\) into one logarithm. Power law first: \(\ln x^2 - \ln(x + 1) + \ln 2\). Then: \(\ln \dfrac{2x^2}{x + 1}\).

Common mistakes, and why they are wrong. \(\log(x + y)\) is not \(\log x + \log y\): try \(x = y = 1\) in base 10: \(\log 2 \approx 0.301\), but \(\log 1 + \log 1 = 0\). And \(\dfrac{\log x}{\log y}\) is not \(\log(x/y)\): in base 10 with \(x = 100\), \(y = 10\), the first is \(2/1 = 2\) and the second is \(\log 10 = 1\). The laws convert products into sums, nothing else.

7.5 Change of base¶

Calculators have only \(\log_{10}\) and \(\ln\). How do we compute \(\log_2 10\)?

Theorem 7.2 (change of base). For positive \(a, b \neq 1\) and \(y > 0\),

\[ \log_a y = \frac{\log_b y}{\log_b a} . \]

Proof. Let \(x = \log_a y\), so \(a^x = y\). Take \(\log_b\) of both sides and use the power law: \(x \log_b a = \log_b y\). Since \(a \neq 1\), \(\log_b a \neq 0\), and dividing gives the result. \(\square\)

So \(\log_2 10 = \dfrac{\ln 10}{\ln 2} \approx \dfrac{2.302585}{0.693147} \approx 3.321928\). Two special cases are worth remembering: \(\log_a b = \dfrac{1}{\log_b a}\) (take \(y = b\)), and every logarithm is a constant multiple of every other: \(\log_a y = (\text{constant}) \cdot \ln y\) with constant \(1/\ln a\). So there is essentially one logarithm function, in different units, just as there is one length measured in metres or feet.

Counting digits. A whole number \(N \ge 1\) has \(d\) digits exactly when \(10^{d-1} \le N < 10^d\), that is \(d - 1 \le \log_{10} N < d\). So

\[ \text{number of digits of } N = \lfloor \log_{10} N \rfloor + 1 . \]

Example 3. How many digits does \(2^{100}\) have? \(\log_{10} 2^{100} = 100 \log_{10} 2 \approx 100 \times 0.30103 = 30.103\). So 31 digits. Check: \(2^{100} = 1\,267\,650\,600\,228\,229\,401\,496\,703\,205\,376\), which has 31 digits. The fractional part tells even more: \(10^{0.103} \approx 1.27\), so the number starts 127.

Try it in Python.

import math
N = 2**100
print(len(str(N)), math.floor(100 * math.log10(2)) + 1)   # 31 31
print(math.log(10) / math.log(2), math.log2(10))          # both 3.3219...

7.6 Solving exponential and logarithmic equations¶

The strategy: when the unknown is in an exponent, take logarithms; when the unknown is inside a logarithm, exponentiate. Then check every answer in the original equation, because the domain of \(\log\) (positive numbers only) can create false solutions.

Example 4. \(3^x = 20\). Take \(\ln\): \(x \ln 3 = \ln 20\), so \(x = \dfrac{\ln 20}{\ln 3} \approx 2.7268\). Check: \(3^{2.7268} \approx 20.0\).

Example 5. Doubling at 7%: \(1.07^t = 2\) gives \(t = \dfrac{\ln 2}{\ln 1.07} \approx 10.24\) years. In general, growth at rate \(r\) per period doubles in \(\ln 2 / \ln(1 + r)\) periods. For small \(r\), \(\ln(1 + r) \approx r\), so the doubling time is about \(0.693/r\); using 72 instead of 69.3 makes the rule easy to divide and slightly more accurate for rates near 8%.

Example 6. \(2^{x+1} = 5^x\). Take \(\ln\): \((x + 1)\ln 2 = x \ln 5\). Collect \(x\): \(x(\ln 5 - \ln 2) = \ln 2\), so \(x = \dfrac{\ln 2}{\ln 2.5} \approx 0.7565\). Check: \(2^{1.7565} \approx 3.378\) and \(5^{0.7565} \approx 3.378\).

Example 7. \(5 \cdot 3^{2x} = 2 \cdot 7^x\). Take \(\ln\): \(\ln 5 + 2x\ln 3 = \ln 2 + x \ln 7\), so \(x(2\ln 3 - \ln 7) = \ln 2 - \ln 5\), that is \(x \ln \tfrac97 = \ln 0.4\), and \(x = \dfrac{\ln 0.4}{\ln(9/7)} \approx -3.646\).

Example 8 (a quadratic in disguise). \(4^x - 3 \cdot 2^x - 4 = 0\). Since \(4^x = (2^x)^2\), put \(u = 2^x\): \(u^2 - 3u - 4 = 0\), \((u - 4)(u + 1) = 0\), so \(u = 4\) or \(u = -1\). But \(u = 2^x > 0\), so only \(2^x = 4\), \(x = 2\). Check: \(16 - 12 - 4 = 0\).

Example 9 (a false solution). \(\log_{10} x + \log_{10}(x - 3) = 1\). Combine: \(\log_{10}(x(x - 3)) = 1\), so \(x(x - 3) = 10\), \(x^2 - 3x - 10 = 0\), \((x - 5)(x + 2) = 0\). Candidates 5 and \(-2\). But \(\log_{10}(-2)\) is undefined, so \(-2\) is rejected. Answer: \(x = 5\). Check: \(\log 5 + \log 2 = \log 10 = 1\).

Where did \(-2\) come from? Combining the logs replaced "\(x > 0\) and \(x - 3 > 0\)" by the weaker "\(x(x - 3) > 0\)", which \(x = -2\) satisfies. This is the general lesson of Chapter 2: a step that is not reversible can add solutions, and only a check catches them.

Example 10 (an inequality). A filter removes 10% of the remaining dust on each pass. How many passes until less than 1% remains? We need \(0.9^n < 0.01\). Take \(\ln\) (an increasing function, so the inequality keeps its direction): \(n \ln 0.9 < \ln 0.01\). Now divide by \(\ln 0.9\), which is negative, so the inequality flips: \(n > \dfrac{\ln 0.01}{\ln 0.9} \approx 43.7\). So 44 passes. Check: \(0.9^{43} \approx 0.0108\), \(0.9^{44} \approx 0.0097\).

7.7 The number \(e\)¶

Level: advanced.

The compound interest question. Put 1 dollar in a bank that pays 100% interest a year. Paid once, at the end of the year, you have 2 dollars. Paid as two half-yearly instalments of 50%, you have \(1.5^2 = 2.25\). Paid \(n\) times, each time \(1/n\) of 100%, you have

\[ a_n = \left(1 + \frac{1}{n}\right)^n . \]

More frequent payments help, because interest earns interest. Does it grow without limit?

\(n\)	1	2	4	12 (monthly)	365 (daily)	8760 (hourly)	1 000 000
\(a_n\)	2	2.25	2.4414	2.6130	2.7146	2.7181	2.7182805

The values climb, but more and more slowly, and seem to settle near 2.71828. Jacob Bernoulli asked this question in 1683 and showed the answer lies between 2 and 3. The limit is called \(e\).

Theorem 7.3. The sequence \(a_n = (1 + 1/n)^n\) is increasing and always less than 3. Hence it has a limit, \(e = \lim_{n \to \infty} a_n\), with \(2 < e \le 3\).

Proof. We need two tools.

Bernoulli's inequality: for \(x \ge -1\) and whole \(m \ge 1\), \((1 + x)^m \ge 1 + mx\). It holds for \(m = 1\), and if it holds for \(m\) then

\[ (1 + x)^{m+1} = (1 + x)^m (1 + x) \ge (1 + mx)(1 + x) = 1 + (m+1)x + mx^2 \ge 1 + (m+1)x , \]

where multiplying by \(1 + x\) was allowed because \(1 + x \ge 0\). So it holds for every \(m\). (This "if for \(m\) then for \(m + 1\)" argument is proof by induction, the subject of Chapter 8.)

Increasing. Compare consecutive terms:

\[ \frac{a_{n+1}}{a_n} = \frac{\left(\frac{n+2}{n+1}\right)^{n+1}}{\left(\frac{n+1}{n}\right)^{n}} = \frac{n+1}{n} \left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1} = \frac{n+1}{n} \left(1 - \frac{1}{(n+1)^2}\right)^{n+1} . \]

By Bernoulli with \(x = -1/(n+1)^2\) and \(m = n + 1\), the bracket raised to the power is at least \(1 - \frac{n+1}{(n+1)^2} = 1 - \frac{1}{n+1} = \frac{n}{n+1}\). So \(\dfrac{a_{n+1}}{a_n} \ge \dfrac{n+1}{n} \cdot \dfrac{n}{n+1} = 1\). (It is in fact strictly bigger, since \(mx^2 > 0\) in the inequality above.)

Bounded. Expand with the binomial theorem (Chapter 8, §8.4):

\[ a_n = \sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k}, \qquad \binom{n}{k}\frac{1}{n^k} = \frac{1}{k!} \cdot \frac{n}{n} \cdot \frac{n-1}{n} \cdots \frac{n-k+1}{n} \le \frac{1}{k!} . \]

And \(k! = 1 \cdot 2 \cdot 3 \cdots k \ge 2^{k-1}\) (each factor after the first is at least 2). So

\[ a_n \le 1 + 1 + \frac12 + \frac14 + \cdots + \frac{1}{2^{n-1}} < 1 + 2 = 3 . \]

An increasing sequence that stays below a fixed number has a limit: this is the completeness of the real numbers, the property that the number line has no gaps (see MAT158H1 Unit 2). And \(a_2 = 2.25 > 2\) with the sequence increasing gives \(e > 2\). \(\square\)

The same comparison, pushed a little further, shows

\[ e = \sum_{k=0}^{\infty} \frac{1}{k!} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \cdots = 2.718281828459045\dots \]

and this series converges very fast: twelve terms give eight correct decimals. \(e\) is irrational (Euler showed this in 1737) and even transcendental (Hermite, 1873), which Chapter 19 of this book discusses.

Continuous growth. With interest at rate \(r\) compounded \(n\) times a year for \(t\) years, 1 dollar becomes \((1 + r/n)^{nt}\). As \(n \to \infty\) this approaches \(e^{rt}\), because \((1 + r/n)^{n} \to e^r\) (substitute \(n = rm\): \((1 + 1/m)^{rm} = \left((1 + 1/m)^m\right)^r \to e^r\)). So continuous compounding gives

\[ A = P e^{rt} . \]

1000 dollars at 5% for 10 years: yearly compounding gives \(1000 \times 1.05^{10} \approx 1628.89\) dollars, monthly \(1000(1 + 0.05/12)^{120} \approx 1647.01\), continuous \(1000 e^{0.5} \approx 1648.72\). The jump from yearly to monthly is large; from monthly to continuous, tiny.

Why \(e\) is "natural". Of all exponential curves \(y = a^x\), the one with \(a = e\) is the one whose slope where it crosses the \(y\)-axis is exactly 1. From that one fact, calculus derives that the rate of change of \(e^x\) is \(e^x\) itself: a quantity growing at a rate equal to its size. That is why \(e\) appears in every law of natural growth and decay, and why \(\ln\) is the logarithm of choice. In the language of half-lives, \(N_0 2^{-t/T} = N_0 e^{-kt}\) with \(k = \ln 2 / T\). The full story belongs to calculus; see Exponential, log and trig derivatives and MAT137 Unit 4.

7.8 Logarithmic scales¶

When a quantity ranges over many powers of ten, a scale that counts factors of ten is more useful than one that counts units. Such a scale takes a logarithm.

Sound: decibels. The intensity of sound we can hear ranges over a factor of about \(10^{12}\). The sound level in decibels (dB) is

\[ L = 10 \log_{10} \frac{I}{I_0}, \]

where \(I_0 = 10^{-12}\) watts per square metre is a reference near the quietest audible sound. Every factor of 10 in intensity adds 10 dB; doubling adds \(10 \log_{10} 2 \approx 3\) dB. A 90 dB sound is not "1.5 times" a 60 dB sound; it is \(10^3 = 1000\) times as intense.

Acidity: pH. \(\mathrm{pH} = -\log_{10} [\mathrm{H}^+]\), where \([\mathrm{H}^+]\) is the concentration of hydrogen ions in moles per litre. Pure water at 25 °C has pH 7. Lemon juice, around pH 2, has \(10^5\) times the hydrogen ion concentration. The minus sign is there only so that the numbers come out positive.

Earthquakes: magnitude. Charles Richter's 1935 scale took the logarithm of the largest swing of a standard seismograph. Today seismologists use the moment magnitude,

\[ M_w = \tfrac23\left(\log_{10} M_0 - 9.1\right), \]

where \(M_0\) is the seismic moment in newton-metres, a measure of the size of the fault slip. One step in magnitude multiplies \(M_0\), and roughly the energy released, by \(10^{1.5} \approx 31.6\); two steps by 1000. NASA's InSight lander recorded more than a thousand marsquakes from 2019 to 2022; the largest, in May 2022, had magnitude about 4.7. An Earth quake of magnitude 6.7 would release about \(10^{1.5 \times 2} = 1000\) times as much energy.

Stars: apparent magnitude. Astronomers grade brightness on a scale inherited from ancient Greek catalogues, fixed precisely by Norman Pogson in 1856: a difference of 5 magnitudes is a factor of exactly 100 in brightness, and brighter objects have smaller magnitudes. So

\[ m_1 - m_2 = -2.5 \log_{10} \frac{F_1}{F_2}, \]

where \(F\) is the light flux received. One magnitude is a factor \(100^{1/5} \approx 2.512\). At its brightest Mars reaches about magnitude \(-2.9\); Sirius, the brightest star, is about \(-1.46\); so Mars at its best outshines Sirius by a factor \(10^{0.4 \times 1.44} \approx 3.8\).

Log-log and semi-log graphs. Plot \(y = C a^x\) with a logarithmic vertical axis and it becomes a straight line, since \(\log y = \log C + x \log a\), a line with slope \(\log a\). A curve that looks straight on such a plot is exponential, and the slope reads off the growth rate. Similarly a power law \(y = Cx^k\) is a straight line when both axes are logarithmic, with slope \(k\). Scientists use these plots to discover laws: Kepler's third law (period squared proportional to orbit size cubed) is a straight line of slope \(3/2\) on a log-log plot of the planets' periods against their distances, Mars included.

A short history. John Napier, a Scottish landowner, published the first table of logarithms in 1614, in Mirifici Logarithmorum Canonis Descriptio; his logarithms were built for trigonometry and did not use any of the bases we use now. Jost Bürgi in Switzerland made a similar table independently, published in 1620. Henry Briggs, after visiting Napier, proposed base 10 and published large tables of common logarithms, notably the Arithmetica Logarithmica of 1624. These tables, and the slide rules built from them (from the 1620s on), were the calculators of astronomers, navigators and engineers for three and a half centuries. Johannes Kepler used logarithms in the calculations for his Rudolphine Tables (1627), the planetary tables built on his laws of planetary motion, which he had found by studying the orbit of Mars. The number \(e\) came later: Bernoulli's interest problem (1683), and Euler, who used the letter \(e\) in his writings from the late 1720s and in his Introductio in analysin infinitorum (1748) set out exponentials and logarithms as inverse functions, much as this chapter does.

Check yourself¶

1. Simplify \(\dfrac{(2x^3)^2 \cdot x^{-4}}{8x^{1/2}}\) for \(x > 0\).

\((2x^3)^2 = 4x^6\). Then \(4x^6 \cdot x^{-4} = 4x^2\), and \(\dfrac{4x^2}{8x^{1/2}} = \tfrac12 x^{3/2}\).

2. Evaluate \(32^{3/5}\), \(\left(\tfrac{4}{9}\right)^{-3/2}\) and \(\log_4 \tfrac18\).

\(32^{1/5} = 2\), so \(32^{3/5} = 8\). \(\left(\tfrac49\right)^{-3/2} = \left(\tfrac94\right)^{3/2} = \left(\tfrac32\right)^3 = \tfrac{27}{8}\). \(\tfrac18 = 2^{-3} = 4^{-3/2}\), so \(\log_4 \tfrac18 = -\tfrac32\).

3. A drug's amount in the blood halves every 6 hours. What fraction is left after 1 day, and after how many hours is 10% left?

After 24 hours, 4 half-lives: \((1/2)^4 = 1/16\). For 10%: \(2^{-t/6} = 0.1\), so \(t/6 = \log_2 10 \approx 3.3219\) and \(t \approx 19.9\) hours.

4. Prove that \(\log_a x \cdot \log_x b = \log_a b\) (for positive \(a, x, b\) with \(a, x \neq 1\)).

Change base to \(\ln\): \(\dfrac{\ln x}{\ln a} \cdot \dfrac{\ln b}{\ln x} = \dfrac{\ln b}{\ln a} = \log_a b\).

5. Solve \(7^{2x - 1} = 3^{x + 2}\).

\((2x - 1)\ln 7 = (x + 2)\ln 3\), so \(x(2\ln 7 - \ln 3) = \ln 7 + 2\ln 3\), that is \(x \ln\tfrac{49}{3} = \ln 63\), and \(x = \dfrac{\ln 63}{\ln(49/3)} \approx 1.4833\). Check: \(7^{1.9666} \approx 45.9\) and \(3^{3.4833} \approx 45.9\).

6. Solve \(\log_2(x + 2) + \log_2(x - 4) = 4\).

\(\log_2((x+2)(x-4)) = 4\) gives \((x + 2)(x - 4) = 16\), \(x^2 - 2x - 24 = 0\), \((x - 6)(x + 4) = 0\). \(x = -4\) makes \(x + 2 < 0\), rejected. \(x = 6\): \(\log_2 8 + \log_2 2 = 3 + 1 = 4\). Answer \(x = 6\).

7. Solve \(9^x - 4 \cdot 3^x + 3 = 0\).

\(u = 3^x\): \(u^2 - 4u + 3 = 0\), \((u - 1)(u - 3) = 0\). \(3^x = 1\) gives \(x = 0\); \(3^x = 3\) gives \(x = 1\). Both check: \(1 - 4 + 3 = 0\) and \(9 - 12 + 3 = 0\).

8. How many digits does \(3^{50}\) have, and what is its first digit?

\(\log_{10} 3^{50} = 50 \log_{10} 3 \approx 23.856\). So 24 digits, and \(10^{0.856} \approx 7.18\), first digit 7. (In fact \(3^{50} = 717\,897\,987\,691\,852\,588\,770\,249\).)

9. 500 dollars is invested at 3% a year. Compare yearly compounding for 5 years with continuous compounding.

Yearly: \(500 \times 1.03^5 \approx 579.64\) dollars. Continuous: \(500 e^{0.15} \approx 580.92\) dollars. The difference is about 1.28 dollars.

10. A concert is 110 dB and a conversation is 60 dB. How many times more intense is the concert?

The difference is 50 dB, that is 5 factors of 10: \(10^5 = 100\,000\) times the intensity.

11. Show that \(\log_2 3\) is irrational.

Suppose \(\log_2 3 = p/q\) with whole numbers \(p, q > 0\). Then \(2^{p/q} = 3\), so \(2^p = 3^q\). The left side is even and the right side is odd: a contradiction. (The same argument shows \(\log_a b\) is irrational whenever \(a\) and \(b\) are powers of different primes.)

12. Using Bernoulli's inequality, show that \((1 + 1/n)^{n+1}\) is decreasing, and conclude that \(e < (1 + 1/n)^{n+1}\) for every \(n\).

Let \(b_n = (1 + 1/n)^{n+1}\). Then \(\dfrac{b_{n-1}}{b_n} = \dfrac{(n/(n-1))^{n}}{((n+1)/n)^{n+1}} = \dfrac{n}{n+1}\left(\dfrac{n^2}{n^2 - 1}\right)^{n} = \dfrac{n}{n+1}\left(1 + \dfrac{1}{n^2 - 1}\right)^{n}\). By Bernoulli, \(\left(1 + \frac{1}{n^2 - 1}\right)^n \ge 1 + \frac{n}{n^2 - 1} > 1 + \frac1n = \frac{n+1}{n}\). So \(b_{n-1}/b_n > 1\): the sequence decreases. Since \(b_n = a_n (1 + 1/n)\) and \(1 + 1/n \to 1\), \(b_n \to e\) too, and a decreasing sequence stays above its limit. For example \(n = 1000\) gives \(2.71692 < e < 2.71964\).

What you learned¶

The meanings of \(a^0 = 1\), \(a^{-n} = 1/a^n\) and \(a^{m/n} = (\sqrt[n]{a})^m\) are forced by keeping the exponent laws; real exponents are defined as limits, for positive bases only.
Exponential functions multiply by a fixed factor at each step; doubling time and half-life describe them, and they eventually beat every power of \(x\).
\(\log_a y\) is the exponent that turns \(a\) into \(y\); it is the inverse function of \(a^x\) and is defined only for \(y > 0\).
The product, quotient and power laws of logarithms are the exponent laws read backwards; change of base shows all logarithms are multiples of one another.
To solve for an exponent, take logs; to solve inside a log, exponentiate; always check against the domain, and flip an inequality when dividing by a negative log.
\(e = \lim (1 + 1/n)^n \approx 2.71828\) exists because the sequence increases and stays below 3; it gives continuous growth \(Pe^{rt}\) and the natural logarithm \(\ln\).
Decibels, pH, earthquake and star magnitudes are logarithmic scales: equal steps mean equal factors.

Next: Chapter 8, sequences and sums, and the method of proof by induction.