Numbers & Functions · 5 Exponentials and Logarithms
5.8 The most important question: which base?
So far we’ve doubled — base 2. But nothing forced that. Every base gives a perfectly good exponential. So here’s a question that cracks open the rest of mathematics: Is there a best base — one that makes the machine behave most simply?
To even ask that we need to say what “simply” means, and the honest answer needs calculus. But I can show you the shape of it. Every exponential curve bx climbs, and at each point it has a steepness. An exponential is special: its steepness is always proportional to its own height. For base 2 the steepness is a bit less than the height; for base 3 a bit more. So somewhere between 2 and 3 there must be a magic base where the steepness is exactly equal to the height. That magic base is an irrational number, a little past 2.7: e = 2.71828 . . .
It is called Euler’s number, and it is one of the two or three most important numbers in all of mathematics, sharing that pedestal with π. The function ex is the purest possible growth machine — the one whose growth rate, at every point, equals itself. Its inverse, loge, gets its own symbol, ln (for natural logarithm). I can’t fully justify e until we have derivatives — but when you meet it again, you’ll already know what it is: the number that was hiding between 2 and 3, waiting for the question “which base is best?”
5.9 Where we are now
We swapped the roles in a power and got the exponential machines, the mathematics of everything that grows or decays on its own terms. We met the runaway x 2x ex 3x tangents at (0, 1): slope < 1 slope = 1 slope > 1 e is the base whose steepness equals its height Figure 5.7. Finding e. At (0, 1), the tangent to 2x is too shallow (slope < 1) and to 3x too steep (slope > 1). The base whose tangent has slope exactly 1 is e ≈2.718 — the curve whose steepness equals its height.
personality of exponential growth and its mirror, decay. We ran the machine backwards and found the logarithm — an exponent in disguise, the tool that flattens multiplication into addition. And by asking “which base is best?” we found e ≈2.718, the base whose growth rate exactly equals itself. We’ve now got two numbers (π and e) and two big promises waiting for calculus. Next we meet machines that go around.
Things worth thinking about 1. We found 21/2 = √ 2. Using the same logic, what should 21/3 be? (What would you multiply together three times to get 2?) Connect this to the cube roots from Chapter 2.
2. The graph of 2x gets closer and closer to zero on the left but never touches it. Why can 2x never actually equal zero, no matter how negative x gets? (Can multiplying and dividing 2s ever land you exactly on nothing?) 3. A logarithm turns multiplication into addition. Before calculators, why would that single property have been worth its weight in gold to a sailor navigating by the stars?
4. (Looking ahead.) We claimed there’s a base e where the curve’s steepness Exponentials and Logarithms exactly equals its height — where the richer it gets, the faster it gets richer, in perfect proportion. Can you think of something in the real world that behaves like that?
Problems to work Problem 1 — Forced meanings.
Using only the rule that exponents add (2a · 2b = 2a+b), work out what each must equal. Reason it out, don’t memorize. (a) 20 (hint: 23 · 20 must equal 23); (b) 2−2 (hint: 22 · 2−2 must equal 1); (c) 21/2 (hint: 21/2 · 21/2 must equal 2). Problem 2 — Simplify with the exponent laws.
Write each as a single power of the base, naming the law(s) used. (a) a4 · a7; (b) a9 a5 ; (c) (a3)4; (d) a2·a6 a3 ; (e) a−3 · a5; (f) (2a2)3.
Problem 3 — Translate between the two languages.
Rewrite each in the other language.
(a) 34 = 81 as a logarithm; (b) log10(1000) = 3 as an exponential; (c) log2 1 4 = −2 as an exponential.
Problem 4 — Read the log as a question.
Answer each by asking “the base, raised to what, gives this?” — no calculator. (a) log2(16); (b) log3(9); (c) log5(1); (d) log2 1 8 .
Problem 5 — Use the log laws.
(a) Expand loga(7x) into a sum. (b) Expand loga x 5 into a difference. (c) Rewrite loga(x3) as a multiple of loga x. (d) Combine loga x + loga y −loga z into a single logarithm. (e) (The payoff.) Solve 2x = 50. (Hint: take log of both sides, then use Law 3 to pull the x down. Leave your answer as a ratio of logs.) ⋆Teach it — explain it to someone Problem 6 — Teach it. Explain to someone the difference between x2 and 2x — why they’re completely different machines even though both have a 2 and an x. Get them to predict which is bigger at x = 10 (102 = 100, but 210 = 1024), then at x = 2 (equal — both 4!). The fact that they cross over is the whole story of why exponential growth eventually wins. Problem 7 — Compounding, and its ceiling.
$1000 sits in an account multiplying by 1.05 each year. (a) Write the exponential function for the balance after x years. (b) Roughly how much after 1 year? After 2? (Notice the second year’s gain is bigger — interest earning interest.) (c) The formula says after 300 years the balance is over $2 billion. Explain why no real deposit has ever done this, and sketch what the balance actually would look like. What is the “ceiling,” and what makes it bite?
Going further ⋆Teach it — explain it to someone Problem 8 — Teach it (the sigmoid decision). Explain to a family member how the curve σ(x) = 1 1+e−x “makes a decision.”
Walk them through three inputs: a big positive number (answer near 1 = “yes”), a big negative one (near 0 = “no”), and zero (exactly 1 2 = “unsure”). Then drop the punchline: this exact curve is how a single artificial neuron decides, and it’s the same shape that makes an epidemic level off. If they see one curve doing both jobs, you’ve taught the chapter’s deepest idea.
Problem 9 — [challenge] Half-life, backwards.
A radioactive sample halves every 10 years: the amount left after t years is A(t) = 100 · (1 2)t/10 grams. (a) How much is left after 30 years — without a calculator? (b) Now the hard direction: after how many years is only 25 grams left? Set A(t) = 25 and solve, thinking in “how many halvings?” (c) Write the exact answer for “when is only 10 grams left” using a logarithm. (This is exactly how carbon dating works.)
Problem 10 — [challenge] The log scale’s superpower.
An earthquake’s energy grows by a factor of about 32 for each whole step up the Richter scale. (a) How many times more energy does a magnitude-7 quake release than a magnitude-5? (b) Explain, using log(xy) = log x + log y, why turning “multiply energy by 32” into “add 1 to the magnitude” is what makes the scale usable by humans. (c) Why would a linear energy scale — plain joules — be hopeless for comparing earthquakes you read about in the news? Problem 11 — [challenge] Stacked exponents.
Which is bigger, 2(32) or (23)2? Compute both. (The lesson: exponent towers are read top-down, and (am)n = amn is not the same as a(mn) — a trap worth feeling once with your own hands.)
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