You put £500 in a savings account. Each year it grows by 6%. After one year you have £530. After two years, £561.80. After twenty years, something like £1604. That’s not 20 × £30 = £600 added — it’s the interest compounding on itself, quietly doubling your money while you sleep.
A phone battery starts fully charged. Every hour it drains by half. An hour in: 50%. Two hours: 25%. Three: 12.5%. That’s the same structure as the savings account, just running backward.
An earthquake measuring 6 on the Richter scale releases ten times more energy than one measuring 5, and a hundred times more than one measuring 4. The scale compresses a range from “barely felt” to “levels cities” into ten manageable numbers.
These are not three different ideas. They’re the same mathematical structure — exponential growth, exponential decay, and the logarithm that reads off what power was used.
22.1 What the notation is saying
22.1.1 The exponent laws
Start with what you already know: \(2^3\) means \(2 \times 2 \times 2 = 8\). The \(3\) is the exponent — it counts how many copies of 2 you multiply. The \(2\) is the base.
The notation \(a^x\) — read “a to the power x” — means the base \(a\) multiplied by itself \(x\) times. When you multiply two of these together with the same base, the exponents add:
\[a^m \times a^n = a^{m+n}\]
because you’re just running \(m\) copies followed by \(n\) copies. Four laws cover everything you need:
A negative exponent means “divide by” instead of “multiply by” — \(2^{-3} = \frac{1}{8}\). A fractional exponent means a root — \(8^{1/3} = \sqrt[3]{8} = 2\). These look odd until you see that they follow directly from the first law above: \(a^3 \times a^{-3} = a^0 = 1\), so \(a^{-3}\) must equal \(\frac{1}{a^3}\).
22.1.2 The exponential function
When the exponent is a variable, you have a function:
\[f(x) = a^x\]
Read it as: “the output is the base \(a\) raised to the power \(x\).”
If \(a > 1\): the function grows as \(x\) increases. Larger \(a\) means faster growth.
If \(0 < a < 1\): the function decays — same shape but reflected.
The y-intercept is always 1: \(f(0) = a^0 = 1\) for any base.
The battery example: \(f(t) = (0.5)^t\) gives the fraction of charge remaining after \(t\) hours. The savings example: \(f(t) = 1.06^t\) gives the growth factor after \(t\) years (multiply by the initial amount to get the actual balance).
22.1.3 The natural base \(e\)
You can build an exponential function with any base. But there is one base that is special.
Consider this question: if a quantity grows at a rate exactly equal to its current size — so the bigger it gets, the faster it grows, and the growth rate is always exactly the current amount — what is the base?
The answer is \(e \approx 2.718\). This number arises whenever a process is described by its own rate of change. You can’t deduce its exact value without calculus, but you can use it freely right now. The function \(f(x) = e^x\) is called the natural exponential. It obeys all the same exponent laws, with \(e\) as the base.
\(e\) is not arbitrary. It’s not a convenient approximation. When you reach differential calculus (Volume 5), you’ll see that \(e^x\) is the only function that is its own derivative — the only curve whose slope at every point equals its own height. That property is what makes it turn up everywhere in science and engineering.
Two special bases appear so often they have short names:
\(\log_{10}(x)\) is often written \(\log(x)\) — the common logarithm
\(\log_e(x)\) is written \(\ln(x)\) — the natural logarithm, read “ell-en of x”
Why this works
Logarithms are the inverse of exponentials the same way division undoes multiplication. If \(2^3 = 8\), then \(\log_2(8) = 3\). One statement, two ways of writing it. The exponential says “start with the power, get the result.” The logarithm says “start with the result, get the power.” They are the same relationship read in opposite directions.
Carbon-14 has a half-life of 5730 years — meaning every 5730 years, half the remaining carbon-14 decays. An organic sample contains 1.0 g of carbon-14 today. How much will remain after 17 190 years?
17 190 years = 3 × 5730, so three half-lives have passed.
where \(I\) is the intensity of the sound and \(I_0 = 10^{-12}\) W/m² is the reference intensity. The reference intensity \(I_0 = 10^{-12}\) W/m² is the quietest sound a human ear can detect — it is the standard baseline from which decibels are measured.
Normal conversation is about \(I = 10^{-6}\) W/m². What is that in dB?
A sound at 70 dB has what intensity ratio compared to 60 dB?
The difference is 10 dB, which corresponds to a factor of \(10^{10/10} = 10^1 = 10\). A 70 dB sound is 10 times as intense as a 60 dB sound. The logarithm compresses very large ratios into small, manageable numbers.
Example 4 — Computing: binary search step count.
Binary search finds a target in a sorted list by repeatedly halving the search space. How many steps to search a list of \(n = 1{,}000{,}000\) items?
After \(k\) steps, the search space has been halved \(k\) times, leaving at most \(n / 2^k\) items. The search ends when \(n / 2^k = 1\):
\[2^k = n \implies k = \log_2(n)\]
For one million items: \(k = \log_2(1{,}000{,}000) \approx 19.93\), so at most 20 steps. That’s why \(O(\log n)\) algorithms are fast — doubling the input adds only one more step.
22.4 Where this goes
The derivative of \(e^x\) is \(e^x\) — the only function that is its own slope. That property, which you can only properly state and prove in differential calculus (Volume 5), is why \(e\) appears in every model of continuous growth and decay. You already have the function; calculus will explain what makes it uniquely suited to describe rates of change.
From there, ordinary differential equations (Volume 7) make the connection concrete. The simplest ODE — “a quantity grows proportionally to itself” — has \(e^{kt}\) as its solution. Radioactive decay, population growth, cooling, charging capacitors: all of them are that one equation in different physical clothes. This chapter is the front end of that pipeline.
Where this shows up
A physicist uses exponential decay to date ancient organic material via carbon-14 ratios.
A financial analyst uses logarithms to find how long an investment takes to reach a target value.
An audio engineer specifies amplifier gain and speaker sensitivity in decibels — a logarithmic unit.
A software engineer analyses why binary search handles a billion records in about 30 comparisons, while linear search needs a billion.
The base changes. The structure does not.
22.5 Exercises
These are puzzles. Each has a clean answer. The interesting part is identifying whether you’re working with an exponential, a log, or both — and setting up the equation before you solve it.
1. Simplify, writing your answer as a single power:
\[\frac{3^8 \times 3^{-3}}{3^2}\]
Code
{const el =makeStepperHTML(1, [ { op:"Write as an equation",eq:"\\frac{3^8 \\times 3^{-3}}{3^2}",note:null }, { op:"Multiply in the numerator — add exponents",eq:"\\frac{3^{8 + (-3)}}{3^2} = \\frac{3^5}{3^2}",note:"Product law: same base, so add the exponents." }, { op:"Divide — subtract exponents",eq:"3^{5 - 2} = 3^3",note:"Quotient law: same base, so subtract the exponents." }, { op:"Evaluate",eq:"3^3 = 27",note:null }, { op:"Check",eq:"\\frac{3^8 \\times 3^{-3}}{3^2} = \\frac{6561 \\times (1/27)}{9} = \\frac{243}{9} = 27 \\checkmark",note:null }, ]);return el;}
2. Solve for \(x\):
\[4^x = 128\]
Code
{const el =makeStepperHTML(2, [ { op:"Write as an equation",eq:"4^x = 128",note:null }, { op:"Express both sides as powers of 2",eq:"(2^2)^x = 2^7",note:"4 = 2² and 128 = 2⁷, so rewrite with a common base." }, { op:"Use the power-of-a-power law",eq:"2^{2x} = 2^7",note:null }, { op:"Match exponents — same base, so the powers must be equal",eq:"2x = 7",note:null }, { op:"Divide both sides by 2",eq:"x = \\frac{7}{2} = 3.5",note:null }, { op:"Check",eq:"4^{3.5} = 4^3 \\times 4^{0.5} = 64 \\times 2 = 128 \\checkmark",note:null }, ]);return el;}
3. A population of bacteria doubles every 3 hours. You start with 400 bacteria. How many hours until there are more than 25 600?
Code
{const el =makeStepperHTML(3, [ { op:"Write as an equation",eq:"400 \\times 2^{t/3} > 25600",note:"Every 3 hours is one doubling cycle, so the exponent is t/3." }, { op:"Divide both sides by 400",eq:"2^{t/3} > 64",note:null }, { op:"Express 64 as a power of 2",eq:"2^{t/3} > 2^6",note:null }, { op:"Match exponents",eq:"\\frac{t}{3} > 6",note:null }, { op:"Multiply both sides by 3",eq:"t > 18 \\text{ hours}",note:null }, { op:"Check",eq:"400 \\times 2^{18/3} = 400 \\times 2^6 = 400 \\times 64 = 25600 \\checkmark \\text{ (exactly at 18 h, so >25600 means after 18 h)}",note:null }, ]);return el;}
4. Use logarithms to solve for \(t\):
\[5^t = 200\]
Give your answer to two decimal places.
Code
{const el =makeStepperHTML(4, [ { op:"Write as an equation",eq:"5^t = 200",note:null }, { op:"Take ln of both sides",eq:"\\ln(5^t) = \\ln(200)",note:"You can use log base 10 equally well — the ratio will be the same." }, { op:"Use the power law — bring t down",eq:"t \\cdot \\ln(5) = \\ln(200)",note:"\\ln(m^n) = n \\cdot \\ln(m)" }, { op:"Divide both sides by ln(5)",eq:"t = \\frac{\\ln(200)}{\\ln(5)} = \\frac{5.2983}{1.6094}",note:null }, { op:"Evaluate",eq:"t \\approx 3.29",note:null }, { op:"Check",eq:"5^{3.29} \\approx 5^3 \\times 5^{0.29} \\approx 125 \\times 1.60 \\approx 200 \\checkmark",note:null }, ]);return el;}
5. Solve the logarithmic equation:
\[\log_2(x+4) + \log_2(x-4) = 5\]
Code
{const el =makeStepperHTML(5, [ { op:"Write as an equation",eq:"\\log_2(x+4) + \\log_2(x-4) = 5",note:null }, { op:"Use the product law — combine the two logs",eq:"\\log_2\\bigl((x+4)(x-4)\\bigr) = 5",note:"\\log_a(m) + \\log_a(n) = \\log_a(mn)" }, { op:"Expand the bracket",eq:"\\log_2(x^2 - 16) = 5",note:"(x+4)(x-4) = x^2 - 16. This uses the difference of two squares shortcut: (a + b)(a − b) = a² − b². You may have seen this in Vol 3 Ch 3." }, { op:"Convert to exponential form",eq:"x^2 - 16 = 2^5 = 32",note:"\\log_2(k) = 5 \\text{ means } 2^5 = k" }, { op:"Solve the quadratic",eq:"x^2 = 48 \\implies x = \\pm\\sqrt{48} = \\pm 4\\sqrt{3} \\approx \\pm 6.93",note:null }, { op:"Reject the negative solution",eq:"x = 4\\sqrt{3} \\approx 6.93",note:"x = -4\\sqrt{3} gives \\log_2(-4\\sqrt{3}+4), which is log of a negative — undefined." }, { op:"Check",eq:"\\log_2(4\\sqrt{3}+4) + \\log_2(4\\sqrt{3}-4) = \\log_2\\bigl((4\\sqrt{3})^2 - 16\\bigr) = \\log_2(48-16) = \\log_2(32) = 5 \\checkmark",note:null }, ]);return el;}
6. The pH of a solution is defined as \(\text{pH} = -\log_{10}[\text{H}^+]\), where \([\text{H}^+]\) denotes the concentration of hydrogen ions in moles per litre — this is standard chemistry notation. Pure water has \([\text{H}^+] = 10^{-7}\) mol/L, giving a pH of 7. A lemon juice sample has \([\text{H}^+] = 10^{-2.4}\) mol/L. What is its pH, and how many times more acidic is it than pure water?
Code
{const el =makeStepperHTML(6, [ { op:"Write as an equation",eq:"\\text{pH} = -\\log_{10}(10^{-2.4})",note:null }, { op:"Evaluate the log",eq:"\\text{pH} = -(-2.4) = 2.4",note:"\\log_{10}(10^k) = k" }, { op:"Find the ratio of hydrogen ion concentrations",eq:"\\frac{[\\text{H}^+]_{\\text{lemon}}}{[\\text{H}^+]_{\\text{water}}} = \\frac{10^{-2.4}}{10^{-7}}",note:null }, { op:"Subtract exponents — dividing with the same base",eq:"= 10^{-2.4 - (-7)} = 10^{4.6}",note:null }, { op:"Evaluate",eq:"10^{4.6} \\approx 39\\,811",note:null }, { op:"Check",eq:"\\text{pH of lemon} = 2.4;\\ 10^{7 - 2.4} = 10^{4.6} \\approx 39\\,811 \\text{ times more acidic than water} \\checkmark",note:null }, ]);return el;}