22  Exponents and logarithms

Growth, decay, and the scale of things

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 times $30, or $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 this chapter helps you do

Symbols to keep handy

These are the bits of notation you'll see a lot. If a line of symbols feels like a fence, read it out loud once, then keep going.

  • log_a(x): log base a of x — the power you raise a to in order to get x

  • eˣ: e to the power x — the natural exponential

  • ln(x): the natural logarithm of x — log base e

  • aˣ: a to the power x — a multiplied by itself x times

  • log(x): common logarithm — log base 10; the subscript 10 is omitted by convention

Here is the main move this chapter is making, in plain terms. You do not need to be fast. You just need to keep the thread.

  • Coming in: You already know that repeated multiplication produces patterns. Some of those patterns grow fast, others shrink fast, and you need a way to describe them precisely — and undo them.

  • Leaving with: An exponential function describes any quantity that grows or shrinks by a constant factor. A logarithm is the operation that undoes that — it asks “what power was used?” Knowing both, you can model the world’s fastest processes and solve for time.

22.2 What the notation is saying

22.2.1 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.

This means exponent notation is a compact way to describe repeated multiplication.

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:

Law What it says Example
a^m \times a^n = a^{m+n} Multiplying: add exponents 2^3 \times 2^4 = 2^7 = 128
a^m \div a^n = a^{m-n} Dividing: subtract exponents 2^5 \div 2^2 = 2^3 = 8
(a^m)^n = a^{mn} Power of a power: multiply exponents (2^3)^2 = 2^6 = 64
a^0 = 1 Zero power: always one 17^0 = 1

Two more for completeness:

a^{-n} = \frac{1}{a^n} \qquad a^{1/n} = \sqrt[n]{a}

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}.

This means the exponent laws extend the same pattern instead of inventing separate rules.

22.2.2 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.

This means the base controls whether the function grows or decays.

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).

This means exponentials model change by repeated scaling, not repeated addition.

22.2.3 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.

This means the special role of e comes from how naturally it matches continuous change.

22.3 The method

22.3.1 Part 1: Working with exponent laws

The laws in the table above are the full toolkit. A typical simplification chains two or three of them together.

Simplify \dfrac{x^5 \cdot x^{-2}}{x^3}.

Top: x^5 \cdot x^{-2} = x^{5 + (-2)} = x^3.

Then: x^3 \div x^3 = x^{3-3} = x^0 = 1.

This means long-looking exponent expressions often collapse quickly once the base is the same.

Simplify (2a^3)^4.

The power applies to everything inside the bracket: (2)^4 \cdot (a^3)^4 = 16 \cdot a^{12} = 16a^{12}.

This means brackets matter: the outside power affects every factor inside.

22.3.2 Part 2: Solving exponential equations by inspection

Some equations can be solved by rewriting both sides as powers of the same base.

Solve 2^x = 16.

Rewrite 16 as a power of 2: 16 = 2^4.

So 2^x = 2^4, which means x = 4.

Solve 9^x = 27.

Both 9 and 27 are powers of 3: 9 = 3^2 and 27 = 3^3.

So (3^2)^x = 3^3, giving 3^{2x} = 3^3, so 2x = 3, and x = \frac{3}{2}.

This only works when you can express both sides with the same base. For 2^x = 10, you can’t. That’s where logarithms come in.

This means logarithms appear when exponentials cannot be solved by inspection.

22.3.3 Part 3: Logarithms

A logarithm is the inverse of an exponential. It answers the question: “what power do I need?”

\log_a(b) = c \quad \text{means} \quad a^c = b

Read it aloud: “log base a of b equals c” means “a raised to the power c gives b.”

Three examples to fix this:

\log_2(8) = 3 \quad \text{because} \quad 2^3 = 8

\log_{10}(1000) = 3 \quad \text{because} \quad 10^3 = 1000

\log_3(9) = 2 \quad \text{because} \quad 3^2 = 9

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”

This means a logarithm always converts an output back into the power that produced it.

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.

22.3.4 Part 4: Log laws and solving logarithmic equations

Logarithms inherit structure from the exponent laws. The three log laws:

Log law What it says Example
\log_a(mn) = \log_a(m) + \log_a(n) Log of a product: add logs \log_2(4 \times 8) = \log_2(4) + \log_2(8) = 2 + 3 = 5
\log_a\!\left(\dfrac{m}{n}\right) = \log_a(m) - \log_a(n) Log of a quotient: subtract logs \log_{10}(1000/10) = 3 - 1 = 2
\log_a(m^n) = n \cdot \log_a(m) Log of a power: bring down the exponent \log_2(8^4) = 4 \cdot \log_2(8) = 4 \times 3 = 12

One more, needed for calculators — the change of base formula:

\log_a(x) = \frac{\ln(x)}{\ln(a)} = \frac{\log(x)}{\log(a)}

Your calculator has \ln and \log_{10} buttons. This formula lets you compute any log from them.

Solving 2^x = 10.

Take \log_{10} (or \ln) of both sides:

\log(2^x) = \log(10)

Use the power law to bring x down:

x \cdot \log(2) = 1

Divide:

x = \frac{1}{\log(2)} = \frac{1}{0.3010\ldots} \approx 3.32

Check: 2^{3.32} \approx 9.97 \approx 10. Yes.

This means logarithms let us solve for the exponent itself.

Solving \log_3(x) + \log_3(x - 2) = 1.

Use the product law to combine the left side:

\log_3\bigl(x(x-2)\bigr) = 1

Convert back to exponential form (3^1 = x(x-2)):

x^2 - 2x = 3

x^2 - 2x - 3 = 0

(x-3)(x+1) = 0

So x = 3 or x = -1. But \log_3(-1) is not defined — you can’t take the log of a negative number. Discard x = -1.

Answer: x = 3. Check: \log_3(3) + \log_3(1) = 1 + 0 = 1. Yes.

This means every logarithmic solution must also pass the domain check.

22.4 Worked examples

Example 1 — Science: radioactive half-life.

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.

\text{Remaining} = 1.0 \times \left(\frac{1}{2}\right)^3 = \frac{1}{8} = 0.125 \text{ g}

If instead you need to find when the sample drops to 0.10 g, set up the equation:

0.10 = 1.0 \times \left(\frac{1}{2}\right)^{t/5730}

Divide both sides by 1.0 to isolate the exponential:

0.10 = (0.5)^{t/5730}

Take \ln of both sides:

\ln(0.10) = \ln\!\left((0.5)^{t/5730}\right)

Apply the power law — bring the exponent down:

\ln(0.10) = \frac{t}{5730} \cdot \ln(0.5)

t = 5730 \times \frac{\ln(0.10)}{\ln(0.5)} = 5730 \times \frac{-2.3026}{-0.6931} \approx 5730 \times 3.322 \approx 19\,030 \text{ years}

This means exponential decay questions often become logarithm questions when time is the unknown.

Example 2 — Finance: compound interest doubling time.

An investment earns 7% per year, compounded annually. How long until it doubles?

Let the initial amount be P. You want the time t when P \times 1.07^t = 2P.

Divide both sides by P:

1.07^t = 2

Take \ln of both sides and bring t down:

t \cdot \ln(1.07) = \ln(2)

t = \frac{\ln(2)}{\ln(1.07)} = \frac{0.6931}{0.06766} \approx 10.2 \text{ years}

The Rule of 72 is a quick mental estimate: divide 72 by the interest rate. 72 / 7 \approx 10.3 — very close. That rule is a log approximation.

This means doubling time depends on the growth rate through a logarithm, not a simple proportion.

Example 3 — Engineering: the decibel.

Sound intensity is measured on a logarithmic scale. The decibel (dB) level is:

L = 10 \cdot \log_{10}\!\left(\frac{I}{I_0}\right)

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?

L = 10 \cdot \log_{10}\!\left(\frac{10^{-6}}{10^{-12}}\right) = 10 \cdot \log_{10}(10^6) = 10 \times 6 = 60 \text{ 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.

This means logarithmic scales are useful when raw ratios grow too quickly to handle comfortably.

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.

This means logarithms measure how many repeated halvings or repeated multiplications are needed.

22.5 Where this goes

This chapter connects repeated multiplication, growth, decay, and the inverse question “what power was used?”

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.

22.6 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}

2. Solve for x:

4^x = 128

3. A population of bacteria doubles every 3 hours. You start with 400 bacteria. How many hours until there are more than 25 600?

4. Use logarithms to solve for t:

5^t = 200

Give your answer to two decimal places.

5. Solve the logarithmic equation:

\log_2(x+4) + \log_2(x-4) = 5

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?