Introduction
If you were told to "just memorize it" without any explanation when you learned this, it's no wonder it never quite clicked.
$$\log(AB) = \log A + \log B$$Why does multiplication suddenly become addition? Memorizing a rule and understanding why it works are two entirely different things.
In one sentence: the logarithm is the function that extracts the exponent, and in the world of exponents, multiplication is addition to begin with. In this article, we'll derive that reason step by step, starting directly from the laws of exponents.
Why This Matters — The Story of the Slide Rule
In the era before electronic calculators, engineers did multiplication with a slide rule. A slide rule is a tool where you slide two rulers against each other and read off the markings — and the markings aren't spaced evenly, they're engraved on a logarithmic scale.
The key idea: on a logarithmic scale, the lengths of two intervals are \(\log A\) and \(\log B\) respectively, so when you butt the rulers together and physically add the two lengths, you get \(\log A + \log B = \log(AB)\). In other words, one sliding motion is one multiplication.
Right up to the 1970s, this tool was actually used for Moon-landing calculations and aircraft design — and it was possible precisely because of this identity.
Step 1 — Start From the Laws of Exponents
First, let's look at exponents. In exponents, multiplication already behaves like addition.
$$10^2 \times 10^3 = 10^{2+3} = 10^5$$This is because \(10^2 = 10\times10\) and \(10^3 = 10\times10\times10\), so multiplying the two means multiplying 10 together five times. The exponents add up.
In general, for the same base \(a\):
$$a^p \times a^q = a^{p+q}$$This is everything in this article. The rest is just another way of writing this single line.
Step 2 — The Logarithm Is the Function That Extracts the Exponent
You only need to remember one thing — the definition of the logarithm.
If \(a^p = M\), then \(\log_a M = p\)
In other words, the logarithm is the function that asks "how many times must I raise a to a power to get M?" Feed an exponent in as the input and you get a number — that's the exponential function; feed a number in as the input and you get an exponent — that's the logarithmic function.
Let's build some intuition with a few examples. We'll fix the base at 10.
- \(10^2 = 100\), so \(\log_{10} 100 = 2\)
- \(10^3 = 1000\), so \(\log_{10} 1000 = 3\)
- \(10^{0.5} = \sqrt{10} \approx 3.162\), so \(\log_{10} 3.162 \approx 0.5\)
- \(10^0 = 1\), so \(\log_{10} 1 = 0\)
Computing a logarithm is complicated, but for now it's enough to hold on to just the definition.
Step 3 — Deriving the Rule Directly
Now let's show \(\log(AB) = \log A + \log B\) directly.
Let \(A\) and \(B\) be any positive numbers. We give each of their logarithm values a name.
$$\log_{10} A = p \qquad \Longleftrightarrow \qquad A = 10^p$$$$\log_{10} B = q \qquad \Longleftrightarrow \qquad B = 10^q$$Now let's compute \(A \times B\).
$$A \times B = 10^p \times 10^q = 10^{p+q}$$The last equality is the law of exponents from Step 1. Now take \(\log_{10}\) of both sides.
$$\log_{10}(A \times B) = \log_{10}(10^{p+q}) = p + q$$But since \(p = \log_{10} A\) and \(q = \log_{10} B\):
$$\boxed{\log_{10}(AB) = \log_{10} A + \log_{10} B}$$That's the end of the derivation. It was never "magic" — it was the law of exponents rewritten in the language of logarithms.
Division derives the same way. Since \(A \div B = 10^p \div 10^q = 10^{p-q}\), taking the logarithm of both sides gives \(\log(A \div B) = p - q\), that is:
$$\log(A \div B) = \log A - \log B$$The Power Rule, One Step at a Time Too
\(\log(A^n) = n\log A\) is likewise not a matter of "it just is" — we can show it without skipping steps, exactly the same way as the product rule. The starting point is the same too — \(A = 10^p\) (that is, \(p = \log_{10} A\)).
First, rewrite \(A^n\) in exponential form.
$$A^n = (10^p)^n$$Here \((10^p)^n = 10^{np}\), but this isn't a new rule — it comes straight out of \(a^p \times a^q = a^{p+q}\) from Step 1. After all, \((10^p)^n\) is just \(10^p\) multiplied by itself \(n\) times.
$$(10^p)^n = \underbrace{10^p \times 10^p \times \cdots \times 10^p}_{n\text{ factors}} = 10^{\,p + p + \cdots + p} = 10^{np}$$Adding \(p\) to itself \(n\) times in the exponent gives \(np\). Therefore
$$A^n = 10^{np}$$Now take \(\log_{10}\) of both sides.
$$\log_{10}(A^n) = \log_{10}\left(10^{np}\right) = np$$Finally, substitute \(p = \log_{10} A\) back in:
$$\boxed{\log_{10}(A^n) = n\log_{10} A}$$Check with numbers — let's try \(A = 100,\ n = 3\). Since \(\log 100 = 2\), the rule says \(\log(100^3) = 3 \times 2 = 6\). And indeed \(100^3 = 1{,}000{,}000 = 10^6\), so \(\log(10^6) = 6\) — an exact match. ✓
All three rules — multiplication, division, and powers — come from the single law of exponents.
Checking Directly With Numbers — 4 × 5 = 20
Once you've understood it abstractly, let's confirm it once more with numbers.
$$\log_{10} 4 \approx 0.602, \qquad \log_{10} 5 \approx 0.699$$According to the rule:
$$\log_{10} 4 + \log_{10} 5 \approx 0.602 + 0.699 = 1.301$$And indeed:
$$\log_{10} 20 \approx 1.301 \checkmark$$On a slide rule it works like this. You line up the "1" (\(=10^0\)) of the sliding ruler with the "4" position of the fixed ruler. Then you read the "5" position on the sliding ruler, and the fixed ruler points to "20." A single physical addition produced \(4 \times 5 = 20\).
Why You Keep Running Into It in Real Life
The reason this rule shows up so often in everyday life is that whenever "multiplicative growth" repeats in nature, the logarithm becomes the more intuitive language.
Decibels (dB): defined so that every tenfold increase in sound energy adds +10 dB.
$$\text{dB} = 10 \log_{10}\left(\frac{I}{I_0}\right)$$If the energy becomes 100 times larger, then since \(\log 100 = 2\) that's +20 dB; a thousandfold is +30 dB — being able to handle multiples as addition is convenient for engineers working with signals.
pH: represents the hydrogen ion concentration \([\text{H}^+]\) of an aqueous solution as a logarithm.
$$\text{pH} = -\log_{10}[\text{H}^+]$$When the ion concentration increases tenfold, the pH drops by 1. Because you can handle changes in concentration with addition and subtraction rather than multiplication, chemists use this scale.
Richter magnitude: earthquake energy is on a logarithmic scale too. A magnitude 6 quake is \(\sqrt{1000} \approx 31.6\) times stronger in energy than a magnitude 5.
All three fields use the same principle: "when magnitudes span an extremely wide range, handling them with logarithms turns the problem into addition and subtraction."
Try It Yourself
In the interactive below, adjusting A and B changes the lengths of the bars on the logarithmic number line. Check whether the sum of the blue bar (\(\log A\)) and the orange bar (\(\log B\)) lands exactly on the green dot (\(A \times B\)). The verification panel at the bottom of the screen lets you confirm the same thing numerically.
Press the preset "10×10 = 100" and you'll see it most intuitively: the blue (log 10 = 1.000) and orange (log 10 = 1.000) each fill exactly half, and their sum becomes 2.000 = log 100.
Key Takeaways
| Rule | Formula | Intuition |
|---|---|---|
| Product → sum | \(\log(AB) = \log A + \log B\) | the exponents add |
| Division → difference | \(\log(A/B) = \log A - \log B\) | the exponents subtract |
| Power → product | \(\log(A^n) = n\log A\) | the exponent is multiplied by n |
All three rules come from the single law of exponents \(a^p \times a^q = a^{p+q}\). If you're going to "memorize" logarithms, memorizing this one line is enough.
A word of caution:
$$\log(A + B) \neq \log A + \log B$$The logarithm of a sum does not split into addition. The logarithm rules work only on the "product," not on the "sum." Many people mistakenly believe the wrong equation above is correct, so be careful.
Studying This With AI
If you ask "prove that log(AB) = log A + log B," AI will usually derive it correctly. But small mistakes often show up in concrete numerical calculations. Used together with a tool like the interactive above that "computes it directly and shows the values of both methods at once," you can verify on the spot whether the numbers AI gives you are right.
Examples of useful questions to ask AI:
- "Why is the log transform used so often in machine learning?" (correcting distributional skew, converting a product of probabilities into a sum)
- "How do you decide between the natural logarithm and the common logarithm in practice?"
Wrapping Up
The logarithm is not a new rule. It's the law of exponents rewritten "in the direction of extracting the exponent." That's why multiplication becomes addition, and thanks to that, slide rules work and decibels and pH become intuitive units.
In the next article, we'll go on to look at why differentiation comes out cleanest precisely when the base of the exponential is \(e = 2.718\ldots\).
Related reading — What on Earth Is the Purpose of Trigonometric Functions · Why Does the Arctangent Get Tacked On · Why Does Cramer's Rule Have That Shape · Logarithm practice: Desmos Graphing Calculator · Wolfram Alpha