Before calculators, every logarithm was found by hand, and the methods still work. Most logs can be found from three things: the powers you already know, the change-of-base rule, and a handful of memorised values. The powers give the exact logs of round numbers. The memorised values — log₁₀(2) ≈ 0.3010, log₁₀(3) ≈ 0.4771, ln(2) ≈ 0.6931, ln(10) ≈ 2.3026 — fill the gaps between them. And the rules of logarithms turn any number into a combination of those pieces. This guide works through the methods with examples, and shows how to check an answer with the antilog.

Start with powers you know

Every exact logarithm by hand starts from a power you already know. In base 10: log₁₀(10) = 1, log₁₀(100) = 2, log₁₀(1000) = 3, log₁₀(0.1) = −1, log₁₀(0.01) = −2. In base 2: log₂(2) = 1, log₂(4) = 2, log₂(8) = 3, log₂(32) = 5, log₂(1024) = 10. In base e: ln(e) = 1, ln(e²) = 2, ln(1) = 0. And in any base, the log of the base itself is 1, and the log of 1 is 0.

Whole powers of the base are the anchors, and everything else is found by moving between them. The number 500 sits between 100 and 1,000, so its base-10 log sits between 2 and 3. The number 50 sits between 10 and 100, so its log sits between 1 and 2. The anchors alone give the answer to one significant figure, and the methods below refine it.

The handful of values worth memorising

Four values cover most hand calculations. In base 10: log₁₀(2) ≈ 0.3010 and log₁₀(3) ≈ 0.4771. In base e: ln(2) ≈ 0.6931 and ln(10) ≈ 2.3026. A fifth is useful for base 2: log₂(10) ≈ 3.3219. These are the values worth memorising, because the rules of logarithms build almost everything else from them.

From log₁₀(2) ≈ 0.3010 alone, the log of 4 follows by the power rule: log₁₀(4) = 2 × 0.3010 = 0.6020. The log of 5 follows by the quotient rule: log₁₀(5) = log₁₀(10) − log₁₀(2) = 1 − 0.3010 = 0.6990. The log of 8 follows by the power rule again: log₁₀(8) = 3 × 0.3010 = 0.9030. With log₁₀(3) ≈ 0.4771 added, the log of 6 follows by the product rule: log₁₀(6) = log₁₀(2) + log₁₀(3) = 0.3010 + 0.4771 = 0.7781, and the log of 9 is 2 × 0.4771 = 0.9542. Two memorised values give the logs of every whole number from 1 to 10.

Splitting numbers into factors

The working method is to split the number into factors whose logs you know, then add. To find log₁₀(50): split 50 into 5 × 10, and add log₁₀(5) ≈ 0.6990 and log₁₀(10) = 1, giving 1.6990. To find log₁₀(500): split 500 into 5 × 100, and add 0.6990 + 2 = 2.6990. To find log₁₀(20): split 20 into 2 × 10, and add 0.3010 + 1 = 1.3010. To find log₁₀(200): split 200 into 2 × 100, and add 0.3010 + 2 = 2.3010.

The same method works in other bases. To find log₂(48): split 48 into 16 × 3, and add log₂(16) = 4 and log₂(3) ≈ 1.5850, giving 5.5850 — and the check confirms it, since 2⁵·⁵⁸⁵ is about 48. To find ln(50): split 50 into 5 × 10, and add ln(5) ≈ 1.6094 and ln(10) ≈ 2.3026, giving 3.9120. The skill is not arithmetic; it is choosing the split. Look for factors of 2, 3, 5, and 10, because those are the logs you know.

Estimating by bracketing

When a number will not split neatly, bracket it between two powers and refine. To estimate log₁₀(50): 50 sits between 10¹ = 10 and 10² = 100, so the log sits between 1 and 2. It is closer to 100 than to 10 in the multiplicative sense — 50 is five times 10 but only twice short of 100 — so the log is closer to 2 than to 1: about 1.7. The exact value is 1.6990, and 10¹·⁷ is about 50, which confirms the estimate.

Bracketing also catches errors. If a calculation gives log₁₀(50) = 0.7, the bracket says it must be between 1 and 2, so the answer is wrong — and indeed 0.6990 is log₁₀(5), not log₁₀(50), a factor-of-ten slip. The bracket is the fastest check in hand calculation, and it should be applied before the arithmetic, not after it: know roughly where the answer sits, then compute, then compare.

Logs of numbers between 0 and 1

Numbers between 0 and 1 have negative logarithms, and the same methods apply. log₁₀(0.01) = −2, because 10⁻² = 0.01. log₁₀(0.5) = −log₁₀(2) = −0.3010, because 0.5 is 1 ÷ 2. log₁₀(0.05) = log₁₀(5) − log₁₀(100) = 0.6990 − 2 = −1.3010. In natural logs, ln(0.5) = −ln(2) = −0.6931.

The sign is the whole story: a negative log means the number is below 1, and the size of the negative number says how far below. A log of −1 means a tenth, −2 means a hundredth, −3 means a thousandth. The pH scale uses exactly this: a pH of 7 means a hydrogen ion concentration of 10⁻⁷, and the minus sign in pH = −log₁₀[H⁺] is what turns that negative log into a positive pH.

Checking with the antilog

Every hand logarithm should be checked with the antilog: raise the base to the answer and see whether the original number comes back. If log₁₀(50) ≈ 1.6990, then 10¹·⁶⁹⁹ should be about 50 — and it is, 49.99 to two decimal places. If log₂(48) ≈ 5.5850, then 2⁵·⁵⁸⁵ should be about 48 — and it is.

The check is approximate by design: a hand logarithm is usually right to three or four decimal places, and the antilog will land within a fraction of a percent of the original number. If it lands far away — if the antilog of your answer is 5 when the number was 50 — the error is a factor of ten, and the bracket step above will show where it entered. The companion guide on logarithms explained gives the rules and the bases this method rests on.

When to reach for the calculator

Hand methods cover round numbers and the handful of memorised values, and they give most logs to three or four decimal places. They do not cover awkward numbers: log₁₀(47) will not split into known factors, and bracketing it between 10 and 100 only says it is between 1 and 2, closer to 2. For such a number, the honest answer is that hand methods give an estimate and the calculator gives the value: log₁₀(47) ≈ 1.6721.

The right division of labour is to do the easy logs by hand, estimate the awkward ones by bracketing, and use the calculator for the exact value — then check the calculator’s answer against the bracket. A calculator that says log₁₀(47) = 2.67 is wrong by a factor of ten, and the bracket catches it. The hand methods are not a replacement for the calculator; they are the understanding that makes the calculator’s answer meaningful, and the check that catches it when it is not.

Sources: Logarithm rules: product, quotient, and power rules; change of base. Memorised values: log₁₀(2) ≈ 0.3010, log₁₀(3) ≈ 0.4771, ln(2) ≈ 0.6931, ln(10) ≈ 2.3026, log₂(10) ≈ 3.3219. Worked values verified by antilog: 10¹·⁶⁹⁹ ≈ 50, 10²·⁶⁹⁹ ≈ 500, 2⁵·⁵⁸⁵ ≈ 48.