A logarithm table can give the correct significant digits while your final answer is ten or a hundred times too large. The missing safeguard is the characteristic: the integer that carries the power of ten. These worked examples using logarithm tables separate the table lookup from the place-value decision, then put them together before taking an antilogarithm.
Throughout this lesson, log means base ten, all original inputs are positive, and ≈ means approximately equal. Four-figure logarithm entries are rounded to four decimal places. The small tables below are author-computed teaching extracts, not reproductions of a commercial table. They provide the entries needed here; they are not a complete examination table.
Characteristic and mantissa do different jobs
Write a positive number as a × 10n, with 1 ≤ a < 10 and integer n. Then log(a × 10n) = n + log a. The characteristic is n; the mantissa is log a, between zero and one. For example, 320 = 3.20 × 102, so log 320 ≈ 2 + 0.5051 = 2.5051. Moving the decimal point changes the characteristic but leaves the mantissa unchanged.
The coefficient a is sometimes loosely called a mantissa in scientific-notation contexts. Here, reserve mantissa for the fractional part of the logarithm. The number 3.20 and the logarithmic mantissa 0.5051 are different things.
| Number | Power-of-ten form | Logarithm, approximately |
|---|---|---|
| 32.0 | 3.20 × 101 | 1 + 0.5051 |
| 3.20 | 3.20 × 100 | 0 + 0.5051 |
| 0.320 | 3.20 × 10−1 | −1 + 0.5051 |
| 0.00320 | 3.20 × 10−3 | −3 + 0.5051 |
A bar characteristic is shorthand for a negative integer followed by a positive mantissa. “Bar 3 point 5051” means −3 + 0.5051, which equals −2.4949. It does not mean −3.5051. Keeping the explicit sum is a reliable alternative whenever an overbar might be lost in handwriting or on a screen.
Read the digits, then restore the scale
In this miniature lookup grid, the row gives the first two significant digits and the column gives the third. Row 12, column 5 represents 1.25 and returns 0.0969. For 125, attach characteristic 2; for 0.0125, attach characteristic −2. Neither the original decimal point nor the number of written zeros selects the table cell.
| First two digits | Third digit 0 | Third digit 5 | Third digit 8 |
|---|---|---|---|
| 12 | 0.0792 | 0.0969 | 0.1072 |
| 24 | 0.3802 | 0.3892 | 0.3945 |
| 32 | 0.5051 | 0.5119 | 0.5159 |
Full tables may include mean differences for an additional significant digit. Follow that table’s printed headings; do not invent an extra column or assume all layouts match. Our examples deliberately use entries supplied directly, so there is no hidden interpolation step.
| Coefficient a | log a, to four decimal places |
|---|---|
| 1.44 | 0.1584 |
| 1.80 | 0.2553 |
| 2.00 | 0.3010 |
| 2.50 | 0.3979 |
| 4.00 | 0.6021 |
| 4.80 | 0.6812 |
| 5.60 | 0.7482 |
| 6.40 | 0.8062 |
Worked case 1: a product and its antilogarithm
Evaluate P = 24.0 × 3.20 using the supplied logarithms. First estimate: 24 × 3 is 72, so the answer should be between 70 and 80. Multiplication becomes addition of logarithms:
log P = log 24.0 + log 3.20 ≈ 1.3802 + 0.5051 = 1.8853.
The characteristic 1 tells us that P lies between 10 and 100. An antilogarithm reverses a logarithm: the coefficient corresponding to mantissa 0.8853 is approximately 7.679. Therefore P ≈ 7.679 × 101 = 76.79. To three significant figures this is 76.8, matching direct multiplication. Writing 7.679 would retain the digits but discard the characteristic.
Worked case 2: a quotient below one
Evaluate Q = 0.0640 ÷ 2.50. Dividing a positive number by 2.50 must make it smaller, so Q is below 0.0640. Because 0.0640 = 6.40 × 10−2, its logarithm is approximately −2 + 0.8062.
log Q ≈ (−2 + 0.8062) − 0.3979 = −2 + 0.4083.
The antilogarithm of mantissa 0.4083 is approximately 2.560. Restore the characteristic to obtain Q ≈ 2.560 × 10−2 = 0.02560. Direct division gives 0.0256. An answer of 0.256 is implausible because it is larger than the dividend.
Worked case 3: divide the whole logarithm for a root
Find R = √0.0144 using logarithms. The number is between 0.01 and 0.04, so its positive square root is between 0.1 and 0.2. We have log 0.0144 ≈ −2 + 0.1584. A square root halves the entire logarithm:
log R ≈ ½(−2 + 0.1584) = −1 + 0.0792.
The lookup grid identifies mantissa 0.0792 with coefficient 1.20. Therefore R ≈ 1.20 × 10−1 = 0.120. Squaring 0.120 gives 0.0144. Halving only the mantissa while keeping characteristic −2 would give 0.0120, whose square is a hundred times too small.
If a characteristic does not divide conveniently, rewrite before division. For a cube root, −2 + m can be written −3 + (1 + m). This keeps the value unchanged and makes the characteristic divisible by three.
Four practice tasks with solved checks
Practice 1: multiply 18.0 by 4.00
log P ≈ 1.2553 + 0.6021 = 1.8574. The corresponding coefficient is approximately 7.201, so P ≈ 72.01, or 72.0 to three significant figures. The exact product is 72. The small difference is table rounding, not a failure of the logarithm law.
Practice 2: divide 0.0480 by 1.20
log Q ≈ (−2 + 0.6812) − 0.0792 = −2 + 0.6020. The corresponding coefficient is approximately 3.999, giving Q ≈ 0.03999, or 0.0400 to three significant figures. Direct division gives 0.04; multiplying that result by 1.20 restores 0.0480.
Practice 3: find the cube root of 0.00800
Since log 8 = 3 log 2 ≈ 0.9030 from the rounded entry, log 0.00800 ≈ −3 + 0.9030. Divide by three: log R ≈ −1 + 0.3010. Thus R ≈ 2.00 × 10−1 = 0.200. Check: 0.200 cubed equals 0.00800. Using a direct four-figure entry for log 8 would introduce a slightly different intermediate rounding, with the same three-significant-figure answer.
Practice 4: interpret a negative characteristic
A positive number N has log N ≈ −3 + 0.7482. The table gives coefficient 5.60, so N ≈ 0.00560. Its size lies between 0.001 and 0.01, exactly as characteristic −3 requires. Changing the expression to −3.7482 would change the given logarithm and solve a different problem.
Logarithmic equations: restrict, solve and check
The table calculations above start with known positive inputs. In an equation, the input may contain an unknown. First require every logarithm argument to be positive. Here log still means base ten. For positive A and B, log A + log B = log(AB), log A − log B = log(A/B), and log A = c means A = 10c. These are exact relationships; no rounded table entry is needed for the equations below.
Worked equation: a product produces two algebraic candidates
Solve log(x − 1) + log(x − 2) = log 6. The original expressions require x − 1 > 0 and x − 2 > 0, so x > 2. Combining the logs gives log[(x − 1)(x − 2)] = log 6. Since a logarithm to a fixed valid base is one-to-one, (x − 1)(x − 2) = 6. Expand: x² − 3x + 2 = 6, hence (x − 4)(x + 1) = 0.
The algebraic candidates are 4 and −1. Only x = 4 belongs to x > 2. In the original equation, log 3 + log 2 = log 6. At x = −1 the original arguments would be −2 and −3, whose real logarithms do not exist. Their product is positive, but that cannot restore the two original logarithms. Keep the domain from before combining.
Worked equation: convert a logarithm to an exponential statement
Solve log(x + 3) − log x = 1. The domain is x > 0. The quotient law gives log[(x + 3)/x] = 1, so (x + 3)/x = 10. Since x is positive, multiplying by x gives x + 3 = 10x. Therefore x = 1/3. Check: [(1/3) + 3]/(1/3) = 10, so the original difference is log 10 = 1. An exact fraction is appropriate; a rounded decimal is unnecessary.
Independent equation practice
- Solve log(x − 2) + log(x − 3) = log 12. State the original domain and test every algebraic candidate.
- Solve log(x + 8) − log x = 1. Explain why the denominator is permitted.
Explained answers
- The domain is x > 3. The product equation is x² − 5x + 6 = 12, so (x − 6)(x + 1) = 0. Reject −1; x = 6 works because log 4 + log 3 = log 12.
- The domain is x > 0, so division by x is valid. (x + 8)/x = 10 gives 8 = 9x and x = 8/9. The quotient is exactly 10 at that value.
Do not turn a log of a sum into a sum of logs
log(x + 3) is not log x + log 3. At x = 1 those would be log 4 and log 3, which differ. Likewise, cancelling the word “log” from a sum is invalid: first apply the product law with positive inputs. A complete solution records the domain, the transformed equation, all algebraic candidates and the candidates that satisfy the original equation.
Frequently asked questions
Can the mantissa be negative in this method?
The conventional table split uses a nonnegative mantissa below one. A calculator may show a negative decimal logarithm; rewrite it as a negative integer plus a positive fractional part.
What is the characteristic of log 1?
It is zero, and the mantissa is also zero, because 10 to power zero equals one.
Can I take the logarithm of zero?
No real base-ten logarithm exists for zero or a negative input. Check the argument before any table lookup.
Does a logarithm table give exact answers?
Usually not. Finite printed entries are rounded, so subsequent arithmetic and inverse lookup give approximations. Keep that distinction visible with ≈.
Why can I not add logarithms for a sum?
The addition rule represents multiplication of inputs. Log(A + B) is generally different from log A + log B.
How many digits should I report?
Follow the requested accuracy. Avoid reporting more precision than the table and inputs justify, and do not round intermediate results repeatedly.
What should I check before accepting an answer?
Check its sign, power-of-ten range and rough size, then reverse the original operation where practical.
Continue with a focused revision task
Practise separating significant digits from their power of ten before mixing products, quotients and roots. The existing Mathematics topical revision collection is a route to further revision resources; access to its contents may require membership. No particular downloadable logarithm file is assumed here.
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