A logarithm is an exponent. That is the whole idea
Logarithm — If ax = y with a > 0 and a ≠ 1, then loga y = x. In words: the logarithm of y to base a is the power to which a must be raised to give y.
If you can read 2⁵ = 32, you already understand logarithms. The equation says three things at once: the base is 2, the exponent is 5, and the result is 32. A logarithm just asks for a different one of the three.
Written as a power we fix the base and the exponent and ask for the result. Written as a logarithm we fix the base and the result and ask for the exponent. So log₂ 32 = 5 and 2⁵ = 32 are the same sentence read from opposite ends.
Why the base cannot be 1, and why y cannot be negative
The two conditions in that definition are not decoration; the paper asks about them.
The base cannot be 1 because 1x = 1 for every x, so log₁ 5 would be asking for a power of 1 that gives 5 — there is none. The base cannot be negative or zero either, because the powers would not behave.
And y must be positive: a positive base raised to any real power is always positive, so no exponent can ever produce a negative answer. log₁₀(−4) does not exist in the real numbers.
Notice the curve never reaches the y-axis and never dips left of it: the domain is x > 0. Every curve passes through (1, 0), because any base to the power 0 is 1.
Common logs, and the characteristic and mantissa
Base 10 is used so often that log x written with no base at all means log₁₀ x. These are the common logarithms, and they are the ones in the four-figure tables. Base e ≈ 2.718 gives the natural logarithms, written ln x.
Every common logarithm splits into two parts. The whole number in front is the characteristic and the decimal part is the mantissa.
| Number | Standard form | Characteristic | Mantissa | log |
|---|---|---|---|---|
| 3.62 | 3.62 × 10⁰ | 0 | 0.5587 | 0.5587 |
| 36.2 | 3.62 × 10¹ | 1 | 0.5587 | 1.5587 |
| 3620 | 3.62 × 10³ | 3 | 0.5587 | 3.5587 |
| 0.362 | 3.62 × 10⁻¹ | −1 (written 1̄) | 0.5587 | 1̄.5587 |
Read the pattern before you memorise the rule
All four numbers have the same digits, so all four have the same mantissa. Only the position of the decimal point changes, and that is exactly what the characteristic records: it equals the power of 10 in standard form. For a number ≥ 1 the characteristic is (digits before the decimal point) − 1; for a number below 1 it is negative, and the bar is written above the digit because only the characteristic is negative, never the mantissa.
The four laws
Every one of these is a law of exponents in disguise. Multiplying powers adds the exponents, so multiplying numbers adds their logarithms — that single sentence is why logarithms were invented, and why they turned four-digit multiplication into addition three hundred years before calculators.
Express log (x³√y / z²) as separate logarithms.
- The whole thing is a quotient, so split it first:
log(x³√y) − log(z²).Work from the outside in — deal with the division before the multiplication inside it. - The first bracket is a product:
log(x³) + log(√y) − log(z²).Product becomes a sum. - Bring every power down as a multiplier, writing
√y = y^(1/2):3 log x + ½ log y − 2 log z.The power law works for fractional powers exactly as it does for whole ones.
3 log x + ½ log y − 2 log z
Three things the laws do NOT say
log(m + n) is not log m + log n — there is no law for the log of a sum. log(m)/log(n) is not log(m/n); the first is a change of base, the second is a difference. And (log m)² is not 2 log m; only a power inside the log comes down.
Solving an equation that has the unknown in the exponent
This is what logarithms are actually for in an exam. If the unknown is stuck upstairs, take the log of both sides and the power law brings it down where you can reach it.
Solve 5x = 400, giving x to three significant figures.
- Take log₁₀ of both sides:
log(5ˣ) = log 400.Whatever you do to one side you do to the other; logging both sides keeps the equation true. - Apply the power law on the left:
x log 5 = log 400.The exponent is now an ordinary multiplier, so x is no longer trapped. x = log 400 / log 5 = 2.6021 / 0.6990.This is a division of two logs, not the log of a quotient.x = 3.72(3 s.f.).Check: 5³·⁷² ≈ 400. A sanity check costs five seconds and catches an inverted fraction.
x ≈ 3.72
Before you leave this chapter
aˣ = y ⟺ logₐ y = x. Converting between the two forms solves most questions on its own.- Base must be positive and ≠ 1; the number itself must be strictly positive.
- Characteristic = the power of 10 in standard form. Mantissa comes from the table and is never negative.
- Product → sum, quotient → difference, power → multiplier. There is no law for log of a sum.
- To free an unknown exponent, take logs of both sides and bring the power down.
e, and the logarithm that goes with it
One base matters more than all the others. The number e ≈ 2.71828 is the base for which the curve y = eˣ has gradient exactly equal to its own height at every point — the rate of growth equals the amount present. That is the defining property of natural growth, which is why e appears in every population, cooling and decay problem you will meet.
The logarithm to base e is written ln x rather than logₑ x. It obeys the same four laws as any other logarithm; nothing new has to be learned about how it behaves.
eˣ and ln x are inverse functions, so each undoes the other: ln(eˣ) = x and e^(ln x) = x. Their graphs are therefore reflections of one another in the line y = x, which fixes their shapes in memory — eˣ never touches the x-axis, so ln x never touches the y-axis, and ln x is undefined for x ⩽ 0.
- e
- about 2.71828the base whose curve has gradient equal to its height
- ln x
- logarithm to base eundefined for x ⩽ 0
- k
- the rate constantits sign decides growth or decay
Taking logs is a legal move on both sides
If two positive quantities are equal, their logarithms are equal. That is what licenses "taking ln of both sides", and it is the standard route whenever the unknown sits in an exponent. The restriction to positive quantities matters: ln of a negative number does not exist, so an equation like eˣ = −2 has no solution at all.
Solving equations with the unknown in the index
When x appears as an exponent, no amount of ordinary rearranging will reach it, because it is not being added or multiplied — it is counting repeated multiplication. Taking logarithms is what brings it down, since the power law turns an exponent into a coefficient.
Solve 5^(2x−1) = 40, giving your answer to 3 significant figures.
- Take logarithms of both sides: ln(5^(2x−1)) = ln 40.Both sides are positive, so this is valid. Either ln or log₁₀ works; the answer is the same.
- Bring the exponent down: (2x − 1) ln 5 = ln 40.The power law log(aⁿ) = n log a is the entire point of the method.
- 2x − 1 = ln 40 / ln 5 = 3.6889 / 1.6094 = 2.2921.A quotient of two logarithms, not the logarithm of a quotient — these are completely different and confusing them is the usual error.
- 2x = 3.2921, so x = 1.65 to 3 s.f.Only the final answer is rounded.
x = 1.65
ln a / ln b is not ln(a/b)
The subtraction law says ln(a/b) = ln a − ln b. A quotient of two logarithms is something else entirely and does not simplify. In the example above, ln 40 / ln 5 = 2.292 while ln(40/5) = ln 8 = 2.079. Any answer that treats them as equal will be wrong.
Reducing a relationship to linear form
Experimental data rarely lies on a straight line, but many relationships become straight once logarithms are taken. The point of doing this is that a straight line has a gradient and an intercept that can be measured off a graph — which is how the unknown constants are found.
Two patterns cover almost every question. Which one applies is decided by where the unknown sits: in the exponent, or as the power of the variable.
| Relationship | Take logs to get | Plot | Gradient / intercept |
|---|---|---|---|
| y = kxⁿ | ln y = ln k + n ln x | ln y against ln x | gradient n, intercept ln k |
| y = k·aˣ | ln y = ln k + x ln a | ln y against x | gradient ln a, intercept ln k |
Getting the constants back out
- For
y = kxⁿ: the power n is the gradient directly, andk = e^(intercept). - For
y = k·aˣ:a = e^(gradient)andk = e^(intercept). - The intercept is always
ln k, never k — exponentiate before quoting it. - Which variable goes on which axis is decided by the form: the equation must read as
Y = mX + c.