How do decibels compare sound intensities using a logarithmic scale?

A sound level can rise by only ten decibels while the sound intensity becomes ten times as large. The small-looking change in the number is not hiding an arithmetic mistake. Decibels use a logarithmic scale, which turns multiplicative changes in intensity into additive steps in the reported level.

Sound intensity describes the rate of sound-energy transfer through a unit area, measured in watts per square meter. To express an intensity level, choose a positive reference intensity Iref and form the ratio I/Iref. The units cancel. The level relative to that reference is L = 10 log₁₀(I/Iref) decibels. A logarithm asks which power of ten produces the ratio, so it must act on a dimensionless positive number.

Suppose three hypothetical signals have intensities ten, one hundred, and one thousand times the same reference. Their levels are 10, 20, and 30 decibels. Each equal ten-decibel step represents another multiplication by ten. Going from the first signal to the third adds twenty decibels but multiplies intensity by one hundred, not by two. These are invented comparisons, not measurements of particular sounds.

The zero point has a precise meaning. If I equals Iref, the ratio is one and its logarithm is zero. Zero decibels therefore means equal to the reference, not complete silence. If I is one-tenth of Iref, the level is minus ten decibels: a positive intensity below the chosen reference. The usual reference for sound intensity level is 10⁻¹² watts per square meter, but a comparison must state which reference and quantity it uses.

The difference between two intensity levels is 10 log₁₀(I₂/I₁), provided they use the same reference. For twice the intensity, the difference is about 3.01 decibels because log₁₀2 is about 0.301. Multiplying intensity by two again gives another 3.01-decibel rise. This illustrates how logarithms turn multiplication into addition, even when the ratio is not an exact power of ten.

Sound pressure is a different quantity. Under the same propagation conditions, intensity is proportional to the square of sound-pressure amplitude. That is why a corresponding pressure-amplitude comparison uses 20 log₁₀ of the pressure ratio instead of 10. Doubling pressure amplitude then gives four times the intensity and about 6.02 decibels. Using the intensity formula directly on a pressure ratio would miss the square.

Neither calculation alone tells us that a sound feels twice as loud. Perceived loudness depends on frequency, duration, listening conditions, and the listener as well as physical level. The mathematical benefit of decibels is more specific: an enormous range of positive intensities becomes a manageable scale of ratios. Equal spaces along that scale mark equal multipliers, so a modest numerical increase can represent a substantial physical change.