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Exponential and Logarithmic Models · Tutorial 112 of 1495

Logarithmic Scales in Real-World Measurements

Interpret pH and decibel values by converting them to concentrations and intensity ratios, then compare what equal changes on each scale mean.

Beginner 9 min read

What You'll Learn

  • Interpret pH as a base-10 logarithmic measure of hydrogen ion concentration.
  • Convert pH values into approximate concentrations and compare those concentrations.
  • Interpret decibels as logarithmic measures of ratios to a reference intensity.
  • Convert decibel differences into multiplicative intensity ratios.
  • Explain why equal intervals on a logarithmic scale represent equal factors.

Why Use a Logarithmic Scale?

A measurement can vary across an enormous range. For example, sound intensities can differ by many powers of ten, while hydrogen ion concentrations in solutions can be very small. A linear scale may make such differences awkward to represent and compare.

A base-10 logarithmic scale expresses the exponent associated with a quantity or ratio. As a result, a fixed step on the scale corresponds to multiplying the underlying quantity by a fixed factor. This tutorial examines that idea through pH and decibels.

Key idea: A logarithmic scale compresses large multiplicative differences into smaller additive differences. Equal changes in the scale value represent equal multiplicative changes in the underlying quantity or ratio.

The pH Scale

For an introductory model, pH is defined using the base-10 logarithm of the hydrogen ion concentration. Write \(c\) for that concentration in moles per litre. In this model, \(c\) is inserted as its numerical value in \(\mathrm{mol/L}\):

$$ \mathrm{pH}=-\log_{10}(c). $$

This convention treats the concentration relative to \(1\ \mathrm{mol/L}\), so the logarithm is applied to a dimensionless numerical ratio. In more precise chemical definitions, pH is defined using hydrogen ion activity rather than concentration. The concentration formula is a useful introductory approximation, and it is the one used in the examples here.

Solving the model for concentration gives:

$$ c=10^{-\mathrm{pH}}\ \mathrm{mol/L}. $$

The minus sign matters. A higher pH corresponds to a smaller hydrogen ion concentration in this model. A difference of one pH unit corresponds to a factor of ten in concentration, not a difference of ten concentration units.

Worked Example: Convert a pH Value to Concentration

Use the introductory concentration model to find the hydrogen ion concentration corresponding to \(\mathrm{pH}=3\).

Start with \(\mathrm{pH}=-\log_{10}(c)\), and substitute \(\mathrm{pH}=3\):

$$ 3=-\log_{10}(c). $$

Multiply both sides by \(-1\):

$$ -3=\log_{10}(c). $$

Convert from logarithmic to exponential form, using the definition of a base-10 logarithm:

$$ c=10^{-3}\ \mathrm{mol/L} =0.001\ \mathrm{mol/L}. $$

Check by substituting the concentration into the original formula:

$$ -\log_{10}(0.001) =-\log_{10}(10^{-3}) =-(-3) =3. $$

The result is \(0.001\ \mathrm{mol/L}\), and it gives the stated pH when substituted back.

Comparing pH Values

To compare two pH values, convert each one to a concentration or compare them through the formula. If their pH values are \(p_1\) and \(p_2\), their model concentrations are \(c_1=10^{-p_1}\) and \(c_2=10^{-p_2}\) in \(\mathrm{mol/L}\). Their concentration ratio is therefore:

$$ \frac{c_1}{c_2} = \frac{10^{-p_1}}{10^{-p_2}} = 10^{p_2-p_1}. $$

Thus, if \(p_1\) is one unit lower than \(p_2\), then \(c_1/c_2=10\): the first solution has ten times the modeled hydrogen ion concentration. A difference of two pH units corresponds to a factor of \(10^2=100\).

Worked Example: Compare Two pH Values

Compare the modeled hydrogen ion concentrations at pH \(2\) and pH \(5\). Find how many times the concentration at pH \(2\) is the concentration at pH \(5\).

For pH \(2\), the concentration is:

$$ c_2=10^{-2}\ \mathrm{mol/L} =0.01\ \mathrm{mol/L}. $$

For pH \(5\), it is:

$$ c_5=10^{-5}\ \mathrm{mol/L} =0.00001\ \mathrm{mol/L}. $$

Divide the first concentration by the second:

$$ \frac{c_2}{c_5} = \frac{10^{-2}}{10^{-5}} = 10^{-2-(-5)} = 10^3 = 1000. $$

So the modeled concentration at pH \(2\) is \(1000\) times the concentration at pH \(5\). Check using the pH difference: \(5-2=3\), and a three-unit difference gives a factor of \(10^3=1000\). The ratio agrees.

The Decibel Scale

Decibels are used to express levels relative to a reference. For sound intensity, let \(I\) be the measured intensity and let \(I_0\) be the reference intensity. The level \(\beta\), measured in decibels (dB), is defined by:

$$ \beta = 10\log_{10}\left(\frac{I}{I_0}\right)\ \mathrm{dB}. $$

The ratio \(I/I_0\) is dimensionless: both quantities are intensities in the same units. For a commonly used reference in air, \(I_0=10^{-12}\ \mathrm{W/m^2}\). The reference must be known to recover an absolute intensity from a decibel level.

To convert from a decibel level to an intensity ratio, divide by \(10\) and use the inverse relationship between logarithms and powers:

$$ \frac{I}{I_0}=10^{\beta/10}. $$

A level of \(0\ \mathrm{dB}\) means \(I=I_0\), not that the intensity is zero. A negative level is also possible: it means that the measured intensity is less than the reference. In this intensity definition, a rise of \(10\ \mathrm{dB}\) means the intensity ratio has been multiplied by \(10\).

Worked Example: Convert a Decibel Level to Intensity

Suppose a sound has an intensity level of \(40\ \mathrm{dB}\) relative to \(I_0=10^{-12}\ \mathrm{W/m^2}\). Find its intensity.

First find the ratio by rearranging the level formula:

$$ \frac{I}{I_0} = 10^{\beta/10} = 10^{40/10} = 10^4 = 10000. $$

Now multiply by the reference intensity:

$$ I = 10000(10^{-12}\ \mathrm{W/m^2}) = 10^4\cdot10^{-12}\ \mathrm{W/m^2} = 10^{-8}\ \mathrm{W/m^2}. $$

Check by putting this intensity back into the formula. Its ratio to the reference is \(10^{-8}/10^{-12}=10^4\), so:

$$ 10\log_{10}(10^4) = 10(4) = 40\ \mathrm{dB}. $$

The intensity is \(10^{-8}\ \mathrm{W/m^2}\), and it reproduces the original level.

Why Equal Scale Intervals Mean Equal Factors

The reason equal intervals correspond to equal multiplicative changes follows directly from the logarithm formulas. Consider two positive intensities \(I_1\) and \(I_2\) measured relative to the same reference \(I_0\). Let their decibel levels be \(\beta_1\) and \(\beta_2\). Subtract the two level formulas:

$$ \begin{aligned} \beta_2-\beta_1 &= 10\log_{10}\left(\frac{I_2}{I_0}\right) - 10\log_{10}\left(\frac{I_1}{I_0}\right)\\ &= 10\log_{10}\left(\frac{I_2/I_0}{I_1/I_0}\right)\\ &= 10\log_{10}\left(\frac{I_2}{I_1}\right). \end{aligned} $$

The common reference cancels. Solving this equation for the intensity ratio gives the general comparison rule:

$$ \frac{I_2}{I_1} = 10^{(\beta_2-\beta_1)/10}. $$

This derivation assumes both levels use the same reference and the intensity values are positive. It shows that the level difference determines the ratio, regardless of the particular reference value.

The same reasoning applies to pH: a difference of \(d\) pH units gives a concentration ratio of \(10^d\), with the direction depending on which concentration is in the numerator. For decibels, a difference of \(d\) dB gives an intensity ratio of \(10^{d/10}\). Therefore, the same numerical interval does not produce the same factor on these two scales: the pH formula has no division by \(10\), while the decibel formula does.

Worked Example: Compare Equal Decibel Intervals

Compare the intensity ratio for a rise from \(20\ \mathrm{dB}\) to \(30\ \mathrm{dB}\) with the ratio for a rise from \(40\ \mathrm{dB}\) to \(50\ \mathrm{dB}\). Use the same reference for all levels.

For the first interval, the difference is \(30-20=10\ \mathrm{dB}\). The intensity ratio is:

$$ \frac{I_{30}}{I_{20}} = 10^{(30-20)/10} = 10^1 = 10. $$

For the second interval, the difference is \(50-40=10\ \mathrm{dB}\). Thus:

$$ \frac{I_{50}}{I_{40}} = 10^{(50-40)/10} = 10^1 = 10. $$

Both equal \(10\ \mathrm{dB}\) intervals correspond to a tenfold increase in intensity. Check directly from the definition: increasing the intensity by a factor of \(10\) adds \(10\log_{10}(10)=10\ \mathrm{dB}\), as expected.

A Common Mistake: Treating Scale Differences as Physical Differences

A frequent error is to read the difference between two scale values as the difference between the underlying measurements. For instance, a \(10\ \mathrm{dB}\) increase does not mean the intensity increases by \(10\) intensity units. It means the intensity is multiplied by \(10\).

Pitfall: From \(20\ \mathrm{dB}\) to \(30\ \mathrm{dB}\), the wrong calculation is to call the change a ten-unit increase in intensity. The correct calculation is \(I_{30}/I_{20}=10^{(30-20)/10}=10\), a tenfold increase. Likewise, a change from pH \(5\) to pH \(4\) is a one-unit decrease in pH, but the modeled concentration increases by a factor of \(10\).

Always identify what the scale measures and write the ratio in the direction required by the question. For pH, lower values mean higher modeled hydrogen ion concentration. For decibels, a higher level means higher intensity when the reference is held fixed.

When These Scales Are Useful

Logarithmic scales make comparisons across large ranges manageable. Instead of reporting an extremely small concentration directly, pH gives a compact value related to its power of ten. Instead of displaying a sound intensity ratio as a long string of zeros, decibels express that ratio on a scale where multiplication becomes addition.

The important information is often the difference between two scale values. A pH difference tells you a concentration factor; a decibel difference tells you an intensity factor. Keep the base and coefficient in each formula visible, since they determine how the scale interval translates to a ratio.

Check Your Understanding

Answer these questions using the formulas in this tutorial. For concentration questions, use the introductory pH model.

  1. What hydrogen ion concentration corresponds to pH \(4\), in \(\mathrm{mol/L}\)?
  2. How many times greater is the modeled hydrogen ion concentration at pH \(3\) than at pH \(6\)?
  3. A sound level rises from \(15\ \mathrm{dB}\) to \(35\ \mathrm{dB}\). By what factor does its intensity increase?
  4. What intensity ratio \(I/I_0\) corresponds to \(0\ \mathrm{dB}\)? Does this mean zero intensity?
  5. Why do two equal intervals of \(10\ \mathrm{dB}\) represent the same intensity factor even when they start at different decibel levels?

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