At Rakhigarhi in Haryana, archaeologists dig out a Harappan layer with the charcoal of an old hearth. The charcoal gives \(8.9\) carbon-14 decays per minute for each gram of carbon. Living wood gives \(15.3\). The chapter Radioactivity says the activity falls as \(A = A_0e^{-\lambda t}\). The age \(t\) is the unknown, and it sits in the power. No amount of dividing or multiplying can free it from the power.
The question "to what power?"
Ask a simple question: \(10\) to what power gives \(1000\)? The answer is \(3\). We write \(\log_{10}1000 = 3\). In the same way \(2\) to the power \(5\) is \(32\), so \(\log_2 32 = 5\). The logarithm of a number to a base \(b\) is the power to which \(b\) must be raised to give that number.
For a base \(b > 0\), \(b \neq 1\), and a number \(y > 0\), the logarithm \(\log_b y\) is the number \(x\) for which \(b^x = y\). The natural logarithm \(\ln y\) has the base \(e = 2.718\). The common logarithm \(\log y\) has the base 10.
The logarithm undoes the exponential, just as a square root undoes a square. Applied to \(e^{-t/\tau} = f\), it gives \(-t/\tau = \ln f\), so \(t = \tau\ln(1/f)\). With \(f = 1/2\) this is the half-life, \(\tau\ln 2\). With \(f = 1/10\) it is \(\tau\ln 10 = 2.3\tau\), the time to fall to a tenth.
Fresh wood gives \(15.3\) carbon-14 decays per minute per gram of carbon, and an old wooden bowl gives \(5.1\). The age is \(t = 8270\,\text{y} \times \ln(A_0/A)\). Find the age, in years. Take \(\ln 3 = 1.099\).
Products become sums
A logarithm is a power, and powers obey the rule \(e^pe^q = e^{p + q}\). So the logarithm of a product is the sum of the logarithms. Three rules follow from this one fact. They are the reason logarithms were invented: they turn multiplication into addition.
\[ a = e^p, \quad b = e^q \]
Why. write each positive number as a power of \(e\); then \(p = \ln a\) and \(q = \ln b\).
\[ ab = e^pe^q = e^{p + q} \]
Why. powers of the same base add when the numbers multiply.
\[ \ln(ab) = p + q = \ln a + \ln b \]
Why. the logarithm of \(e^{p + q}\) is its power, \(p + q\).
\[ a^n = (e^p)^n = e^{np} \quad\Rightarrow\quad \ln a^n = n\ln a \]
Read it. a power of a power multiplies the powers. Dividing gives \(\ln(a/b) = \ln a - \ln b\) in the same way.
Before calculators, engineers multiplied with a slide rule. Its scales are marked so that the number \(k\) stands at a distance proportional to \(\log k\). Sliding one scale along the other adds two lengths. Adding \(\log a\) and \(\log b\) gives \(\log ab\), so the product can be read off the scale.
Which of these equals \(\ln 2 + \ln 3\)?
Two bases, one conversion
Physics meets two bases. Exponential laws in \(e^{-t/\tau}\) call for the natural logarithm. Scales that count powers of ten, such as the decibel, call for the common logarithm. The two are always in the same ratio. From \(x = b^{\log_b x}\), take the natural logarithm of both sides: \(\ln x = \log_b x \times \ln b\).
So \(\log_b x = \ln x/\ln b\), the change of base. With \(b = 10\), \(\ln x = \ln 10 \times \log x = 2.303\log x\). A few values are worth knowing by heart: \(\ln 2 = 0.693\), \(\ln 3 = 1.099\), \(\ln 10 = 2.303\), \(\log 2 = 0.301\) and \(\log 3 = 0.477\).
Move the slider in the figure above. The logarithm curve crosses zero at \(x = 1\), because \(e^0 = 1\). It is negative for every number between \(0\) and \(1\), since those need a negative power of \(e\). And it has no value at all for zero or for a negative number, because no power of \(e\) is zero or negative.
A sound level needs \(\log(5.0 \times 10^9)\), the common logarithm. Find it, to two decimal places. Take \(\log 5 = 0.70\).
Two moles of an ideal gas at \(300\,\text{K}\) are compressed isothermally from \(20\,\text{L}\) to \(5\,\text{L}\). The work done by the gas is \(W = nRT\ln(V_2/V_1)\), with \(R = 8.31\,\text{J/(mol K)}\). How much heat does the gas give out, in joules? Take \(\ln 4 = 1.3863\).
A rock holds three uranium-238 atoms for every lead-206 atom. Assuming it formed with no lead-206, find its age. Take \(T_{1/2} = 4.47 \times 10^9\) years, \(\ln 2 = 0.693\) and \(\ln(4/3) = 0.288\).
Of every four uranium atoms the rock began with, three remain and one has become lead. So the fraction of uranium left is \(N_U/N_0 = 3/4\).
\(N_U/N_0 = e^{-\lambda t}\), so \(t = \dfrac{1}{\lambda}\ln\dfrac{N_0}{N_U} = \dfrac{T_{1/2}}{\ln 2}\ln\dfrac43\).
\(t = \dfrac{4.47 \times 10^9}{0.693} \times 0.288 = 6.45 \times 10^9 \times 0.288 = 1.86 \times 10^9\) years.
The rock is about \(1.86 \times 10^9\) years old. The answer assumes no lead-206 at the start and none gained or lost since.
Now the hearth. The fraction left is \(A/A_0 = 8.9/15.3\), so \(t = (1/\lambda)\ln(A_0/A)\). For carbon-14, \(1/\lambda = 5730/0.693 = 8270\) years. Then \(\ln(15.3/8.9) = \ln 1.72 = 0.54\), and \(t = 8270 \times 0.54 = 4500\) years, to two figures. The logarithm freed the time from the power.
An iodine-131 source, half-life 8 days, has an activity of \(3200\,\text{Bq}\). After how many days is its activity \(100\,\text{Bq}\)? The number of half-lives is \(\log_2(3200/100)\).
A logarithm exists only for a positive pure number. So always take the logarithm of a ratio, such as \(A_0/A\) or \(V_2/V_1\), in which the units cancel. The logarithm of zero does not exist: a decaying sample never reaches zero, and no time makes \(e^{-t/\tau} = 0\). There is also no rule for \(\ln(a + b)\); only products and quotients become sums and differences.
The logarithm dates rocks and charcoal in the chapter Radioactivity, and gives the half-life from the decay constant there. It gives the work and the heat of an isothermal process, \(nRT\ln(V_2/V_1)\), and the entropy changes of the chapter on the second law. It measures sound in decibels, and it gives the logarithmic decrement of a damped oscillator. It sets the number of generations a chain reaction needs, and the rocket equation carries it too.
- Taking the logarithm of a difference, such as \(\ln(15.3 - 5.1)\).The decay law gives a ratio. Take \(\ln(A_0/A)\).
- Writing \(\ln(a + b) = \ln a + \ln b\).The sum rule is for products: \(\ln(ab) = \ln a + \ln b\). A sum inside a logarithm cannot be split.
- Mixing \(\ln\) and \(\log\).\(\ln\) is base \(e\), \(\log\) is base 10, and \(\ln x = 2.303\log x\). Keep to one per calculation.