We look at the origins of a logarithm.
1 Why would someone care about logarithms?
Logarithms were originally developed as a computational tool. The key fact that made
this possible is that:
\[ \log _b(xy) = \log _b(x)+\log _b(y). \]
Before the days of calculators and computers, this was critical knowledge for anyone
in a computational discipline.
Compute
\(138\cdot 23.4\) using logarithms.
Start by writing both numbers in scientific
notation
\[ \left (1.38\cdot 10^{\answer [given]{2}}\right )\cdot \left (2.34 \cdot 10^{\answer [given]{1}}\right ). \]
Next we use a log-table, which gives
\(\log _{10}(N)\) for values of
\(N\) ranging
between
\(0\) and
\(9\). We’ve reproduced part of such a table below.
From the table, we see that
\[ \log _{10}(1.38) \approx \answer [given]{0.1399}\qquad \text {and}\qquad \log _{10}(2.34)\approx \answer [given]{0.3692} \]
Add these numbers together to get
\(\answer [given]{0.5091}\). Essentially, we
know the following at this point:
Using the table again, we see that \(\log _{10}(\answer [given]{3.23})\approx 0.5091\). Since we were working in scientific notation, we
need to multiply this by \(10^3\). Our final answer is
\[ 3230 \approx 138\cdot 23.4 \]
Since
\(138\cdot 23.4 = 3229.2\), this is a good approximation.
The moral is:
Logarithms allow us to use addition in place of
multiplication.
2 How does the logarithm work?
Still thinking about our work above, consider this: If we want a function
\[ \log (x)\qquad \text {(nevermind the base)} \]
to have the
property
\[ \log (a\cdot b) = \log (a) + \log (b), \]
then how is a logarithm actually defined?
The
natural logarithm is the accumulation function defined by
\[ \ln (x) = \int _1^x \frac {1}{t} \d t \]
The domain of the
natural logarithm is
\(\RR ^+\) (all positive real numbers), and its range is
\(\RR \) (all real numbers).
Finally, we’ll show that the natural logarithm has the property that transforms
multiplication into addition.
Show that:
\[ \ln (a\cdot b) = \ln (a) + \ln (b) \]
Write with me:
\begin{align*} \ln (a\cdot b) &= \int _1^{a\cdot b} \frac {1}{t}\d t\\ &= \int _1^{a} \frac {1}{t}\d t + \int _a^{a\cdot b} \frac {1}{t}\d t\\ &= \ln (a) + \int _a^{a\cdot b} \frac {1}{t}\d t \end{align*}
Looking just at the last integral, set \(g= t/a\), so
\[ \frac {1}{t} = \answer [given]{\frac {1}{a\cdot g}} \]
now
\(\d g = \d t/a\), and so
\(\d t= a \d g\). Substituting in we find
\begin{align*} \int _a^{a\cdot b} \frac {1}{t}\d t &= \int _{a/a}^{a\cdot b/a} \frac {1}{a \cdot g} \cdot a \d g\\ &= \int _{1}^{b} \frac {1}{g}\d g\\ &= \ln (b). \end{align*}
Hence
\[ \ln (a\cdot b) = \ln (a) + \ln (b). \]
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