The origins of a logarithm

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We look at the origins of a logarithm.

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.

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 $\ln (a\cdot b) = \ln (a) + \ln (b).$