history - Geschichte
logarithms were originally developed to improve calculations with trigonometric values of sine as in astronomy at the time it was necessary to make calculations to a fraction of a degree (minutes and seconds, actually).
- Bürgi - history of Bürgi’s logarithms
- Napier history
- Briggs
definition
If $${y = a^x }$$ then the logarithm is defined as
$${x = \log_a y }$$, where $${y \gt 0}$$
We read it as
“x is log to the base a of y”
where the logarithm is equivalent to the exponent x of the exponential equation.
laws - Rechengesetze
The advantage of logarithms is that a multiplication problem is converted into an addition problem which made calculations easier at the time especially when dealing with 8 digits of accuracy (8 significant figures)
logarithm of a product
$${\log_c \left(a \cdot b \right) = \log_c {a} + \log_c {b} }$$
logarithm of a quotient
$${\log_c \left(\frac {a}{b}\right) = \log_c {a} - \log_c {b} }$$
logarithm of a power
$${\log_c (a)^b = b \cdot \log_c {a} }$$
proof
$$\begin{align}\text{Let } M = \log_c{a} \Leftrightarrow c^M & = a \newline \Leftrightarrow (c^M)^b & = a^b \newline \Leftrightarrow c^{bM} & = a^b \newline \Leftrightarrow bM & = \log_c a^b \newline \Leftrightarrow b \cdot \log_c a & = \log_c{ a^b} \newline \end{align}$$
example
$${\log_5 6 = \log_5 (2 \cdot 3) = \log_5 2 + \log_5 3 }$$ $${\lg (\frac {3}{4}) = \lg 3 - \lg 4 }$$
Note that $\log_{10}$ can be abreviated as $\lg $ and the $\log_{e}$ is abbreviated as $\ln $ (natural logarithm). On calculators $\log $ means $\lg $ ($\log_{10}$).
By definition
$${\log_a a = 1}$$
my logarithmic table base 10
usage
Left column are the numbers from 1.0 to 9.9 without the decimal, so 10 to 99. The columns to the right “0 to 9” is the extra decimal place value giving us all the values from 1.00 to 9.99 .
example 2.34 for is read from row 23 and column 4 which is 3692 in the table and reads 0.3692.
use logarithms to calculate a square root
example
find the square root of 3457.
$${\begin{align} y & = \sqrt{3457} \newline y & = {3457}^{\frac{1}{2}} \newline \log_{10} {y} & = \log_{10} \left({3.457} \cdot 10^3\right)^{\frac{1}{2}} \newline & = \frac{1}{2} \cdot \left(\log_{10} {3.457} + \log_{10} 10^3\right) \newline & = \frac{1}{2} \cdot \left(\log_{10} {3.457} + 3\right) \newline & = \frac{1}{2} \cdot \left(0.5386994 + 3\right) \newline & = \frac{1}{2} \cdot 3.5386993795 \newline & = 1.7693496898 \newline & = 0.7693496898 + 1 \end{align}}$$
Rounding the ${0.7693496898 + 1}$ to ${0.7693497 + 1}$
Reading off from the log tables 7693497 gives the value 58796 which is equivalent to 5.8796 and the “+1” means multiply by 10 as ${\log_{10} {10}= 1}$.
The answer is ${58.796}$.
$${\Rightarrow y = 10^{1.76934969} \approx 58.796 ~ ,\text{ (using log tables)}}$$
converting from one base to another
Usually, you will only find log tables for base 10 and the base e, so it is necessary to convert from one base to the other.
$${\log_b (a) = \frac{ \log_c {a}}{\log_c {b}} }$$
example
$${\log_2 (7) = \frac{ \log_{10} {7}}{\log_{10} {2}} = \frac{ \log\ {7}}{\log {2}} }$$
log tables - Logarithmustafeln
historical log tables - LOCOMAT
slide rule - Der Rechenschieber
a “mechanical” calculator based on logarithms