Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/261

Placing $M=3$ and $N=2$ in (E), we get

$\log 3=\log 2+2\left[{\frac {1}{5}}+{\frac {1}{3}}\cdot {\frac {1}{5^{8}}}+{\frac {1}{5}}\cdot {\frac {1}{5^{5}}}+\cdots \right]=1.09861229\cdots .$

It is only necessary to compute the logarithms of prime numbers in this way, the logarithms of composite numbers being then found by using theorems 7-10, §1. Thus

${\begin{aligned}\log 8&=\log 2{^{3}}&=3\log 2&=2.07944154\cdots ,\\\log 6&=\log 3&+\log 2&=1.79175947\cdots .\\\end{aligned}}$

All the above are Napierian or natural logarithms, i.e. the base is $e=2.7182818$. If we wish to find Briggs's or common logarithms, where the base 10 is employed, all we need to do is to change the base by means of the formula

$\log _{10}n={\frac {\log _{e}n}{\log _{e}10}}.$

Thus

$\log _{10}2={\frac {\log _{e}2}{\log _{e}10}}={\frac {0.693\cdots }{2.302\cdots }}=0.301\cdots .$

In the actual computation of a table of logarithms only a few of the tabulated values are calculated from series, all the rest being found by employing theorems in the theory of logarithms and various ingenious devices designed for the purpose of saving work.

EXAMPLES

Calculate by the methods of this article the following logarithms:

1.	$\log _{e}5=1.6094\cdots .$	2.	$\log _{e}24=3.1781\cdots .$
2.	$\log _{e}10=2.3025\cdots .$	4.	$\log _{10}5=0.6990\cdots .$

147. Approximate formulas derived from series. Interpolation. In the two preceding sections we evaluated a function from its equivalent power series by substituting the given value of $x$ in a certain number of the first terms of that series, the number of terms taken depending on the degree of accuracy required. It is of great practical importance to note that this really means that we are considering the function as approximately equal to an ordinary polynomial with constant coefficients. For example, consider the series

(A)	$\sin x=x+{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots .$