Introductio in analysin infinitorum

86 DE QUANTITATUM EXPONENTIALIUM LIB. I. rantis, puta $1 + \frac{1}{1000000}$, ita ut fit $k \omega = \frac{1}{1000000}$; erit $l(1 + \frac{1}{1000000}) = l\frac{1000001}{1000000} = 0,00000043429 = \omega$. Hinc, ob $k \omega = 0,00000100000$, erit $\frac{1}{k} = \frac{43429}{100000}$ & $k = \frac{100000}{43429} = 2,30258$: unde patet $k$ esse numerum finitum pendentem a valore basis $a$. Si enim alius numerus pro basi $a$ statuatur, tum Logarithmus ejusdem numeri $1+k\omega$ ad priorem datam tenebit rationem, unde simul alius valor litteræ $k$ prodiret. 115. Cum fit $a^\omega = 1+k\omega$, erit $a^{i\omega} = (1+k\omega)^i$, quicunque numerus loco $i$ substituatur. Erit ergo $a^{i\omega [TRUNCATED]