Introductio in analysin infinitorum

AC LOGARITHM. PER SERIES EXPLICAT. 91 CAP. VII. $ \frac{2}{7 \cdot 9^7} + \&c. $, facto $ x = \frac{1}{9} $, erit $ l \frac{5}{4} = \frac{2}{1 \cdot 9} + \frac{2}{3 \cdot 9^3} + \frac{2}{5 \cdot 9^5} + $ $ \frac{2}{7 \cdot 9^7} + \&c.. $ Ex Logarithmis vero harum fractionum reperientur Logarithmi numerorum integrorum, erit enim ex natura Logarithmorum $ l \frac{3}{2} + l \frac{4}{3} = l 2 $; tum $ l \frac{3}{2} + l 2 = l 3 $; & $ 2 l 2 = l 4 $; porro $ l \frac{5}{4} + l 4 = l 5 $; $ l 2 + l 3 = l 6 $; $ 3 l 2 = l 8 $; $ 2 l 3 = l 9 $; & $ l 2 + l 5 = l 10 $. EXEMPLUM. Hinc Logarithmi hyperbolici numerorum ab 1 usque ad 10 ita se habebunt, ut sit $ l 1 = 0, [TRUNCATED]