Introductio in analysin infinitorum

92 DE QUANTITATUM EXPONENTIALIUM LIB. I. 124. Ponatur Logarithmus hyperbolicus ipsius $1+x$ seu $l(1+x) = y$; erit $y = \frac{x}{1} - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \&c..$ Sumto autem numero $a$ pro basi Logarithmica, sit numeri ejusdem $1+x$ Logarithmus $=v$; erit, ut vidimus, $v = \frac{1}{k}(x - \frac{xx}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \&c.) = \frac{y}{k}$; hincque $k = \frac{y}{v}$; ex quo commodissime valor ipsius $k$ basi $a$ respondens ita definitur ut sit æqualis cujusvis numeri Logarithmo hyperbolico diviso per Logarithmum ejusdem numeri ex basi $a$ formati. Posito ergo numero hoc $=a$, erit $v=1$, hincque fit $k = \text{Logarithmo hyperbolico basis } a$. In systemate ergo Logarithmorum communium, ubi est $a=10$, erit $k = \text{Logarithmo hyperbolico ipsius } 10$, unde fit $k [TRUNCATED]