Steps / Historical AI

Leonhard Euler · 1748 · Internet Archive

Introductio in analysin infinitorum

All 320 leaves · leaf 105

Transcription

EXPONENTIALIBUS AC LOGARITHMIS. 77 tematum Logarithmicorum erit infinitus. Perpetuo autem in CAP. VI duobus systematis Logarithmi ejusdem numeri eandem inter se servant rationem. Sit basis unius systematis $= a$, alterius $= b$, atque numeri $n$ Logarithmus in priori systemate $= p$, in posteriori $= q$; erit $a^p = n$ & $b^q = n$; unde $a^p = b^q$; ideoque $a = b^{\frac{q}{p}}$. Oportet ergo ut Fractio $\frac{q}{p}$ constantem obtineat valorem, quicunque numerus pro $n$ fuerit assumtus. Quod si ergo pro uno systemate Logarithmi omnium numerorum fuerint computati, hinc facili negotio per regulam auream Logarithmi pro quovis alio systemate reperiri possunt. Sic, cum dentur Logarithmi pro basi 10, hinc Logarithmi pro quavis alia basi, puta 2, inveniri possunt; quæratur enim Logarithmus numeri $n$ pro basi 2, qui fit $= q$, cum ejusdem numeri $n$ Logarithmus fit $= p$ pro basi 10. Quoniam pro basi 10 est $l 2 = 0,3010300$, & pro basi 2, est $l 2 = 1$, erit $0,3010300 : 1 = p : q$ ideoque $q = \frac{p}{0,3010300} = 3,3219277. p$, si ergo omnes Logarithmi communes multiplicentur per numerum $3,3219277$, prodibit tabula Logarithmorum pro basi 2. 108. Hinc sequitur duorum numerorum Logarithmos in quocunque systemate eandem tenere rationem Sint enim duo numeri $M$ & $N$, quorum pro basi $a$ Logarithmi sint $m$ & $n$, erit $M = a^m$ & $N = a^n$: hinc fiet $M^n = a^{mn} = N^m$, ideoque $M^n = N^m$; in qua æquatione cum basis $a$ non amplius insit, perspicuum est Fractionem $\frac{m}{n}$ habere valorem à basi $a$ non pendentem. Sint enim pro alia basi $b$ numerorum eorundem $M$ & $N$ Logarithmi $\mu$ & $\nu$, pari modo colligetur fore $M^\nu = N^\mu$. Erit ergo $N^{\frac{m}{n}} = N^{\frac{\mu}{\nu}}$, hincque $\frac{m}{n} = \frac{\mu}{\nu}$, seu $m:n = \mu:\nu$. Ita jam vidimus in K 3
confidence 0.88 41 expressions source scan at the Internet Archive

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