Hans Georg Schaathun
September 2016
\(x = \ln y \quad \quad y > 0\)
\(y = \exp x \)
| \(\ln 1 = 0\) | \(\exp 0 = 1\) |
| \(\ln (xy) = \ln x + \ln y\) | \(\exp (x+y) = (\exp x)(\exp y)\) |
| \(\ln \frac1x = - \ln x \) | \(\exp (-x) = \frac1{\exp x} \) |
| \(\ln \frac xy = \ln x - \ln y \) | \(\exp (x-y) = \frac{\exp x}{\exp y}\) |
| \(\ln (x^y) = y\ln x\) | \((\exp x)^y = y\exp x\) |
$$e = \exp 1 \approx 2{,}718$$
Definisjon: \(e = \exp 1\)
$$\exp x = \exp ( 1\cdot x) = (\exp 1)^x = e^x$$