From Förberedande kurs i matematik 1

Revision as of 09:27, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.3:5a moved to Solution 3.3:5a: Robot: moved page) ← Previous diff		Current revision (07:19, 2 October 2008) (edit) (undo) Tek (Talk | contribs) m
(One intermediate revision not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	The notation ln is used for the logarithm with the natural base <math>e = 2\textrm{.}7182818\ldots\,,</math> and we therefore have that <math>\ln e = 1\,</math>, which together with the logarithm law, <math>b\lg a=\lg a^b</math>, gives that
-	<center> [[Image:3_3_5a.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\ln e^3 + \ln e^2 = 3\cdot\ln e + 2\cdot \ln e = 3\cdot 1 + 2\cdot 1 = 5\,\textrm{.}</math>}}

---

Current revision

The notation ln is used for the logarithm with the natural base \displaystyle e = 2\textrm{.}7182818\ldots\,, and we therefore have that \displaystyle \ln e = 1\,, which together with the logarithm law, \displaystyle b\lg a=\lg a^b, gives that

\displaystyle \ln e^3 + \ln e^2 = 3\cdot\ln e + 2\cdot \ln e = 3\cdot 1 + 2\cdot 1 = 5\,\textrm{.}