From Förberedande kurs i matematik 1

Revision as of 09:23, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.3:3a moved to Solution 3.3:3a: Robot: moved page) ← Previous diff		Current revision (06:24, 2 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	By writing the argument <math>8</math> as <math>8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^3</math>, the logarithm law, <math>\lg a^b = b\lg a</math>, gives
-	<center> [[Image:3_3_3a.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\log _{2}8 = \log _{2} 2^3 = 3\cdot\log _{2}2 = 3\cdot 1 = 3\,,</math>}}
		+
		+	where we have used <math>\log _{2}2 = 1\,</math>.

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Current revision

By writing the argument \displaystyle 8 as \displaystyle 8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^3, the logarithm law, \displaystyle \lg a^b = b\lg a, gives

\displaystyle \log _{2}8 = \log _{2} 2^3 = 3\cdot\log _{2}2 = 3\cdot 1 = 3\,,

where we have used \displaystyle \log _{2}2 = 1\,.