From Förberedande kurs i matematik 1

Revision as of 09:23, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.3:2h moved to Solution 3.3:2h: Robot: moved page) ← Previous diff		Current revision (06:22, 2 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	The argument in the logarithm can be rewritten as <math>\frac{1}{10^{2}} = 10^{-2}</math> and then the log law <math>\lg a^b = b\lg a</math> gives the rest
-	<center> [[Image:3_3_2h.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\lg \frac{1}{10^2} = \lg 10^{-2} = (-2)\cdot \lg 10 = (-2)\cdot 1 = -2\,\textrm{.}</math>}}

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Current revision

The argument in the logarithm can be rewritten as \displaystyle \frac{1}{10^{2}} = 10^{-2} and then the log law \displaystyle \lg a^b = b\lg a gives the rest

\displaystyle \lg \frac{1}{10^2} = \lg 10^{-2} = (-2)\cdot \lg 10 = (-2)\cdot 1 = -2\,\textrm{.}