From Förberedande kurs i matematik 1

Revision as of 09:22, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.3:2g moved to Solution 3.3:2g: Robot: moved page) ← Previous diff		Current revision (06:20, 2 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	We know that <math>10^{\lg x} = x</math>, so therefore we rewrite the exponent as
-	<center> [[Image:3_3_2g.gif]] </center>	+	<math>-\lg 0\textrm{.}1 = (-1)\cdot\lg 0\textrm{.}1 = \lg 0\textrm{.}1^{-1}</math>
-	{{NAVCONTENT_STOP}}	+	by using the log law <math>b\lg a = \lg a^b</math>. This gives
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		+	{{Displayed math||<math>10^{-\lg 0\textrm{.}1}=10^{\lg 0\textrm{.}1^{-1}}=0\textrm{.}1^{-1}=\frac{1}{0\textrm{.}1}=10\,\textrm{.}</math>}}

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Current revision

We know that \displaystyle 10^{\lg x} = x, so therefore we rewrite the exponent as \displaystyle -\lg 0\textrm{.}1 = (-1)\cdot\lg 0\textrm{.}1 = \lg 0\textrm{.}1^{-1} by using the log law \displaystyle b\lg a = \lg a^b. This gives

\displaystyle 10^{-\lg 0\textrm{.}1}=10^{\lg 0\textrm{.}1^{-1}}=0\textrm{.}1^{-1}=\frac{1}{0\textrm{.}1}=10\,\textrm{.}