Solution 3.3:2g - Revision history

Tek at 06:20, 2 October 2008

2008-10-02T06:20:10Z

←Older revision		Revision as of 06:20, 2 October 2008
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-	We know that	+	We know that <math>10^{\lg x} = x</math>, so therefore we rewrite the exponent as
-	<math>10^{\lg x}=x</math>, so therefore we rewrite the exponent as	+	<math>-\lg 0\textrm{.}1 = (-1)\cdot\lg 0\textrm{.}1 = \lg 0\textrm{.}1^{-1}</math>
-	<math>-\lg 0.1=~~\left~~( -1 ~~\right~~)\~~centerdot~~ \lg 0.1=\lg 0.1^{-1}</math>	+	by using the log law <math>b\lg a = \lg a^b</math>. This gives
-	by using the log law	+
-	<math>b\lg a=\lg a^{b}</math>. This gives	+

-		+	{{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>}}
-	<math>10^{-\lg 0.1}=10^{\lg 0.1^{-1}}=0.1^{-1}=\frac{1}{0.1}=10</math>	+

2008-09-25T13:31:37Z

←Older revision		Revision as of 13:31, 25 September 2008
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-	{~~{NAVCONTENT_START}~~}	+	We know that
-	<~~center~~> ~~[[Image:3_3_2g~~.~~gif]]~~ </~~center~~>	+	<math>10^{\lg x}=x</math>, so therefore we rewrite the exponent as
-	{{~~NAVCONTENT_STOP~~}}	+	<math>-\lg 0.1=\left( -1 \right)\centerdot \lg 0.1=\lg 0.1^{-1}</math>
		+	by using the log law
		+	<math>b\lg a=\lg a^{b}</math>. This gives
		+
		+
		+	<math>10^{-\lg 0.1}=10^{\lg 0.1^{-1}}=0.1^{-1}=\frac{1}{0.1}=10</math>

2008-09-09T09:22:43Z

←Older revision

Revision as of 09:22, 9 September 2008

2008-08-21T06:58:29Z

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2008-04-02T09:32:46Z

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