Solution 3.3:3a - Revision history

Tek at 06:24, 2 October 2008

2008-10-02T06:24:18Z

←Older revision		Revision as of 06:24, 2 October 2008
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-	By writing the argument	+	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
-	<math>~~\text{~~8}</math>	+
-	as	+
-	<math>8=2\~~centerdot~~ 4=2\~~centerdot~~ 2\~~centerdot~~ 2=2^{3}</math>, the logarithm law,	+
-	<math>\lg a^{b}=b\lg a</math>, gives	+

-		+	{{Displayed math\|\|<math>\log _{2}8 = \log _{2} 2^3 = 3\cdot\log _{2}2 = 3\cdot 1 = 3\,,</math>}}
-	<math>\log _{2}8=\log _{2}2^{3}=3\~~centerdot~~ \log _{2}2=3\~~centerdot~~ 1=3</math>	+

-		+	where we have used <math>\log _{2}2 = 1\,</math>.
-	where we have used	+
-	<math>\log _{2}2=1</math>.	+

2008-09-25T13:57:40Z

←Older revision		Revision as of 13:57, 25 September 2008
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-	{{~~NAVCONTENT_START}~~}	+	By writing the argument
-	<~~center~~> ~~[[Image:3_3_3a.gif]]~~ </~~center~~>	+	<math>\text{8}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	as
		+	<math>8=2\centerdot 4=2\centerdot 2\centerdot 2=2^{3}</math>, the logarithm law,
		+	<math>\lg a^{b}=b\lg a</math>, gives
		+
		+
		+	<math>\log _{2}8=\log _{2}2^{3}=3\centerdot \log _{2}2=3\centerdot 1=3</math>
		+
		+
		+	where we have used
		+	<math>\log _{2}2=1</math>.

2008-09-09T09:23:23Z

←Older revision

Revision as of 09:23, 9 September 2008

2008-08-21T06:58:49Z

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2008-04-02T09:34:12Z

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