Solution 3.3:3h - Revision history

Tek at 07:04, 2 October 2008

2008-10-02T07:04:04Z

←Older revision		Revision as of 07:04, 2 October 2008
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-	Because	+	Because <math>a^{2}\sqrt{a} = a^{2}a^{1/2} = a^{2+1/2} = a^{5/2}</math>, the logarithm law, <math>b\lg a = \lg a^b</math>, gives that
-	<math>a^{2}\sqrt{a}=a^{2}a^~~{\frac~~{1}{2}}=a^{2+~~\frac{~~1}{2}}=a^~~{\frac~~{5}{2}}</math>, the logarithm law,	+
-	<math>b\lg a=\lg a^{b}</math>, gives that	+

		+	{{Displayed math\|\|<math>\log_{a} \bigl(a^{2}\sqrt{a}\,\bigr) = \log_{a}a^{5/2} = \frac{5}{2}\cdot\log_{a}a = \frac{5}{2}\cdot 1 = \frac{5}{2}\,,</math>}}

-	<math>\~~log _~~{a}a~~^{2}\sqrt{a}~~=~~\log _{a}a^{\frac{5}{2}}=\frac{5}{2}\centerdot \log _{a}a=\frac{5}{2}\centerdot~~ 1=\~~frac{5}{2}~~,</math>	+	where we have used that <math>\log_{a}a = 1\,</math>.


-	~~where we have used that~~	+	Note: In this exercise, we assume, implicitly, that <math>a > 0</math> and <math>a\ne 1\,</math>.
-	~~<math>\log _{a}a=1</math>.~~	+
-		+
-	~~NOTE~~: In this exercise, we assume, implicitly, that	+
-	<math>~~\text{~~a}>0~~\text{ }~~</math>	+
-	and	+
-	<math>~~\text{~~a}\ne ~~\text{~~1}</math>.	+

2008-09-25T14:41:14Z

←Older revision		Revision as of 14:41, 25 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	Because
-	<~~center~~> ~~[[Image:3_3_3h.gif]]~~ </~~center~~>	+	<math>a^{2}\sqrt{a}=a^{2}a^{\frac{1}{2}}=a^{2+\frac{1}{2}}=a^{\frac{5}{2}}</math>, the logarithm law,
-	{{~~NAVCONTENT_STOP~~}}	+	<math>b\lg a=\lg a^{b}</math>, gives that
		+
		+
		+	<math>\log _{a}a^{2}\sqrt{a}=\log _{a}a^{\frac{5}{2}}=\frac{5}{2}\centerdot \log _{a}a=\frac{5}{2}\centerdot 1=\frac{5}{2},</math>
		+
		+
		+	where we have used that
		+	<math>\log _{a}a=1</math>.
		+
		+	NOTE: In this exercise, we assume, implicitly, that
		+	<math>\text{a}>0\text{ }</math>
		+	and
		+	<math>\text{a}\ne \text{1}</math>.

2008-09-09T09:25:43Z

←Older revision

Revision as of 09:25, 9 September 2008

2008-08-21T06:59:59Z

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2008-04-02T09:48:07Z

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