Solution 3.1:8d - Revision history

Tek at 14:41, 30 September 2008

2008-09-30T14:41:36Z

←Older revision		Revision as of 14:41, 30 September 2008
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	In power form, the expressions become		In power form, the expressions become

		+	{{Displayed math\|\|<math>\begin{align}
		+	\sqrt{2}\bigl(\sqrt[4]{3}\bigr)^{3}
		+	&= 2^{1/2}\bigl(3^{1/4}\bigr)^{3}
		+	= 2^{1/2}3^{3/4},\\[5pt]
		+	\sqrt[3]{2}\cdot 3
		+	&= 2^{1/3}3^{1}\,\textrm{.}
		+	\end{align}</math>}}

-	<math>~~\begin{align}~~	+	Admittedly, it is true that <math>2^{1/2} > 2^{1/3}</math> and <math>3^1 > 3^{3/4}</math>, but this does not help us to say anything about how the products are related to each other. Instead, we observe that the exponents 1/2, 3/4, 1/3 and 1 have <math>3\cdot 4 = 12</math> as the lowest common denominator which we can take out
-	~~& \sqrt{~~2~~}\left( \sqrt[4]{3} \right)~~^{3}=2^{{1}/~~{2}\;}\left(~~ 3~~^{{1~~}/~~{4}\;} \right)^{~~3~~}=2~~^{{1~~}/{2}\;}~~3^{{3}/{4}~~\;}~~, \\	+
-	~~& \sqrt[3]{~~2~~}\centerdot~~ 3~~=2^{{~~1}/{3~~}\;}~~3~~^{1}~~ \\	+
-	~~\end{align}~~</math>	+
-		+

-	~~Admittedly, it is true that~~	+	{{Displayed math\|\|<math>\begin{align}
-	<math>2^{{1}/{2}\;}>2^{{1}/{3}\;}~~</math>~~	+	2^{1/2}3^{3/4}
-	~~and~~	+	&= 2^{6/12}3^{(3\cdot 3)/12}
-	~~<math>3~~^{1}>3^{~~{3}/{4~~}\~~;}</math>, but this does not help us to say anything about how the products are related to each other. Instead, we observe that the exponents~~	+	= \bigl(2^{6}\cdot 3^{9}\bigr)^{1/12},\\[5pt]
-	~~<math>\frac~~{1~~}{2~~},\ \ ~~\frac~~{3}{4}~~,\ \ \frac~~{1}{3}</~~math>~~	+	2^{1/3}3^{1}
-	~~and~~	+	&= 2^{4/12}3^{12/12}
-	~~<math>~~\~~text~~{1}~~</math>~~	+	= \bigl(2^{4}\cdot 3^{12}\bigr)^{1/12}\,\textrm{.}
-	~~have~~	+	\end{align}</math>}}
-	~~<math>~~\~~text~~{3}\~~centerdot~~ \~~text~~{4}=\~~text~~{12 }</math>	+
-	~~as the lowest common denominator which we can take out:~~	+

-		+	Now, we can compare the bases <math>2^6\cdot 3^9</math> and <math>2^4\cdot 3^{12}</math> with each other and so decide which number is larger.
-	~~<math>\begin{align}~~	+
-	~~& 2^{\frac{1}{2}}3^{\frac{3}{4}}=2^{\frac{6}{12}}3^{\frac{3\centerdot 3}{12}}=\left( 2^{6}\centerdot 3^{9} \right)^{\frac{1}{12}}, \\~~	+
-	~~& 2^{\frac{1}{3}}3^{1}=2^{\frac{4}{12}}3^{\frac{12}{12}}=\left( 2^{4}\centerdot 3^{12} \right)^{\frac{1}{12}}. \\~~	+
-	~~\end{align}</math>~~	+
-		+
-	Now, we can compare the bases	+
-	<math>~~\text{~~2}^~~{\text{~~6}}\~~centerdot \text{~~3}^~~{\text{~~9}}</math>	+
-	and	+
-	<math>~~\text{~~2}^~~{\text{~~4}}\~~centerdot \text{~~3}^~~{\text~~{12}}</math>	+
-	with each other and so decide which number is larger.	+

	Because		Because

		+	{{Displayed math\|\|<math>\frac{2^6\cdot 3^9}{2^4\cdot 3^{12}} = 2^{6-4}3^{9-12} = 2^{2}3^{-3} = \frac{2^{2}}{3^{3}} = \frac{4}{27} < 1</math>}}

-	~~<math>\frac{\text{2}^{\text{6}}\centerdot \text{3}^{\text{9}}}{\text{2}^{\text{4}}\centerdot \text{3}^{\text{12}}}=2^{6-4}3^{9-12}=2^{2}3^{-3}=\frac{2^{2}}{3^{3}}=\frac{4}{27}<1</math>~~	+	the denominator <math>2^{4}\cdot 3^{12}</math> is larger than the numerator
-		+	<math>2^6\cdot 3^9</math>, which means that <math>\sqrt[3]{2}\cdot 3</math>
-	the denominator	+	is larger than <math>\sqrt{2}\bigl(\sqrt[4]{3}\bigr)^{3}</math>.
-	<math>~~\text{~~2}^~~{\text~~{4}}\~~centerdot \text{~~3}^~~{\text~~{12}}</math>	+
-	is larger than the numerator	+
-	<math>~~\text{~~2}^~~{\text{~~6}}\~~centerdot \text{~~3}^~~{\text{~~9}}</math>, which means that	+
-	<math>\sqrt[3]{2}\~~centerdot~~ 3</math>	+
-	is larger than	+
-	<math>\sqrt{2}\~~left~~( \sqrt[4]{3} \~~right~~)^{3}</math>.	+

Ian at 11:39, 23 September 2008

2008-09-23T11:39:58Z

←Older revision		Revision as of 11:39, 23 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	In power form, the expressions become
-	<~~center~~> ~~[[Image~~:~~3_1_8d-~~1(2).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>\begin{align}
-	<~~center~~> ~~[[Image:3_1_8d~~-2(2~~).gif]~~] </~~center~~>	+	& \sqrt{2}\left( \sqrt[4]{3} \right)^{3}=2^{{1}/{2}\;}\left( 3^{{1}/{4}\;} \right)^{3}=2^{{1}/{2}\;}3^{{3}/{4}\;}, \\
-	{{~~NAVCONTENT_STOP~~}}	+	& \sqrt[3]{2}\centerdot 3=2^{{1}/{3}\;}3^{1} \\
		+	\end{align}</math>
		+
		+
		+	Admittedly, it is true that
		+	<math>2^{{1}/{2}\;}>2^{{1}/{3}\;}</math>
		+	and
		+	<math>3^{1}>3^{{3}/{4}\;}</math>, but this does not help us to say anything about how the products are related to each other. Instead, we observe that the exponents
		+	<math>\frac{1}{2},\ \ \frac{3}{4},\ \ \frac{1}{3}</math>
		+	and
		+	<math>\text{1}</math>
		+	have
		+	<math>\text{3}\centerdot \text{4}=\text{12 }</math>
		+	as the lowest common denominator which we can take out:
		+
		+
		+	<math>\begin{align}
		+	& 2^{\frac{1}{2}}3^{\frac{3}{4}}=2^{\frac{6}{12}}3^{\frac{3\centerdot 3}{12}}=\left( 2^{6}\centerdot 3^{9} \right)^{\frac{1}{12}}, \\
		+	& 2^{\frac{1}{3}}3^{1}=2^{\frac{4}{12}}3^{\frac{12}{12}}=\left( 2^{4}\centerdot 3^{12} \right)^{\frac{1}{12}}. \\
		+	\end{align}</math>
		+
		+	Now, we can compare the bases
		+	<math>\text{2}^{\text{6}}\centerdot \text{3}^{\text{9}}</math>
		+	and
		+	<math>\text{2}^{\text{4}}\centerdot \text{3}^{\text{12}}</math>
		+	with each other and so decide which number is larger.
		+
		+	Because
		+
		+
		+	<math>\frac{\text{2}^{\text{6}}\centerdot \text{3}^{\text{9}}}{\text{2}^{\text{4}}\centerdot \text{3}^{\text{12}}}=2^{6-4}3^{9-12}=2^{2}3^{-3}=\frac{2^{2}}{3^{3}}=\frac{4}{27}<1</math>
		+
		+	the denominator
		+	<math>\text{2}^{\text{4}}\centerdot \text{3}^{\text{12}}</math>
		+	is larger than the numerator
		+	<math>\text{2}^{\text{6}}\centerdot \text{3}^{\text{9}}</math>, which means that
		+	<math>\sqrt[3]{2}\centerdot 3</math>
		+	is larger than
		+	<math>\sqrt{2}\left( \sqrt[4]{3} \right)^{3}</math>.

Tekbot: Lösning 3.1:8d moved to Solution 3.1:8d: Robot: moved page

2008-09-09T09:17:03Z

Lösning 3.1:8d moved to Solution 3.1:8d: Robot: moved page

←Older revision

Revision as of 09:17, 9 September 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

2008-08-21T06:55:39Z

Robot: Automated text replacement (-[[Bild: +[[Image:)

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2008-04-01T14:16:32Z

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