Solution 1.3:5f - Revision history

Tek at 14:42, 22 September 2008

Tek — Mon, 22 Sep 2008 14:42:53 GMT

←Older revision		Revision as of 14:42, 22 September 2008
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-	The whole expression is quite complicated, so it can be useful to simplify the terms	+	The whole expression is quite complicated, so it can be useful to simplify the terms <math>\bigl(125^{\frac{1}{3}}\bigr)^{2}</math> and <math>\bigl(27^{\frac{1}{3}}\bigr)^{-2}</math> first,
-	<math>\~~left~~( 125^{\frac{1}{3}} \~~right~~)^{2}</math>	+
-	and	+
-	<math>\~~left~~( 27^{\frac{1}{3}} \~~right~~)^{-2}</math>	+
-	first:	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\bigl(125^{\frac{1}{3}}\bigr)^{2} &= 125^{\frac{1}{3}\cdot 2} = 125^{\frac{2}{3}}\,,\\[5pt]
		+	\bigl(27^{\frac{1}{3}}\bigr)^{-2} &= 27^{\frac{1}{3}\cdot (-2)} = 27^{-\frac{2}{3}}\,\textrm{.}\end{align}</math>}}

-	~~<math>\left( 125^{\frac{1}{3}} \right)^{2}=125^{\frac{1}{3}\centerdot 2}=125^{\frac{2}{3}}</math>~~	+	Then, the bases 125, 27 and 9 can be rewritten as
-		+
-		+
-		+
-	~~<math>\left( 27^{\frac{1}{3}} \right)^{-2}=27^{\frac{1}{3}\centerdot \left( -2 \right)}=27^{-\frac{2}{3}}</math>~~	+
-		+
-		+
-	Then, the bases	+
-	~~<math>~~125,~~\ \~~ 27~~</math>~~	+
-	and	+
-	~~<math>~~9~~</math>~~	+
-	can be rewritten as	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& 125=5\centerdot 25=5\centerdot 5\centerdot 5=5^{3} \\~~	+
-	~~& \\~~	+
-	~~& 27=3\centerdot 9=3\centerdot 3\centerdot 3=3^{3} \\~~	+
-	~~& \\~~	+
-	~~& 9=3\centerdot 3=3^{2} \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	125 &= 5\cdot 25 = 5\cdot 5\cdot 5 = 5^{3},\\
		+	27 &= 3\cdot 9 = 3\cdot 3\cdot 3 = 3^{3},\\
		+	9 &= 3\cdot 3 = 3^{2}\textrm{.}
		+	\end{align}</math>}}

	With the help of the power rules,		With the help of the power rules,

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\bigl(125^{\frac{1}{3}}\bigr)^{2}\cdot\bigl(27^{\frac{1}{3}}\bigr)^{-2}\cdot 9^{\frac{1}{2}} &= 125^{\frac{2}{3}}\cdot 27^{-\frac{2}{3}}\cdot 9^{\frac{1}{2}}\\[5pt]
-	& \~~left~~( 125^{\frac{1}{3}} \~~right~~)^{2}\~~centerdot~~ \~~left~~( 27^{\frac{1}{3}} \~~right~~)^{-2}\~~centerdot~~ 9^{\frac{1}{2}}=125^{\frac{2}{3}}\~~centerdot~~ 27^{-\frac{2}{3}}\~~centerdot~~ 9^{\frac{1}{2}} \\	+	&= \bigl(5^{3}\bigr)^{\frac{2}{3}}\cdot \bigl(3^{3}\bigr)^{-\frac{2}{3}}\cdot \bigl(3^{2}\bigr)^{\frac{1}{2}}\\[5pt]
-	~~& \\~~	+	&= 5^{3\cdot\frac{2}{3}}\cdot 3^{3\cdot (-\frac{2}{3})}\cdot 3^{2\cdot\frac{1}{2}}\\[5pt]
-	& =\~~left~~( 5^{3} \~~right~~)^{\frac{2}{3}}\~~centerdot~~ \~~left~~( 3^{3} \~~right~~)^{-\frac{2}{3}}\~~centerdot~~ \~~left~~( 3^{2} \~~right~~)^{\frac{1}{2}}=5^{3\~~centerdot~~ \frac{2}{3}}\~~centerdot~~ 3^{3\~~centerdot \left~~( -\frac{2}{3} ~~\right~~)}\~~centerdot~~ 3^{2\~~centerdot~~ \frac{1}{2}} \\	+	&= 5^{2}\cdot 3^{-2}\cdot 3^{1}\\[5pt]
-	~~& \\~~	+	&= 5^{2}\cdot 3^{-2+1}\\[5pt]
-	& =5^{2}\~~centerdot~~ 3^{-2}\~~centerdot~~ 3^{1}=5^{2}\~~centerdot~~ 3^{-2+1}=5^{2}\~~centerdot~~ 3^{-1}=5\~~centerdot~~ 5\~~centerdot~~ \frac{1}{3} \\	+	&= 5^{2}\cdot 3^{-1}\\[5pt]
-	& \\	+	&= 5\cdot 5\cdot \frac{1}{3}\\[5pt]
-	& \frac{25}{3} \\	+	&= \frac{25}{3}\,\textrm{.}
-	\end{align}</math>	+	\end{align}</math>}}

Ian at 12:49, 15 September 2008

Ian — Mon, 15 Sep 2008 12:49:17 GMT

←Older revision		Revision as of 12:49, 15 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	The whole expression is quite complicated, so it can be useful to simplify the terms
-	<~~center~~> ~~[[Image~~:~~1_3_5f.gif]]~~ </~~center~~>	+	<math>\left( 125^{\frac{1}{3}} \right)^{2}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
		+	<math>\left( 27^{\frac{1}{3}} \right)^{-2}</math>
		+	first:
		+
		+
		+	<math>\left( 125^{\frac{1}{3}} \right)^{2}=125^{\frac{1}{3}\centerdot 2}=125^{\frac{2}{3}}</math>
		+
		+
		+
		+	<math>\left( 27^{\frac{1}{3}} \right)^{-2}=27^{\frac{1}{3}\centerdot \left( -2 \right)}=27^{-\frac{2}{3}}</math>
		+
		+
		+	Then, the bases
		+	<math>125,\ \ 27</math>
		+	and
		+	<math>9</math>
		+	can be rewritten as
		+
		+
		+	<math>\begin{align}
		+	& 125=5\centerdot 25=5\centerdot 5\centerdot 5=5^{3} \\
		+	& \\
		+	& 27=3\centerdot 9=3\centerdot 3\centerdot 3=3^{3} \\
		+	& \\
		+	& 9=3\centerdot 3=3^{2} \\
		+	\end{align}</math>
		+
		+
		+	With the help of the power rules,
		+
		+
		+	<math>\begin{align}
		+	& \left( 125^{\frac{1}{3}} \right)^{2}\centerdot \left( 27^{\frac{1}{3}} \right)^{-2}\centerdot 9^{\frac{1}{2}}=125^{\frac{2}{3}}\centerdot 27^{-\frac{2}{3}}\centerdot 9^{\frac{1}{2}} \\
		+	& \\
		+	& =\left( 5^{3} \right)^{\frac{2}{3}}\centerdot \left( 3^{3} \right)^{-\frac{2}{3}}\centerdot \left( 3^{2} \right)^{\frac{1}{2}}=5^{3\centerdot \frac{2}{3}}\centerdot 3^{3\centerdot \left( -\frac{2}{3} \right)}\centerdot 3^{2\centerdot \frac{1}{2}} \\
		+	& \\
		+	& =5^{2}\centerdot 3^{-2}\centerdot 3^{1}=5^{2}\centerdot 3^{-2+1}=5^{2}\centerdot 3^{-1}=5\centerdot 5\centerdot \frac{1}{3} \\
		+	& \\
		+	& \frac{25}{3} \\
		+	\end{align}</math>

Tekbot: Lösning 1.3:5f moved to Solution 1.3:5f: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:28:03 GMT

Lösning 1.3:5f moved to Solution 1.3:5f: Robot: moved page

←Older revision

Revision as of 08:28, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:31:09 GMT

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

←Older revision		Revision as of 06:31, 21 August 2008
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Lina: Ny sida: {{NAVCONTENT_START}} Bild:1_3_5f.gif {{NAVCONTENT_STOP}}

Lina — Wed, 26 Mar 2008 15:11:30 GMT

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