Solution 1.3:6f - Revision history

Tek at 07:33, 23 September 2008

2008-09-23T07:33:43Z

←Older revision		Revision as of 07:33, 23 September 2008
Line 1:		Line 1:
-	We can factorize the exponents	+	We can factorize the exponents 40 and 56 as
-	~~<math>~~40~~</math>~~	+
-	and	+
-	~~<math>~~56~~</math>~~	+
-	as	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	40 &= 4\cdot 10 = 2\cdot 2\cdot 2\cdot 5 = 2^{3}\cdot 5 \\[3pt]
		+	56 &= 7\cdot 8 = 7\cdot 2\cdot 4 = 7\cdot 2\cdot 2\cdot 2 = 2^{3}\cdot 7
		+	\end{align}</math>}}

-	<math>~~\begin{align}~~	+	and we then see that they have <math>2^{3} = 8</math> as a common factor. We can take this factor out as an "outer" exponent
-	~~& 40=4\centerdot 10=2\centerdot 2\centerdot 2\centerdot 5=~~2^{3}~~\centerdot 5 \\~~	+
-	~~& \\~~	+
-	~~& 56~~=~~7\centerdot~~ 8~~=7\centerdot 2\centerdot 4=7\centerdot 2\centerdot 2\centerdot 2=2^{3}\centerdot 7 \\~~	+
-	~~\end{align}~~</math>	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	3^{40} &= 3^{5\cdot 8} = \bigl(3^{5}\bigr)^{8} = (3\cdot 3\cdot 3\cdot 3\cdot 3)^{8} = 243^{8}\,,\\[3pt]
		+	2^{56} &= 2^{7\cdot 8} = \bigl(2^{7}\bigr)^{8} = (2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2)^{8} = 128^{8}\,\textrm{.}
		+	\end{align}</math>}}

-	~~and we then see that they have~~	+	This shows that <math>3^{40} = 243^{8}</math> is bigger than <math>2^{56} = 128^{8}</math>.
-	~~<math>2^{3}=8</math>~~	+
-	~~as a common factor. We can take this factor out as an "outer" exponent:~~	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& 3^{40}=3^{5\centerdot 8}=\left( 3^{5} \right)^{8}=\left( 3\centerdot 3\centerdot 3\centerdot 3\centerdot 3 \right)^{8}=243^{8} \\~~	+
-	~~& \\~~	+
-	~~& 2^{56}=2^{7\centerdot 8}=\left( 2^{7} \right)^{8}=\left( 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2 \right)^{8}=128^{8} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	This shows that	+
-	<math>3^{40}=243^{8}</math>	+
-	is bigger than	+
-	<math>2^{56}=128^{8}</math>	+

Ian at 13:01, 15 September 2008

2008-09-15T13:01:37Z

←Older revision		Revision as of 13:01, 15 September 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START~~}}	+	We can factorize the exponents
-	<~~center~~> ~~[[Image:1_3_6f~~.~~gif]]~~ </~~center~~>	+	<math>40</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
		+	<math>56</math>
		+	as
		+
		+
		+	<math>\begin{align}
		+	& 40=4\centerdot 10=2\centerdot 2\centerdot 2\centerdot 5=2^{3}\centerdot 5 \\
		+	& \\
		+	& 56=7\centerdot 8=7\centerdot 2\centerdot 4=7\centerdot 2\centerdot 2\centerdot 2=2^{3}\centerdot 7 \\
		+	\end{align}</math>
		+
		+
		+	and we then see that they have
		+	<math>2^{3}=8</math>
		+	as a common factor. We can take this factor out as an "outer" exponent:
		+
		+
		+	<math>\begin{align}
		+	& 3^{40}=3^{5\centerdot 8}=\left( 3^{5} \right)^{8}=\left( 3\centerdot 3\centerdot 3\centerdot 3\centerdot 3 \right)^{8}=243^{8} \\
		+	& \\
		+	& 2^{56}=2^{7\centerdot 8}=\left( 2^{7} \right)^{8}=\left( 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2 \right)^{8}=128^{8} \\
		+	\end{align}</math>
		+
		+
		+	This shows that
		+	<math>3^{40}=243^{8}</math>
		+	is bigger than
		+	<math>2^{56}=128^{8}</math>

Tekbot: Lösning 1.3:6f moved to Solution 1.3:6f: Robot: moved page

2008-09-09T08:30:03Z

Lösning 1.3:6f moved to Solution 1.3:6f: Robot: moved page

←Older revision

Revision as of 08:30, 9 September 2008

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

2008-08-21T06:32:09Z

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

←Older revision		Revision as of 06:32, 21 August 2008
Line 1:		Line 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:1_3_6f.gif]] </center>	+	<center> [[Image:1_3_6f.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Lina: Ny sida: {{NAVCONTENT_START}} Bild:1_3_6f.gif {{NAVCONTENT_STOP}}

2008-03-26T15:21:59Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:1_3_6f.gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:1_3_6f.gif]] </center>
{{NAVCONTENT_STOP}}