Solution 3.3:3f - Revision history

Tek at 06:56, 2 October 2008

Tek — Thu, 02 Oct 2008 06:56:15 GMT

←Older revision		Revision as of 06:56, 2 October 2008
Line 1:		Line 1:
-	If we write	+	If we write 4 and 16 as
-	~~<math>\text{~~4~~}</math>~~	+
-	and	+
-	~~<math>\text{~~16~~}</math>~~	+
-	as	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \text{4}=2\centerdot 2=2^{2} \\~~	+
-	~~& 16=2\centerdot 8=2\centerdot 2\centerdot 4=2\centerdot 2\centerdot 2\centerdot 2=2^{4} \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	4 &= 2\cdot 2 = 2^2\,,\\[5pt]
		+	16 &= 2\cdot 8 = 2\cdot 2\cdot 4 = 2\cdot 2\cdot 2\cdot 2 = 2^4\,,
		+	\end{align}</math>}}

	we obtain		we obtain

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\log_2 4 + \log_2\frac{1}{16}
-	& \~~log _{2}~~4+\~~log _{2}~~\frac{1}{16}=\~~log _{2}~~2^{2}+\~~log _{2}~~\frac{1}{2^{4}} \\	+	&= \log_2 2^2 + \log_2\frac{1}{2^4}\\[5pt]
-	& =\~~log _{2}~~2^{2}+\~~log _{2}~~2^{-4}=2\~~centerdot~~ \~~log _{2}~~2+~~\left~~( -4 ~~\right~~)\~~centerdot~~ \~~log _{2}~~2 \\	+	&= \log_2 2^2 + \log_2 2^{-4}\\[5pt]
-	& =2\~~centerdot~~ 1+~~\left~~( -4 ~~\right~~)\~~centerdot~~ 1=-2 \\	+	&= 2\cdot\log_2 2 + (-4)\cdot\log_2 2\\[5pt]
-	\end{align}</math>	+	&= 2\cdot 1 + (-4)\cdot 1\\[5pt]
		+	&= -2\,\textrm{.}
		+	\end{align}</math>}}

Ian — Thu, 25 Sep 2008 14:28:03 GMT

←Older revision		Revision as of 14:28, 25 September 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START}~~}	+	If we write
-	<~~center~~> ~~[[Image:3_3_3f.gif]]~~ </~~center~~>	+	<math>\text{4}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
		+	<math>\text{16}</math>
		+	as
		+
		+
		+	<math>\begin{align}
		+	& \text{4}=2\centerdot 2=2^{2} \\
		+	& 16=2\centerdot 8=2\centerdot 2\centerdot 4=2\centerdot 2\centerdot 2\centerdot 2=2^{4} \\
		+	\end{align}</math>
		+
		+
		+	we obtain
		+
		+
		+	<math>\begin{align}
		+	& \log _{2}4+\log _{2}\frac{1}{16}=\log _{2}2^{2}+\log _{2}\frac{1}{2^{4}} \\
		+	& =\log _{2}2^{2}+\log _{2}2^{-4}=2\centerdot \log _{2}2+\left( -4 \right)\centerdot \log _{2}2 \\
		+	& =2\centerdot 1+\left( -4 \right)\centerdot 1=-2 \\
		+	\end{align}</math>

Tekbot — Tue, 09 Sep 2008 09:25:03 GMT

←Older revision

Revision as of 09:25, 9 September 2008

Tekbot — Thu, 21 Aug 2008 06:59:39 GMT

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

Oskar — Wed, 02 Apr 2008 09:41:58 GMT

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