Solution 3.3:6a - Revision history

Tek at 07:53, 2 October 2008

Tek — Thu, 02 Oct 2008 07:53:14 GMT

←Older revision		Revision as of 07:53, 2 October 2008
Line 1:		Line 1:
-	The calculator does not have button for	+	The calculator does not have button for <math>\log_{3}</math>, but it does have one for the natural logarithm ln, so we need to rewrite <math>\log_{3}4</math> in terms of ln.
-	<math>\~~log _~~{3}</math>, but it does have one for the natural logarithm ln, so we need to rewrite	+
-	<math>\~~log _~~{3}4</math>	+
-	in terms of ln.	+

-	If we go back to the definition of the logarithm, we see that	+	If we go back to the definition of the logarithm, we see that <math>\log _{3}4</math> is that number which satisfies
-	<math>\log _{3}4</math>	+
-	is that number which satisfies	+
-		+
-		+
-	~~<math>3^{\log _{3}4}=4</math>~~	+

		+	{{Displayed math\|\|<math>3^{\log _{3}4} = 4\,\textrm{.}</math>}}

	Now, take the natural logarithm of both sides,		Now, take the natural logarithm of both sides,

		+	{{Displayed math\|\|<math>\ln 3^{\log _{3}4}=\ln 4\,\textrm{.}</math>}}

-	~~<math>\ln 3^{\log _{3}4}=\ln 4</math>~~	+	Using the logarithm law, <math>\lg a^b = b\lg a</math>, the left-hand side can be written as <math>\log_{3}4\cdot\ln 3</math> and the relation is
-		+
-		+
-	Using the logarithm law,	+
-	<math>\lg a^{b}=b\lg a</math>, the left-hand side can be written as	+
-	<math>\~~log _~~{3}4\~~centerdot~~ \ln 3</math>	+
-	and the relation is	+
-		+
-		+
-	~~<math>\log _{3}4\centerdot \ln 3=\ln 4</math>~~	+
-		+
-		+
-	~~Thus, after dividing by~~	+
-	~~<math>\text{ln 3}</math>, we have~~	+

		+	{{Displayed math\|\|<math>\log_{3}4\cdot \ln 3 = \ln 4\,\textrm{.}</math>}}

-	<math>~~\log _{3}4=\frac{\ln 4}{~~\ln 3~~}=\frac{1.386294...}{1.098612...}=1.2618595~~</math>	+	Thus, after dividing by <math>\ln 3</math>, we have

		+	{{Displayed math\|\|<math>\log_{3}4 = \frac{\ln 4}{\ln 3} = \frac{1\textrm{.}386294\,\ldots}{1\textrm{.}098612\,\ldots} = 1\textrm{.}2618595\,\ldots</math>}}

	which gives 1.262 as the rounded-off answer.		which gives 1.262 as the rounded-off answer.

-	NOTE: on a calculator, the answer is obtained by pressing the buttons

		+	Note: On the calculator, the answer is obtained by pressing the buttons

-	<~~math~~>~~\left[~~ 4 ~~\right]\quad \left[ \text~~{LN} ~~\right]\quad \left[ \div \right]\quad \left[~~ 3 ~~\right]\quad \left[ \text~~{LN} ~~\right]\quad \left[~~ = ~~\right]~~</~~math~~>	+	<center>
		+	{\|
		+	\|\|
		+	{\| border="1" cellpadding="3" cellspacing="0"
		+	\|width="30px" align="center"\|4
		+	\|}
		+	\|\|
		+	\|\|
		+	{\| border="1" cellpadding="3" cellspacing="0"
		+	\|width="30px" align="center"\|LN
		+	\|}
		+	\|\|
		+	\|\|
		+	{\| border="1" cellpadding="3" cellspacing="0"
		+	\|width="30px" align="center"\|÷
		+	\|}
		+	\|\|
		+	\|\|
		+	{\| border="1" cellpadding="3" cellspacing="0"
		+	\|width="30px" align="center"\|3
		+	\|}
		+	\|\|
		+	\|\|
		+	{\| border="1" cellpadding="3" cellspacing="0"
		+	\|width="30px" align="center"\|LN
		+	\|}
		+	\|\|
		+	\|\|
		+	{\| border="1" cellpadding="3" cellspacing="0"
		+	\|width="30px" align="center"\|=
		+	\|}
		+	\|}
		+	</center>

Ian at 09:25, 26 September 2008

Ian — Fri, 26 Sep 2008 09:25:11 GMT

←Older revision		Revision as of 09:25, 26 September 2008
Line 39:		Line 39:


-	<math>~~\begin{align}~~	+	<math>\left[ 4 \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ = \right]</math>
-	& \left[ 4 \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ = \right] \\	+
-	~~& \quad \left[ 4 \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ = \right] \\~~	+
-	~~\end{align}~~</math>	+

Ian at 09:08, 26 September 2008

Ian — Fri, 26 Sep 2008 09:08:20 GMT

←Older revision		Revision as of 09:08, 26 September 2008
Line 1:		Line 1:
-	{~~{NAVCONTENT_START}~~}	+	The calculator does not have button for
-	<~~center~~> ~~[[Image:3_3_6a-1(2)~~.~~gif]]~~ </~~center~~>	+	<math>\log _{3}</math>, but it does have one for the natural logarithm ln, so we need to rewrite
-	{{~~NAVCONTENT_STOP~~}}	+	<math>\log _{3}4</math>
-	{{~~NAVCONTENT_START~~}}	+	in terms of ln.
-	<~~center~~> ~~[[Image:3_3_6a~~-~~2(2)~~.~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	If we go back to the definition of the logarithm, we see that
-	[[~~Image:3_3_6_a.gif\|center~~]]	+	<math>\log _{3}4</math>
		+	is that number which satisfies
		+
		+
		+	<math>3^{\log _{3}4}=4</math>
		+
		+
		+	Now, take the natural logarithm of both sides,
		+
		+
		+	<math>\ln 3^{\log _{3}4}=\ln 4</math>
		+
		+
		+	Using the logarithm law,
		+	<math>\lg a^{b}=b\lg a</math>, the left-hand side can be written as
		+	<math>\log _{3}4\centerdot \ln 3</math>
		+	and the relation is
		+
		+
		+	<math>\log _{3}4\centerdot \ln 3=\ln 4</math>
		+
		+
		+	Thus, after dividing by
		+	<math>\text{ln 3}</math>, we have
		+
		+
		+	<math>\log _{3}4=\frac{\ln 4}{\ln 3}=\frac{1.386294...}{1.098612...}=1.2618595</math>
		+
		+
		+	which gives 1.262 as the rounded-off answer.
		+
		+	NOTE: on a calculator, the answer is obtained by pressing the buttons
		+
		+
		+	<math>\begin{align}
		+	& \left[ 4 \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ = \right] \\
		+	& \quad \left[ 4 \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ = \right] \\
		+	\end{align}</math>

Tekbot: Lösning 3.3:6a moved to Solution 3.3:6a: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:29:03 GMT

Lösning 3.3:6a moved to Solution 3.3:6a: Robot: moved page

←Older revision

Revision as of 09:29, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 07:01:39 GMT

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

←Older revision		Revision as of 07:01, 21 August 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_3_6a-1(2).gif]] </center>	+	<center> [[Image:3_3_6a-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_3_6a-2(2).gif]] </center>	+	<center> [[Image:3_3_6a-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
-	[[~~Bild~~:3_3_6_a.gif\|center]]	+	[[Image:3_3_6_a.gif\|center]]

Madde at 12:15, 15 August 2008

Madde — Fri, 15 Aug 2008 12:15:20 GMT

←Older revision		Revision as of 12:15, 15 August 2008
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	<center> [[Bild:3_3_6a-2(2).gif]] </center>		<center> [[Bild:3_3_6a-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+	[[Bild:3_3_6_a.gif\|center]]

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_3_6a-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:3_3_6a-2(2).gif {{NAVCONTENT_STOP}}

Oskar — Wed, 02 Apr 2008 09:55:55 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:3_3_6a-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:3_3_6a-2(2).gif </center> {{NAVCONTENT_STOP}}

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