3.3 Logarithms - Revision history

Jdinglis at 22:30, 28 October 2008

Jdinglis — Tue, 28 Oct 2008 22:30:16 GMT

←Older revision		Revision as of 22:30, 28 October 2008
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	You may need to spend some time studying logarithms.		You may need to spend some time studying logarithms.

-	Logarithms ~~usually~~ are not usually dealt with in detail in high school. Therefore, many college students tend to encounter problems when it comes to calculations with logarithms.	+	Logarithms are not usually dealt with in detail in high school. Therefore, many college students tend to encounter problems when it comes to calculations with logarithms.

Jdinglis at 20:44, 28 October 2008

Jdinglis — Tue, 28 Oct 2008 20:44:47 GMT

←Older revision		Revision as of 20:44, 28 October 2008
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	</div>		</div>

-	The general formula for changing from one base <math>a</math> to another base <math>b</math> ~~can be derived in the same way~~	+	The general formula for changing from one base <math>a</math> to another base <math>b</math> is

	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
	\log_{\scriptstyle\,a} x		\log_{\scriptstyle\,a} x
	= \frac{\log_{\scriptstyle\, b} x}{\log_{\scriptstyle\, b} a}		= \frac{\log_{\scriptstyle\, b} x}{\log_{\scriptstyle\, b} a}
-	\,\mbox{.}</math>}}	+	\,\mbox{,}</math>}}

-	~~If one wants to~~ change the base of a power~~, one can do this by~~ using logarithms. For instance, if we want to write <math> 2^5 </math> using the base 10 ~~one~~ first ~~writes~~ 2 as a power with the base 10;	+	and can be derived in the same way. We can also change the base of a power using logarithms. For instance, if we want to write <math> 2^5 </math> using the base 10 first write 2 as a power with the base 10:
-	{{Displayed math\|\|<math>2 = 10^{\lg 2}</math>}}	+	{{Displayed math\|\|<math>2 = 10^{\lg 2}</math>.}}

-	~~and then~~ using one of the laws of exponents	+	Then, using one of the laws of exponents,
	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
	2^5 = (10^{\lg 2})^5 = 10^{5\cdot \lg 2}		2^5 = (10^{\lg 2})^5 = 10^{5\cdot \lg 2}
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	<br>		<br>
	First, we write 10 as a power of ''e'',		First, we write 10 as a power of ''e'',
-	{{Displayed math\|\|<math>10 = e^{\ln 10}</math>}}	+	{{Displayed math\|\|<math>10 = e^{\ln 10}</math>.}}

-	~~and then use~~ the laws of exponents	+	Using the the laws of exponents we can then see that
	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
	10^x = (e^{\ln 10})^x = e^{\,x \cdot \ln 10}		10^x = (e^{\ln 10})^x = e^{\,x \cdot \ln 10}
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-	'''Keep in mind that: '''	+	'''Keep in mind that... '''

-	You may need to spend ~~much~~ time studying logarithms.	+	You may need to spend some time studying logarithms.

-	Logarithms usually are dealt with ~~summarily~~ in high school. Therefore, many college students tend to encounter problems when it comes to calculations with logarithms.	+	Logarithms usually are not usually dealt with in detail in high school. Therefore, many college students tend to encounter problems when it comes to calculations with logarithms.


	'''Reviews'''		'''Reviews'''

-	For those of you who want to deepen your ~~studies~~ or need more detailed explanations consider the following references:	+	For those of you who want to deepen your understanding or need more detailed explanations consider the following references:

-	[http://en.wikipedia.org/wiki/Logarithm Learn more about logarithms ~~in English~~ Wikipedia ]	+	[http://en.wikipedia.org/wiki/Logarithm Learn more about logarithms on Wikipedia ]

	[http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/e.html Learn more about the number ''e'' in The MacTutor History of Mathematics archive ]		[http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/e.html Learn more about the number ''e'' in The MacTutor History of Mathematics archive ]

Jdinglis at 20:38, 28 October 2008

Jdinglis — Tue, 28 Oct 2008 20:38:14 GMT

←Older revision		Revision as of 20:38, 28 October 2008
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	{{Displayed math\|\|<math>10^{\lg 5} = 5\,\mbox{.}</math>}}		{{Displayed math\|\|<math>10^{\lg 5} = 5\,\mbox{.}</math>}}

-	~~Take~~ the natural logarithm ( ln ) of both sides.	+	Taking the natural logarithm ( <math>\ln</math>) of both sides yields
	{{Displayed math\|\|<math>\ln 10^{\lg 5} = \ln 5\,\mbox{.}</math>}}		{{Displayed math\|\|<math>\ln 10^{\lg 5} = \ln 5\,\mbox{.}</math>}}

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	{{Displayed math\|\|<math>\lg 5 \cdot \ln 10 = \ln 5\,\mbox{.}</math>}}		{{Displayed math\|\|<math>\lg 5 \cdot \ln 10 = \ln 5\,\mbox{.}</math>}}

-	Now divide both sides by <math>\ln 10</math> ~~giving~~ the answer	+	Now divide both sides by <math>\ln 10</math> to get the answer
	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
	\lg 5 = \frac{\ln 5}{\ln 10}		\lg 5 = \frac{\ln 5}{\ln 10}

Jdinglis at 20:36, 28 October 2008

Jdinglis — Tue, 28 Oct 2008 20:36:20 GMT

←Older revision		Revision as of 20:36, 28 October 2008
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	<br>		<br>
	<br>		<br>
-	By definition, ~~the~~ <math>\lg 5</math> is a number that satisfies the equality	+	By definition, <math>\lg 5</math> is a number that satisfies the equality
	{{Displayed math\|\|<math>10^{\lg 5} = 5\,\mbox{.}</math>}}		{{Displayed math\|\|<math>10^{\lg 5} = 5\,\mbox{.}</math>}}

Jdinglis at 20:35, 28 October 2008

Jdinglis — Tue, 28 Oct 2008 20:35:01 GMT

←Older revision		Revision as of 20:35, 28 October 2008
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	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
	a\cdot b = 10^{\textstyle\log a} \cdot 10^{\textstyle\log b}		a\cdot b = 10^{\textstyle\log a} \cdot 10^{\textstyle\log b}
-	= \left\{ \mbox{laws of exponents} \right\}
	= 10^{\,\bbox[#AAEEFF,1pt]{\,\log a+\log b\,}}</math>}}		= 10^{\,\bbox[#AAEEFF,1pt]{\,\log a+\log b\,}}</math>}}

Jdinglis at 20:33, 28 October 2008

Jdinglis — Tue, 28 Oct 2008 20:33:09 GMT

←Older revision		Revision as of 20:33, 28 October 2008
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	==The natural logarithms ==		==The natural logarithms ==

-	In practice there are two bases that are commonly used for logarithms, 10 and the number <math>e</math> <math>({}\approx 2\textrm{.}71828 \ldots\,)</math>. Logarithms using the base ''e'' are called '' natural logarithms'' and we use the notation <math>ln</math> instead of <math>\log_{\,e}</math>.	+	In practice there are two bases that are commonly used for logarithms, 10 and the number <math>e</math> <math>({}\approx 2\textrm{.}71828 \ldots\,)</math>. Logarithms using the base ''e'' are called '' natural logarithms'' and we use the notation <math>\ln</math> instead of <math>\log_{\,e}</math>.

	<div class="exempel">		<div class="exempel">

Jdinglis at 20:32, 28 October 2008

Jdinglis — Tue, 28 Oct 2008 20:32:28 GMT

←Older revision		Revision as of 20:32, 28 October 2008
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	</div>		</div>

-	If the base 10 is used, one rarely writes <math>\log_{\,10}</math>, but as we have previously seen we use the notation <math> lg </math>, or simply <math>log</math>. These symbols appear on many calculators. We should also note that it makes no sense to define logarithms with base <math>1</math>.	+	If the base 10 is used, one rarely writes <math>\log_{\,10}</math>, but as we have previously seen we use the notation <math> \lg </math>, or simply <math>\log</math>. These symbols appear on many calculators. We should also note that it makes no sense to define logarithms with base <math>1</math>.

Jdinglis at 20:30, 28 October 2008

Jdinglis — Tue, 28 Oct 2008 20:30:10 GMT

←Older revision		Revision as of 20:30, 28 October 2008
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	:::"the exponent of 0.01 is -2".		:::"the exponent of 0.01 is -2".

-	This is how logarithms are defined. We ~~formalises~~ this as follows:	+	This is how logarithms are defined. We formalise this as follows:

	:::"''The logarithm'' of 1000 is 3",		:::"''The logarithm'' of 1000 is 3",

Jdinglis at 20:28, 28 October 2008

Jdinglis — Tue, 28 Oct 2008 20:28:34 GMT

←Older revision		Revision as of 20:28, 28 October 2008
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	*To calculate simple logarithmic expressions using the definition of a logarithm.		*To calculate simple logarithmic expressions using the definition of a logarithm.
	*That logarithms are only defined for positive numbers.		*That logarithms are only defined for positive numbers.
-	* The ~~meaning~~ of the number <math>e</math>.	+	* The value of the number <math>e</math>.
	* To use the laws of logarithms to simplify logarithmic expressions.		* To use the laws of logarithms to simplify logarithmic expressions.
	* To know when the laws of logarithms are valid.		* To know when the laws of logarithms are valid.

Jdinglis at 20:08, 26 October 2008

Jdinglis — Sun, 26 Oct 2008 20:08:21 GMT

←Older revision		Revision as of 20:08, 26 October 2008
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	{{Displayed math\|\|<math>10^{\ \bbox[#AAEEFF,2pt]{\,x\,}} = y\,\mbox{.} </math>}}		{{Displayed math\|\|<math>10^{\ \bbox[#AAEEFF,2pt]{\,x\,}} = y\,\mbox{.} </math>}}

-	Note that <math>y</math> must be a positive number for the logarithm <math>\lg y</math> to be defined, since there is no power of 10 that evaluates to a negative number or for that matter zero .	+	Note that <math>y</math> must be a positive number for the logarithm <math>\lg y</math> to be defined, since there is no power of 10 that evaluates to a negative number, or for that matter zero .

	<div class="exempel">		<div class="exempel">
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	</div>		</div>

-	In the penultimate example, one can easily understand that <math>\lg 50</math> must lie somewhere between 1 and 2 since <math>10^1 < 50 < 10^2</math>, ~~but~~ to obtain a more precise value of the irrational number <math>\lg 50 = 1\textrm{.}69897\ldots</math> one needs ~~in practice,~~ a calculator (or table.)	+	In the penultimate example, one can easily understand that <math>\lg 50</math> must lie somewhere between 1 and 2 since <math>10^1 < 50 < 10^2</math>. However, in practice to obtain a more precise value of the irrational number <math>\lg 50 = 1\textrm{.}69897\ldots</math> one needs a calculator (or table).

	<div class="exempel">		<div class="exempel">
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	== Different bases ==		== Different bases ==

-	~~One~~ can ~~imagine~~ logarithms~~, which use~~ a base other than 10 ~~(except 1!)~~. ~~One~~ must clearly indicate which number is used as a base for a logarithm. If ~~one uses~~ a base such as 2 ~~one uses the notation~~ <math>\log_{\,2}</math> ~~for a "~~base-2 ~~logarithm"~~.	+	We can define logarithms with respect to a base other than 10. We must clearly indicate which number is used as the base for the logarithm. If we use a base such as 2 we write <math>\log_{\,2}</math> to mean a logarithm with base 2.

	<div class="exempel">		<div class="exempel">
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	</div>		</div>

-	~~One deals~~ with logarithms which have other bases in the same way.	+	We deal with logarithms which have other bases in the same way.

	<div class="exempel">		<div class="exempel">
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	</div>		</div>

-	If the base 10 is used, one rarely writes <math>\log_{\,10}</math>, but as we have previously seen ~~one uses~~ the notation lg, or simply log~~, which appears~~ on many calculators.	+	If the base 10 is used, one rarely writes <math>\log_{\,10}</math>, but as we have previously seen we use the notation <math> lg </math>, or simply <math>log</math>. These symbols appear on many calculators. We should also note that it makes no sense to define logarithms with base <math>1</math>.


	==The natural logarithms ==		==The natural logarithms ==

-	In practice there are two bases that are commonly used for logarithms, 10 and the number <math>e</math> <math>({}\approx 2\textrm{.}71828 \ldots\,)</math>. Logarithms using the base ''e'' are called '' natural logarithms'' and ~~one uses~~ the notation ln instead of <math>\log_{\,e}</math>.	+	In practice there are two bases that are commonly used for logarithms, 10 and the number <math>e</math> <math>({}\approx 2\textrm{.}71828 \ldots\,)</math>. Logarithms using the base ''e'' are called '' natural logarithms'' and we use the notation <math>ln</math> instead of <math>\log_{\,e}</math>.

	<div class="exempel">		<div class="exempel">
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	<ol type="a">		<ol type="a">
-	<li><math> \ln 10 \approx 2{,}3\quad</math> because <math>	+	<li><math> \ln 10 \approx 2{\textrm{.}}3\quad</math> because <math>
-	e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2{,}3\,}} \approx 10</math>.</li>	+	e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2{\textrm{.}}3\,}} \approx 10</math>.</li>
	<li><math> \ln e = 1\quad</math> because <math>		<li><math> \ln e = 1\quad</math> because <math>
	e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = e</math>.</li>		e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = e</math>.</li>
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	</div>		</div>

-	Most advanced calculators ~~usually~~ have buttons for 10-logarithms and natural logarithms.	+	Most advanced calculators have buttons for 10-logarithms and natural logarithms.


	== Laws of Logarithms ==		== Laws of Logarithms ==

-	Between the years 1617 and 1624 Henry Biggs published a table of logarithms for all integers up to 20 000, and in 1628 Adriaan Vlacq expanded the table for all integers up to 100 000. The reason such an enormous amount of work was invested in producing these tables is that with the help of logarithms one can multiply numbers together just by adding their logarithms (addition ~~goes~~ much faster ~~to perform~~ than multiplication).	+	Between the years 1617 and 1624 Henry Biggs published a table of logarithms for all integers up to 20 000, and in 1628 Adriaan Vlacq expanded the table for all integers up to 100 000. The reason such an enormous amount of work was invested in producing these tables is that with the help of logarithms one can multiply numbers together just by adding their logarithms (addition is a much faster calculation than multiplication).

	<div class="exempel">		<div class="exempel">
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	35 \cdot 54 \approx 10^{\,1\textrm{.}5441} \cdot 10^{\,1\textrm{.}7324}		35 \cdot 54 \approx 10^{\,1\textrm{.}5441} \cdot 10^{\,1\textrm{.}7324}
	= 10^{\,1\textrm{.}5441 + 1\textrm{.}7324}		= 10^{\,1\textrm{.}5441 + 1\textrm{.}7324}
-	= 10^{\,3\textrm{.}2765}</math>}}	+	= 10^{\,3\textrm{.}2765}</math>.}}

-	~~and we then know that~~ <math>10^{\,3\textrm{.}2765} \approx 1890</math> (i.e. <math>\lg 1890 \approx 3\textrm{.}2765</math>) ~~thus~~ we have managed to calculate the product	+	Since <math>10^{\,3\textrm{.}2765} \approx 1890</math> (i.e. <math>\lg 1890 \approx 3\textrm{.}2765</math>), we have thus managed to calculate the product

	{{Displayed math\|\|<math>35 \cdot 54 = 1890</math>}}		{{Displayed math\|\|<math>35 \cdot 54 = 1890</math>}}

-	~~and this~~ just by adding together exponents <math>1\textrm{.}5441</math> and <math>1\textrm{.}7324</math>.	+	just by adding together the exponents <math>1\textrm{.}5441</math> and <math>1\textrm{.}7324</math>.
	</div>		</div>

-	~~This is an~~ example of a logarithmic law which ~~says~~ that	+	In the above example we have used a logarithmic law which states that

-	{{Displayed math\|\|<math>\log (ab) = \log a + \log b</math>}}	+	{{Displayed math\|\|<math>\log (ab) = \log a + \log b</math>,}}

-	~~This~~ ~~stems from the fact~~ that on the ~~one hand~~,	+	for all <math>a,b>0</math>. We can see that this is true by using the laws of exponents. Indeed,

	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
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	a\cdot b = 10^{\,\bbox[#AAEEFF,1pt]{\,\log (ab)\,}}\,\mbox{.}</math>}}		a\cdot b = 10^{\,\bbox[#AAEEFF,1pt]{\,\log (ab)\,}}\,\mbox{.}</math>}}

-	By exploiting the laws of exponents in this way ~~can~~ we obtain the corresponding ''laws of logarithms'':	+	so that we can see <math>\log (ab) = \log a + \log b</math>. By exploiting the laws of exponents in this way we can obtain the corresponding ''laws of logarithms'':

	<div class="regel">		<div class="regel">
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	\log(ab) &= \log a + \log b,\\[4pt]		\log(ab) &= \log a + \log b,\\[4pt]
	\log\frac{a}{b} &= \log a - \log b,\\[4pt]		\log\frac{a}{b} &= \log a - \log b,\\[4pt]
-	\log a^b &= b\cdot \log a\,\mbox{.}\\	+	\log a^b &= b\cdot \log a\,\mbox{,}\\
	\end{align*}</math>}}		\end{align*}</math>}}
	</div>		</div>
-		+	for <math>a,b>0</math>. These laws hold regardless of the base we are working with.
-	~~The~~ laws ~~of logarithms apply~~ regardless of base.	+

	<div class="exempel">		<div class="exempel">
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	<ol type="a">		<ol type="a">
-	<li><math>\lg 9 + \lg 1000 - \lg 3 + \lg 0{,}001 = \lg 9 + 3 - \lg 3 - 3	+	<li><math>\lg 9 + \lg 1000 - \lg 3 + \lg 0\textrm{.}001 = \lg 9 + 3 - \lg 3 - 3
	= \lg 9- \lg 3 = \lg \displaystyle \frac{9}{3} = \lg 3</math></li>		= \lg 9- \lg 3 = \lg \displaystyle \frac{9}{3} = \lg 3</math></li>
	<li><math>\ln\frac{1}{e} + \ln \sqrt{e}		<li><math>\ln\frac{1}{e} + \ln \sqrt{e}
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	== Changing the base ==		== Changing the base ==

-	It sometimes ~~can be~~ a good idea to express a logarithm as a logarithm ~~having~~ another base.	+	It is sometimes a good idea to express a logarithm as a logarithm with respect to another base.

	<div class="exempel">		<div class="exempel">
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	{{Displayed math\|\|<math>\ln 10^{\lg 5} = \ln 5\,\mbox{.}</math>}}		{{Displayed math\|\|<math>\ln 10^{\lg 5} = \ln 5\,\mbox{.}</math>}}

-	With the help of the logarithm law <math>\ln a^b = b \ln a</math> the left-hand side can be written as <math>\lg 5 \cdot \ln 10</math> and the equality becomes	+	With the help of the logarithm law <math>\ln a^b = b \ln a</math>, the left-hand side can be written as <math>\lg 5 \cdot \ln 10</math> and the equality becomes
	{{Displayed math\|\|<math>\lg 5 \cdot \ln 10 = \ln 5\,\mbox{.}</math>}}		{{Displayed math\|\|<math>\lg 5 \cdot \ln 10 = \ln 5\,\mbox{.}</math>}}

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	<br>		<br>
	Using the definition of a logarithm one has that <math>\log_2 100</math> formally satisfies		Using the definition of a logarithm one has that <math>\log_2 100</math> formally satisfies
-	{{Displayed math\|\|<math>2^{\log_{\scriptstyle 2} 100} = 100</math>}}	+	{{Displayed math\|\|<math>2^{\log_{\scriptstyle 2} 100} = 100</math>.}}

-	~~and taking~~ the 10-logarithm ~~(lg)~~ of both sides, ~~one gets~~	+	Taking the 10-logarithm of both sides, we get
	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
	\lg 2^{\log_{\scriptstyle 2} 100} = \lg 100\,\mbox{.}</math>}}		\lg 2^{\log_{\scriptstyle 2} 100} = \lg 100\,\mbox{.}</math>}}

-	Since <math>\lg a^b = b \lg a</math> ~~one gets~~ <math>\lg 2^{\log_2 100} = \log_{\scriptstyle 2} 100 \cdot \lg 2</math> and the right-hand side can be simplified to <math>\lg 100 = 2</math>. ~~This gives the equality~~	+	Since <math>\lg a^b = b \lg a</math> we get <math>\lg 2^{\log_2 100} = \log_{\scriptstyle 2} 100 \cdot \lg 2</math>, and moreover the right-hand side can be simplified to <math>\lg 100 = 2</math>. Thus we see that
	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
	\log_{\scriptstyle 2} 100 \cdot \lg 2 = 2\,\mbox{.}</math>}}		\log_{\scriptstyle 2} 100 \cdot \lg 2 = 2\,\mbox{.}</math>}}

-	Finally, dividing by <math>\lg 2</math> gives ~~that~~	+	Finally, dividing by <math>\lg 2</math> gives
	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
	\log_{\scriptstyle 2} 100 = \frac{2}{\lg 2}		\log_{\scriptstyle 2} 100 = \frac{2}{\lg 2}