3.4 Logarithmic equations - Revision history

Phil: Some changes to language, notation and terminology. NB: I've replaced "first and second order" w/ "linear and quadratic", throughout, but it should in any case have read "first and second degree".

2009-01-02T17:08:36Z

Some changes to language, notation and terminology. NB: I've replaced "first and second order" w/ "linear and quadratic", throughout, but it should in any case have read "first and second degree".

(Difference between revisions)

Jdinglis at 22:29, 28 October 2008

2008-10-28T22:29:57Z

←Older revision		Revision as of 22:29, 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.
	</div>		</div>

Jdinglis at 22:28, 28 October 2008

2008-10-28T22:28:32Z

←Older revision		Revision as of 22:28, 28 October 2008
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	== Spurious roots ==		== Spurious roots ==

-	When you solve equations you should also bear in mind that the arguments of logarithms have to be positive and that terms of the type <math>e^{(\ldots)}</math> can only have positive values. ~~We may generate spurious roots, so~~ we must ~~check~~ that our answer makes sense.	+	When you solve equations you should also bear in mind that the arguments of logarithms have to be positive and that terms of the type <math>e^{(\ldots)}</math> can only have positive values. In other words we must be careful to make sure that our answer makes sense.

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

Jdinglis at 22:26, 28 October 2008

2008-10-28T22:26:02Z

←Older revision		Revision as of 22:26, 28 October 2008
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	{{Displayed math\|\|<math> \Rightarrow (\ln x)^2 + \ln x - 1 = 0\,\mbox{.}</math>}}		{{Displayed math\|\|<math> \Rightarrow (\ln x)^2 + \ln x - 1 = 0\,\mbox{.}</math>}}

-	Completing the square on the left-hand side	+	Completing the square on the left-hand side we see that

	{{Displayed math\|\|<math>\begin{align*}		{{Displayed math\|\|<math>\begin{align*}
	\textstyle (\ln x)^2 + \ln x -1		\textstyle (\ln x)^2 + \ln x -1
	&= \bigl( \ln x + \frac{1}{2} \bigr)^2 - \bigl(\frac{1}{2} \bigr)^2 - 1\\		&= \bigl( \ln x + \frac{1}{2} \bigr)^2 - \bigl(\frac{1}{2} \bigr)^2 - 1\\
-	&= \bigl( \ln x + \frac{1}{2} \bigr)^2 - \frac{5}{4}\\	+	&= \bigl( \ln x + \frac{1}{2} \bigr)^2 - \frac{5}{4} \mbox{.}\\
	\end{align*}</math>}}		\end{align*}</math>}}

-	~~We continue~~ by taking ~~the root giving~~	+	Then by taking roots,

	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
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	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
	x= e^{(-1 + \sqrt{5})/2}		x= e^{(-1 + \sqrt{5})/2}
-	\quad \mbox{~~och~~} \quad	+	\quad \mbox{or} \quad
	x= e^{-(1+\sqrt{5})/2}\,\mbox{.}</math>}}		x= e^{-(1+\sqrt{5})/2}\,\mbox{.}</math>}}
	</div>		</div>
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	== Spurious roots ==		== Spurious roots ==

-	When you solve equations you should also bear in mind that the arguments of logarithms have to be positive and that terms of the type <math>e^{(\ldots)}</math> can only have positive values. ~~The risk is otherwise that you get~~ spurious roots.	+	When you solve equations you should also bear in mind that the arguments of logarithms have to be positive and that terms of the type <math>e^{(\ldots)}</math> can only have positive values. We may generate spurious roots, so we must check that our answer makes sense.

	<div class="exempel">		<div class="exempel">
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	<br>		<br>
	<br>		<br>
-	For the equation to be satisfied the arguments <math>4x^2-2x</math> and <math>1-2x</math> must be equal,	+	For the equation to be satisfied the arguments <math>4x^2-2x</math> and <math>1-2x</math> must be be positive and equal i.e.
-	{{Displayed math\|\|<math>4x^2 - 2x = 1 - 2x\,,</math>\|<math>(*)</math>}}	+	{{Displayed math\|\|<math>4x^2 - 2x = 1 - 2x\mbox{.}\,,</math>\|<math>(*)</math>}}

-	~~and also be positive.~~ We solve the equation <math>(*)</math> by moving all of the terms to one side	+	We solve the equation <math>(*)</math> by moving all of the terms to one side

	{{Displayed math\|\|<math>4x^2 - 1= 0</math>}}		{{Displayed math\|\|<math>4x^2 - 1= 0</math>}}

-	and ~~take~~ the root. This gives ~~that~~	+	and taking the root. This gives

	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
	\textstyle x= -\frac{1}{2}		\textstyle x= -\frac{1}{2}
-	\quad\mbox{~~and~~}\quad	+	\quad\mbox{or}\quad
	x = \frac{1}{2} \; \mbox{.}</math>}}		x = \frac{1}{2} \; \mbox{.}</math>}}

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	<br>		<br>
	<br>		<br>
-	The first term can be written as <math>e^{2x} = (e^x)^2</math>. The whole equation is a quadratic with <math>e^x</math> as the unknown	+	The first term can be written as <math>e^{2x} = (e^x)^2</math>. The whole equation is a quadratic with <math>e^x</math> as the unknown i.e.
	{{Displayed math\|\|<math>(e^x)^2 - e^x = \tfrac{1}{2}\,\mbox{.}</math>}}		{{Displayed math\|\|<math>(e^x)^2 - e^x = \tfrac{1}{2}\,\mbox{.}</math>}}

-	The equation can be a little easier to manage if we write <math>t</math> instead of <math>e^x</math>,	+	The equation can be a little easier to manage if we write <math>t</math> instead of <math>e^x</math>, so that we try and solve

	{{Displayed math\|\|<math>t^2 -t = \tfrac{1}{2}\,\mbox{.}</math>}}		{{Displayed math\|\|<math>t^2 -t = \tfrac{1}{2}\,\mbox{.}</math>}}

-	~~Complete~~ the square ~~for~~ the left-hand side.	+	Completing the square on the left-hand side gives

	{{Displayed math\|\|<math>\begin{align*}		{{Displayed math\|\|<math>\begin{align*}
	\textstyle \bigl(t-\frac{1}{2}\bigr)^2 - \bigl(\frac{1}{2}\bigr)^2		\textstyle \bigl(t-\frac{1}{2}\bigr)^2 - \bigl(\frac{1}{2}\bigr)^2
-	&= \frac{1}{2}\,~~\mbox{,}~~\\	+	&= \frac{1}{2}\,\\
-	\bigl(t-\frac{1}{2}\bigr)^2	+	\Rightarrow \bigl(t-\frac{1}{2}\bigr)^2
-	&= \frac{3}{4}\,\mbox{,}\\	+	&= \frac{3}{4}\,\mbox{.}\\
	\end{align*}</math>}}		\end{align*}</math>}}

-	~~which gives solutions~~	+	so that

	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
	t=\frac{1}{2} - \frac{\sqrt{3}}{2}		t=\frac{1}{2} - \frac{\sqrt{3}}{2}
-	\quad\mbox{~~and~~}\quad	+	\quad\mbox{or}\quad
	t=\frac{1}{2} + \frac{\sqrt{3}}{2} \, \mbox{.}</math>}}		t=\frac{1}{2} + \frac{\sqrt{3}}{2} \, \mbox{.}</math>}}

-	Since <math>\sqrt3 > 1</math> ~~then~~ <math>\frac{1}{2}-\frac{1}{2}\sqrt3 <0</math> ~~and~~ it is only <math>t= \frac{1}{2}+\frac{1}{2}\sqrt3</math> that provides a solution to the original equation because <math>e^x</math> is always positive. Taking logarithms finally gives ~~that~~	+	Since <math>\sqrt3 > 1</math>, <math>\frac{1}{2}-\frac{1}{2}\sqrt3 <0</math>. Therefore it is only <math>t= \frac{1}{2}+\frac{1}{2}\sqrt3</math> that provides a solution to the original equation because <math>e^x</math> is always positive. Taking logarithms finally gives

	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>

Jdinglis at 21:35, 28 October 2008

2008-10-28T21:35:03Z

Jdinglis at 21:30, 28 October 2008

2008-10-28T21:30:40Z

←Older revision		Revision as of 21:30, 28 October 2008
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	Take logarithms of both sides and use the laws of logarithms to get		Take logarithms of both sides and use the laws of logarithms to get
	{{Displayed math\|\|<math>\eqalign{(2x+1)\lg 5 &= 5x \cdot \lg 3\,\cr \Rightarrow 2x \cdot \lg 5 + \lg 5 &= 5x \cdot \lg 3\,\mbox{.}\cr}</math>}}		{{Displayed math\|\|<math>\eqalign{(2x+1)\lg 5 &= 5x \cdot \lg 3\,\cr \Rightarrow 2x \cdot \lg 5 + \lg 5 &= 5x \cdot \lg 3\,\mbox{.}\cr}</math>}}
-	~~Collect~~ <math>x</math> to one side	+	Collecting <math>x</math> to one side gives
-	{{Displayed math\|\|<math>\eqalign{\lg 5 &= 5x \cdot \lg 3 -2x \cdot \lg 5\,\~~mbox{,}~~\cr \lg 5 &= x\,(5 \lg 3 -2 \lg 5)\,\mbox{.}\cr}</math>}}	+	{{Displayed math\|\|<math>\eqalign{\lg 5 &= 5x \cdot \lg 3 -2x \cdot \lg 5\,\cr \Rightarrow \lg 5 &= x\,(5 \lg 3 -2 \lg 5)\,\mbox{.}\cr}</math>}}
-	The solution is	+	The solution is then
	{{Displayed math\|\|<math>x = \frac{\lg 5}{5 \lg 3 -2 \lg 5}\,\mbox{.}</math>}}</li>		{{Displayed math\|\|<math>x = \frac{\lg 5}{5 \lg 3 -2 \lg 5}\,\mbox{.}</math>}}</li>
	</ol>		</ol>

Jdinglis at 21:28, 28 October 2008

2008-10-28T21:28:55Z

←Older revision		Revision as of 21:28, 28 October 2008
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	{{Displayed math\|\|<math>\begin{align*}		{{Displayed math\|\|<math>\begin{align*}
	10^x = y\quad&\Leftrightarrow\quad x = \lg y\\		10^x = y\quad&\Leftrightarrow\quad x = \lg y\\
-	e^x = y\quad&\Leftrightarrow\quad x = \ln y \\	+	e^x = y\quad&\Leftrightarrow\quad x = \ln y \mbox{.}\\
-	\end{align*}.</math>}}	+	\end{align*}</math>}}

	We consider only 10-logarithms or natural logarithms, though the methods can just as easily be applied in the case of logarithms with an arbitrary base.		We consider only 10-logarithms or natural logarithms, though the methods can just as easily be applied in the case of logarithms with an arbitrary base.

Jdinglis at 21:28, 28 October 2008

2008-10-28T21:28:01Z

←Older revision		Revision as of 21:28, 28 October 2008
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	{{Displayed math\|\|<math>\begin{align*}		{{Displayed math\|\|<math>\begin{align*}
	10^x = y\quad&\Leftrightarrow\quad x = \lg y\\		10^x = y\quad&\Leftrightarrow\quad x = \lg y\\
-	e^x = y\quad&\Leftrightarrow\quad x = \ln y ~~\rm{.}~~\\	+	e^x = y\quad&\Leftrightarrow\quad x = \ln y \\
-	\end{align*}</math>}}	+	\end{align*}.</math>}}

	We consider only 10-logarithms or natural logarithms, though the methods can just as easily be applied in the case of logarithms with an arbitrary base.		We consider only 10-logarithms or natural logarithms, though the methods can just as easily be applied in the case of logarithms with an arbitrary base.

Jdinglis at 21:27, 28 October 2008

2008-10-28T21:27:07Z

←Older revision		Revision as of 21:27, 28 October 2008
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	{{Displayed math\|\|<math>\begin{align*}		{{Displayed math\|\|<math>\begin{align*}
	10^x = y\quad&\Leftrightarrow\quad x = \lg y\\		10^x = y\quad&\Leftrightarrow\quad x = \lg y\\
-	e^x = y\quad&\Leftrightarrow\quad x = \ln y.\\	+	e^x = y\quad&\Leftrightarrow\quad x = \ln y \rm{.}\\
	\end{align*}</math>}}		\end{align*}</math>}}

Jdinglis at 21:26, 28 October 2008

2008-10-28T21:26:38Z

←Older revision		Revision as of 21:26, 28 October 2008
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	{{Displayed math\|\|<math>\begin{align*}		{{Displayed math\|\|<math>\begin{align*}
	10^x = y\quad&\Leftrightarrow\quad x = \lg y\\		10^x = y\quad&\Leftrightarrow\quad x = \lg y\\
-	e^x = y\quad&\Leftrightarrow\quad x = \ln y\\	+	e^x = y\quad&\Leftrightarrow\quad x = \ln y.\\
-	\end{align*}</math>.}}	+	\end{align*}</math>}}

	We consider only 10-logarithms or natural logarithms, though the methods can just as easily be applied in the case of logarithms with an arbitrary base.		We consider only 10-logarithms or natural logarithms, though the methods can just as easily be applied in the case of logarithms with an arbitrary base.

←Older revision		Revision as of 21:35, 28 October 2008
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	<br>		<br>
	<br>		<br>
-	Multiply both sides by <math>3e^x+1</math> and <math>e^{-x}+2</math> to eliminate the denominators	+	Multiply both sides by <math>3e^x+1</math> and <math>e^{-x}+2</math> to eliminate the denominators, so that

	{{Displayed math\|\|<math>6e^x(e^{-x}+2) = 5(3e^x+1)\,\mbox{.}</math>}}		{{Displayed math\|\|<math>6e^x(e^{-x}+2) = 5(3e^x+1)\,\mbox{.}</math>}}

-	~~Note that since <math>e^x</math> and <math>e^{-x}</math> are always positive regardless of the value of <math>x</math>, in~~ this ~~latest~~ step we have multiplied the equation by factors <math>3e^x+1</math> and <math>e^{-x} +2</math>. Both of these factors are different from zero, so this step cannot introduce new (spurious) roots of the equation.	+	In this last step we have multiplied the equation by factors <math>3e^x+1</math> and <math>e^{-x} +2</math>. Both of these factors are different from zero, so this step cannot introduce new (spurious) roots of the equation.

-	Simplify both sides of the equation	+	Simplify both sides of the equation to get
-	{{Displayed math\|\|<math>6+12e^x = 15e^x+5\,\mbox{,}</math>}}	+	{{Displayed math\|\|<math>6+12e^x = 15e^x+5\,\mbox{.}</math>}}
-	~~where~~ we used <math>e^{-x} \cdot e^x = e^{-x + x} = e^0 = 1</math>. If we treat <math>e^x</math> as the unknown variable, the equation is essentially a first order equation which has a solution	+	Here we have used <math>e^{-x} \cdot e^x = e^{-x + x} = e^0 = 1</math>. If we treat <math>e^x</math> as the unknown variable, the equation is essentially a first order equation which has a solution
	{{Displayed math\|\|<math>e^x=\frac{1}{3}\,\mbox{.}</math>}}		{{Displayed math\|\|<math>e^x=\frac{1}{3}\,\mbox{.}</math>}}

-	Taking logarithms then gives the answer	+	Taking logarithms then gives the answer:
	{{Displayed math\|\|<math>x=\ln\frac{1}{3}= \ln 3^{-1} = -1 \cdot \ln 3 = -\ln 3\,\mbox{.}</math>}}		{{Displayed math\|\|<math>x=\ln\frac{1}{3}= \ln 3^{-1} = -1 \cdot \ln 3 = -\ln 3\,\mbox{.}</math>}}
	</div>		</div>
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	<br>		<br>
	The term <math>\ln\frac{1}{x}</math> can be written as <math>\ln\frac{1}{x} = \ln x^{-1} = -1 \cdot \ln x = - \ln x</math> and then the equation becomes		The term <math>\ln\frac{1}{x}</math> can be written as <math>\ln\frac{1}{x} = \ln x^{-1} = -1 \cdot \ln x = - \ln x</math> and then the equation becomes
-	{{Displayed math\|\|<math>\frac{1}{\ln x} - \ln x = 1\,\mbox{,}</math>}}	+	{{Displayed math\|\|<math>\frac{1}{\ln x} - \ln x = 1\,\mbox{.}</math>}}
-	~~where we can consider <math>\ln x</math> as a new unknown.~~ We multiply both sides by <math>\ln x</math> (which is different from zero when <math>x \neq 1</math>) and this gives us a quadratic equation in <math>\ln x</math>	+	We multiply both sides by <math>\ln x</math> (which is different from zero when <math>x \neq 1</math>, and <math>x = 1</math> is clearly not a solution) and this gives us a quadratic equation in <math>\ln x</math>:
-	{{Displayed math\|\|<math>1 - (\ln x)^2 = \ln x\,~~\mbox{,}~~</math>}}	+	{{Displayed math\|\|<math>1 - (\ln x)^2 = \ln x\,</math>}}
-	{{Displayed math\|\|<math> (\ln x)^2 + \ln x - 1 = 0\,\mbox{.}</math>}}	+	{{Displayed math\|\|<math> \Rightarrow (\ln x)^2 + \ln x - 1 = 0\,\mbox{.}</math>}}

	Completing the square on the left-hand side		Completing the square on the left-hand side