3.4 Logarithmic equations

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{{Info|
{{Info|
'''Innehåll:'''
'''Innehåll:'''
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* Logaritmekvationer
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* Logarithmic Equations
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* Exponentialekvationer
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* Exponential equations
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* Falska rötter.
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* Spurious roots
}}
}}
{{Info|
{{Info|
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'''Lärandemål:'''
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'''Learning outcomes::'''
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Efter detta avsnitt ska du ha lärt dig att:
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After this section, you will have learned to:
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* Lösa ekvationer som innehåller logaritm- eller exponentialuttryck och som kan reduceras till första- eller andragradsekvationer.
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* Solve equations that contain logarithm or exponential expressions and which can be reduced to first or second order equations.
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* Hantera falska rötter och veta när de uppstår.
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* Deal with spurious roots, and know when they arise.
}}
}}
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== Grundekvationer ==
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== Basic Equations ==
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Ekvationer där logaritmer behövs eller är inblandade förekommer i många olika fall. Först ges några exempel där lösningen ges nästan direkt genom definitionen av logaritm, dvs.
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Equations where logarithms appear can vary a lot. Here are some examples where the solution is given almost immediately by the definition of a logarithm, that is.
{{Fristående formel||<math>\begin{align*}
{{Fristående formel||<math>\begin{align*}
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\end{align*}</math>}}
\end{align*}</math>}}
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(Vi använder oss här enbart av 10-logaritmer eller naturliga logaritmer.)
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(We consider only 10-logarithms or natural logarithms.)
<div class="exempel">
<div class="exempel">
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'''Exempel 1'''
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''' Example 1'''
Lös ekvationerna
Lös ekvationerna
<ol type="a">
<ol type="a">
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<li><math>10^x = 537\quad</math> har lösningen <math>x = \lg 537</math>.</li>
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<li><math>10^x = 537\quad</math> has a solution <math>x = \lg 537</math>.</li>
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<li><math>10^{5x} = 537\quad</math> ger att <math>5x
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<li><math>10^{5x} = 537\quad</math> gives <math>5x
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= \lg 537</math>, dvs. <math>x=\frac{1}{5} \lg 537</math>.</li>
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= \lg 537</math>, i.e. <math>x=\frac{1}{5} \lg 537</math>.</li>
<li><math>\frac{3}{e^x} = 5 \quad
<li><math>\frac{3}{e^x} = 5 \quad
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</math> Multiplikation av båda led med <math>e^x
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</math> Multiplication of both sides with <math>e^x
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</math> och division med 5 ger att <math>\tfrac{3}{5}=e^x
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</math> and division by 5 gives <math>\tfrac{3}{5}=e^x
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</math>, vilket betyder att <math>x=\ln\tfrac{3}{5}</math>.</li>
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</math>, which means that <math>x=\ln\tfrac{3}{5}</math>.</li>
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<li><math>\lg x = 3 \quad</math> Definitionen ger direkt att <math>
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<li><math>\lg x = 3 \quad</math> The definition gives directly <math>
x=10^3 = 1000</math>.</li>
x=10^3 = 1000</math>.</li>
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<li><math>\lg(2x-4) = 2 \quad</math> Från definitionen har vi att <math>
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<li><math>\lg(2x-4) = 2 \quad</math> From the definition we have <math>
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2x-4 = 10^2 = 100</math> och då följer att <math>x = 52</math>.</li>
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2x-4 = 10^2 = 100</math> and it follows that <math>x = 52</math>.</li>
</ol>
</ol>
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 2'''
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''' Example 2'''
<ol type="a">
<ol type="a">
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<li> Lös ekvationen <math>\,(\sqrt{10}\,)^x = 25</math>.
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<li> Solve the equation <math>\,(\sqrt{10}\,)^x = 25</math>.
<br>
<br>
<br>
<br>
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Eftersom <math>\sqrt{10} = 10^{1/2}</math> är vänsterledet lika med <math>(\sqrt{10}\,)^x = (10^{1/2})^x = 10^{x/2}</math> och ekvationen lyder
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Since <math>\sqrt{10} = 10^{1/2}</math> the left-hand side is equal to <math>(\sqrt{10}\,)^x = (10^{1/2})^x = 10^{x/2}</math> and the equation becomes
{{Fristående formel||<math>10^{x/2} = 25\,\mbox{.}</math>}}
{{Fristående formel||<math>10^{x/2} = 25\,\mbox{.}</math>}}
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Denna grundekvation har lösningen <math>\frac{x}{2} = \lg 25</math>, dvs. <math>x = 2 \lg 25</math>.</li>
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This equation has a solution <math>\frac{x}{2} = \lg 25</math>, ie. <math>x = 2 \lg 25</math>.</li>
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<li>Lös ekvationen <math>\,\frac{3 \ln 2x}{2} + 1 = \frac{1}{2}</math>.
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<li>Solve the equation <math>\,\frac{3 \ln 2x}{2} + 1 = \frac{1}{2}</math>.
<br>
<br>
<br>
<br>
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Multiplicera båda led med 2 och subtrahera sedan 2 från båda led
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Multiply both sides by 2 and then subtracting 2 from both sides
{{Fristående formel||<math> 3 \ln 2x = -1\,\mbox{.}</math>}}
{{Fristående formel||<math> 3 \ln 2x = -1\,\mbox{.}</math>}}
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Dividera båda led med 3
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Divide both sides by 3
{{Fristående formel||<math> \ln 2x = -\frac{1}{3}\,\mbox{.}</math>}}
{{Fristående formel||<math> \ln 2x = -\frac{1}{3}\,\mbox{.}</math>}}
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Nu ger definitionen direkt att <math>2x = e^{-1/3}</math>, vilket betyder att
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Now, the definition directly gives <math>2x = e^{-1/3}</math>, which means that
{{Fristående formel||<math> x = {\textstyle\frac{1}{2}} e^{-1/3} = \frac{1}{2e^{1/3}}\,\mbox{.} </math>}}</li>
{{Fristående formel||<math> x = {\textstyle\frac{1}{2}} e^{-1/3} = \frac{1}{2e^{1/3}}\,\mbox{.} </math>}}</li>
</ol>
</ol>
</div>
</div>
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I många praktiska tillämpningar rörande exponentiell tillväxt eller avtagande dyker det upp ekvationer av typen
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In many practical applications of exponential growth or decline there appear equations of the type
{{Fristående formel||<math>a^x = b\,\mbox{,}</math>}}
{{Fristående formel||<math>a^x = b\,\mbox{,}</math>}}
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där <math>a</math> och <math>b</math> är positiva tal. Dessa ekvationer löses enklast genom att ta logaritmen för båda led
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where <math>a</math> and <math>b</math> are positive numbers. These equations are best solved by taking the logarithm of both sides
{{Fristående formel||<math>\lg a^x = \lg b</math>}}
{{Fristående formel||<math>\lg a^x = \lg b</math>}}
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och använda logaritmlagen för potenser
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and use the law of logarithms for powers
{{Fristående formel||<math>x \cdot \lg a = \lg b</math>}}
{{Fristående formel||<math>x \cdot \lg a = \lg b</math>}}
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vilket ger lösningen <math>\ x = \displaystyle \frac{\lg b}{\lg a}</math>.
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which gives the solution <math>\ x = \displaystyle \frac{\lg b}{\lg a}</math>.
<div class="exempel">
<div class="exempel">
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'''Exempel 3'''
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''' Example 3'''
<ol type="a">
<ol type="a">
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<li>Lös ekvationen <math>\,3^x = 20</math>.
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<li>Solve the equation <math>\,3^x = 20</math>.
<br>
<br>
<br>
<br>
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Logaritmera båda led
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Take logarithms of both sides
{{Fristående formel||<math>\lg 3^x = \lg 20\,\mbox{.}</math>}}
{{Fristående formel||<math>\lg 3^x = \lg 20\,\mbox{.}</math>}}
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Vänsterledet kan skrivas som <math>\lg 3^x = x \cdot \lg 3</math> och då får vi att
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The left-hand side can be written as <math>\lg 3^x = x \cdot \lg 3</math> giving
{{Fristående formel||<math>x = \displaystyle \frac{\lg 20}{\lg 3} \quad ({}\approx 2{,}727)\,\mbox{.}</math>}}</li>
{{Fristående formel||<math>x = \displaystyle \frac{\lg 20}{\lg 3} \quad ({}\approx 2{,}727)\,\mbox{.}</math>}}</li>
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<li>Lös ekvationen <math>\ 5000 \cdot 1{,}05^x = 10\,000</math>.
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<li>Solve the equation <math>\ 5000 \cdot 1{,}05^x = 10\,000</math>.
<br>
<br>
<br>
<br>
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Dividera båda led med 5000
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Divide both sides by 5000
{{Fristående formel||<math>1{,}05^x = \displaystyle \frac{ 10\,000}{5\,000} = 2\,\mbox{.}</math>}}
{{Fristående formel||<math>1{,}05^x = \displaystyle \frac{ 10\,000}{5\,000} = 2\,\mbox{.}</math>}}
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Denna ekvation löser vi genom att logaritmera båda led med lg och skriva om vänsterledet som <math>\lg 1{,}05^x = x\cdot\lg 1{,}05</math>,
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This equation can be solved by taking the lg logarithm of both sides of and rewriting the left-hand side as <math>\lg 1{,}05^x = x\cdot\lg 1{,}05</math>,
{{Fristående formel||<math>x = \frac{\lg 2}{\lg 1{,}05} \quad ({}\approx 14{,}2)\,\mbox{.}</math>}}</li>
{{Fristående formel||<math>x = \frac{\lg 2}{\lg 1{,}05} \quad ({}\approx 14{,}2)\,\mbox{.}</math>}}</li>
</ol>
</ol>
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<div class="exempel">
<div class="exempel">
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'''Exempel 4'''
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''' Example 4'''
<ol type="a">
<ol type="a">
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<li>Lös ekvationen <math>\ 2^x \cdot 3^x = 5</math>.
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<li>Solve the equation <math>\ 2^x \cdot 3^x = 5</math>.
<br>
<br>
<br>
<br>
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Vänsterledet kan skrivas om med potenslagarna till <math>2^x\cdot 3^x=(2 \cdot 3)^x</math> och ekvationen blir
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The left-hand side can be rewritten using the laws of powers giving <math>2^x\cdot 3^x=(2 \cdot 3)^x</math> and the equation becomes
{{Fristående formel||<math>6^x = 5\,\mbox{.}</math>}}
{{Fristående formel||<math>6^x = 5\,\mbox{.}</math>}}
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Denna ekvation löser vi på vanligt sätt med logaritmering och får att
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This equation is solved in the usual way by taking logarithms giving
{{Fristående formel||<math>x = \frac{\lg 5}{\lg 6}\quad ({}\approx 0{,}898)\,\mbox{.}</math>}}</li>
{{Fristående formel||<math>x = \frac{\lg 5}{\lg 6}\quad ({}\approx 0{,}898)\,\mbox{.}</math>}}</li>
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<li>Lös ekvationen <math>\ 5^{2x + 1} = 3^{5x}</math>.
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<li>Solve the equation <math>\ 5^{2x + 1} = 3^{5x}</math>.
<br>
<br>
<br>
<br>
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Logaritmera båda led och använd logaritmlagen <math>\lg a^b = b \cdot \lg a</math>
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Take logarithms of both sides and use the laws of logarithms <math>\lg a^b = b \cdot \lg a</math>
{{Fristående formel||<math>\eqalign{(2x+1)\lg 5 &= 5x \cdot \lg 3\,\mbox{,}\cr 2x \cdot \lg 5 + \lg 5 &= 5x \cdot \lg 3\,\mbox{.}\cr}</math>}}
{{Fristående formel||<math>\eqalign{(2x+1)\lg 5 &= 5x \cdot \lg 3\,\mbox{,}\cr 2x \cdot \lg 5 + \lg 5 &= 5x \cdot \lg 3\,\mbox{.}\cr}</math>}}
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Samla <math>x</math> i ena ledet
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Collect <math>x</math> to one side
{{Fristående formel||<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>}}
{{Fristående formel||<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>}}
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Lösningen är
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The solution is
{{Fristående formel||<math>x = \frac{\lg 5}{5 \lg 3 -2 \lg 5}\,\mbox{.}</math>}}</li>
{{Fristående formel||<math>x = \frac{\lg 5}{5 \lg 3 -2 \lg 5}\,\mbox{.}</math>}}</li>
</ol>
</ol>
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== Några mer komplicerade ekvationer ==
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== Some more complicated equations ==
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Ekvationer som innehåller exponential- eller logaritmuttryck kan ibland behandlas som förstagrads- eller andragradsekvationer genom att betrakta "<math>\ln x</math>" eller "<math>e^x</math>" som obekant.
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Equations containing exponential or logarithmic expressions can sometimes be treated as first order or second order equations by considering "<math>\ln x</math>" or "<math>e^x</math>" as the unknown variable.
<div class="exempel">
<div class="exempel">
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'''Exempel 5'''
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''' Example 5'''
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Lös ekvationen <math>\,\frac{6e^x}{3e^x+1}=\frac{5}{e^{-x}+2}</math>.
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Solve the equation <math>\,\frac{6e^x}{3e^x+1}=\frac{5}{e^{-x}+2}</math>.
<br>
<br>
<br>
<br>
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Multiplicera båda led med <math>3e^x+1</math> och <math>e^{-x}+2</math> för att få bort nämnarna
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Multiply both sides by <math>3e^x+1</math> och <math>e^{-x}+2</math> to eliminate the denominators
{{Fristående formel||<math>6e^x(e^{-x}+2) = 5(3e^x+1)\,\mbox{.}</math>}}
{{Fristående formel||<math>6e^x(e^{-x}+2) = 5(3e^x+1)\,\mbox{.}</math>}}
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Notera att eftersom <math>e^x</math> och <math>e^{-x}</math> alltid är positiva oavsett värdet på <math>x</math> så multiplicerar vi alltså ekvationen med faktorer <math>3e^x+1</math> och <math>e^{-x} +2</math> som är skilda från noll, så detta steg riskerar inte att introducera nya (falska) rötter till ekvationen.
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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 multiply the equation by factors <math>3e^x+1</math> and <math>e^{-x} +2</math>both of which are different from zero, so this step cannot introduce new (spurious) roots of the equation.
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Förenkla båda led
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Simplify both sides of the equation
{{Fristående formel||<math>6+12e^x = 15e^x+5\,\mbox{,}</math>}}
{{Fristående formel||<math>6+12e^x = 15e^x+5\,\mbox{,}</math>}}
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där vi använt att <math>e^{-x} \cdot e^x = e^{-x + x} = e^0 = 1</math>. Betraktar vi nu <math>e^x</math> som obekant är ekvationen väsentligen en förstagradsekvation som har lösningen
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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
{{Fristående formel||<math>e^x=\frac{1}{3}\,\mbox{.}</math>}}
{{Fristående formel||<math>e^x=\frac{1}{3}\,\mbox{.}</math>}}
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En logaritmering ger sedan svaret
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Taking logarithms then gives the answer
{{Fristående formel||<math>x=\ln\frac{1}{3}= \ln 3^{-1} = -1 \cdot \ln 3 = -\ln 3\,\mbox{.}</math>}}
{{Fristående formel||<math>x=\ln\frac{1}{3}= \ln 3^{-1} = -1 \cdot \ln 3 = -\ln 3\,\mbox{.}</math>}}
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 6'''
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''' Example 6'''
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Lös ekvationen <math>\,\frac{1}{\ln x} + \ln\frac{1}{x} = 1</math>.
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Solve the equation <math>\,\frac{1}{\ln x} + \ln\frac{1}{x} = 1</math>.
<br>
<br>
<br>
<br>
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Termen <math>\ln\frac{1}{x}</math> kan skrivas som <math>\ln\frac{1}{x} = \ln x^{-1} = -1 \cdot \ln x = - \ln x</math> och då blir ekvationen
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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
{{Fristående formel||<math>\frac{1}{\ln x} - \ln x = 1\,\mbox{,}</math>}}
{{Fristående formel||<math>\frac{1}{\ln x} - \ln x = 1\,\mbox{,}</math>}}
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där vi kan betrakta <math>\ln x</math> som en ny obekant. Multiplicerar vi båda led med <math>\ln x</math> (som är skild från noll när <math>x \neq 1</math>) får vi en andragradsekvation i <math>\ln x</math>
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where we can consider <math>\ln x</math> as a new unknown. We multiply both sides with <math>\ln x</math> (which is different from zero when <math>x \neq 1</math>) gives us a quadratic equation in <math>\ln x</math>
{{Fristående formel||<math>1 - (\ln x)^2 = \ln x\,\mbox{,}</math>}}
{{Fristående formel||<math>1 - (\ln x)^2 = \ln x\,\mbox{,}</math>}}
{{Fristående formel||<math> (\ln x)^2 + \ln x - 1 = 0\,\mbox{.}</math>}}
{{Fristående formel||<math> (\ln x)^2 + \ln x - 1 = 0\,\mbox{.}</math>}}
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Kvadratkomplettering av vänsterledet
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Completing the square on the left-hand side
{{Fristående formel||<math>\begin{align*}
{{Fristående formel||<math>\begin{align*}
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\end{align*}</math>}}
\end{align*}</math>}}
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följt av rotutdragning ger att
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We continue by taking the root giving
{{Fristående formel||<math>
{{Fristående formel||<math>
\ln x = -\frac{1}{2} \pm \frac{\sqrt{5}}{2} \,\mbox{.}</math>}}
\ln x = -\frac{1}{2} \pm \frac{\sqrt{5}}{2} \,\mbox{.}</math>}}
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Detta betyder att ekvationen har två lösningar
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This means that the equation has two solutions
{{Fristående formel||<math>
{{Fristående formel||<math>
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== Falska rötter ==
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== Spurious roots ==
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När man löser ekvationer gäller det också att tänka på att argument till logaritmer måste vara positiva och att uttryck av typen <math>e^{(\ldots)}</math> bara kan anta positiva värden. Risken är annars att man får med falska rötter.
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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.
<div class="exempel">
<div class="exempel">
-
'''Exempel 7'''
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''' Example 7'''
-
Lös ekvationen <math>\,\ln(4x^2 -2x) = \ln (1-2x)</math>.
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Solve the equation <math>\,\ln(4x^2 -2x) = \ln (1-2x)</math>.
<br>
<br>
<br>
<br>
-
För att ekvationen ska vara uppfylld måste argumenten <math>4x^2-2x</math> och <math>1-2x</math> vara lika,
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For the equation to be satisfied the arguments <math>4x^2-2x</math> and <math>1-2x</math> must be equal,
-
 
+
{{Fristående formel||<math>4x^2 - 2x = 1 - 2x\,,</math>|<math>(*)</math>}}
{{Fristående formel||<math>4x^2 - 2x = 1 - 2x\,,</math>|<math>(*)</math>}}
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och dessutom positiva. Vi löser ekvationen <math>(*)</math> genom att flytta över alla termer i ena ledet
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and also positive. We solve the equation <math>(*)</math> by moving of all the terms to one side
{{Fristående formel||<math>4x^2 - 1= 0</math>}}
{{Fristående formel||<math>4x^2 - 1= 0</math>}}
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och använder rotutdragning. Detta ger att
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and take the root. This gives that
{{Fristående formel||<math>
{{Fristående formel||<math>
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x = \frac{1}{2} \; \mbox{.}</math>}}
x = \frac{1}{2} \; \mbox{.}</math>}}
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Vi kontrollerar nu om båda led i <math>(*)</math> är positiva
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We now check if both sides of <math>(*)</math> are positive
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* Om <math>x= -\tfrac{1}{2}</math> blir båda led lika med <math>4x^2 - 2x = 1-2x = 1-2 \cdot \bigl(-\tfrac{1}{2}\bigr) = 1+1 = 2 > 0</math>.
+
* If <math>x= -\tfrac{1}{2}</math> then both are sides are equal to <math>4x^2 - 2x = 1-2x = 1-2 \cdot \bigl(-\tfrac{1}{2}\bigr) = 1+1 = 2 > 0</math>.
-
* Om <math>x= \tfrac{1}{2}</math> blir båda led lika med <math>4x^2 - 2x = 1-2x = 1-2 \cdot \tfrac{1}{2} = 1-1 = 0 \not > 0</math>.
+
* If <math>x= \tfrac{1}{2}</math> then both are sides are equal to <math>4x^2 - 2x = 1-2x = 1-2 \cdot \tfrac{1}{2} = 1-1 = 0 \not > 0</math>.
-
Alltså har logaritmekvationen bara en lösning <math>x= -\frac{1}{2}</math>.
+
So the logarithmic equation has only one solution <math>x= -\frac{1}{2}</math>.
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 8'''
+
''' Example 8'''
-
Lös ekvationen <math>\,e^{2x} - e^{x} = \frac{1}{2}</math>.
+
Solve the equation <math>\,e^{2x} - e^{x} = \frac{1}{2}</math>.
<br>
<br>
<br>
<br>
-
Den första termen kan vi skriva som <math>e^{2x} = (e^x)^2</math>. Hela ekvationen är alltså en andragradsekvation med <math>e^x</math> som obekant
+
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
{{Fristående formel||<math>(e^x)^2 - e^x = \tfrac{1}{2}\,\mbox{.}</math>}}
{{Fristående formel||<math>(e^x)^2 - e^x = \tfrac{1}{2}\,\mbox{.}</math>}}
-
Ekvationen kan vara lite enklare att hantera om vi skriver <math>t</math> istället för <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>,
{{Fristående formel||<math>t^2 -t = \tfrac{1}{2}\,\mbox{.}</math>}}
{{Fristående formel||<math>t^2 -t = \tfrac{1}{2}\,\mbox{.}</math>}}
-
Kvadratkomplettera vänsterledet
+
Complete the square for the left-hand side.
{{Fristående formel||<math>\begin{align*}
{{Fristående formel||<math>\begin{align*}
Line 248: Line 247:
\end{align*}</math>}}
\end{align*}</math>}}
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vilket ger lösningarna
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which gives solutions
{{Fristående formel||<math>
{{Fristående formel||<math>
Line 255: Line 254:
t=\frac{1}{2} + \frac{\sqrt{3}}{2} \, \mbox{.}</math>}}
t=\frac{1}{2} + \frac{\sqrt{3}}{2} \, \mbox{.}</math>}}
-
Eftersom <math>\sqrt3 > 1</math> så är <math>\frac{1}{2}-\frac{1}{2}\sqrt3 <0</math> och det är bara <math>t= \frac{1}{2}+\frac{1}{2}\sqrt3</math> som ger en lösning till den ursprungliga ekvationen eftersom <math>e^x</math> alltid är positiv. Logaritmering ger slutligen att
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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 gives finally that
{{Fristående formel||<math>
{{Fristående formel||<math>
x = \ln \Bigl(\,\frac{1}{2}+\frac{\sqrt3}{2}\,\Bigr)</math>}}
x = \ln \Bigl(\,\frac{1}{2}+\frac{\sqrt3}{2}\,\Bigr)</math>}}
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är den enda lösningen till ekvationen.
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as the only solution to the equation.
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[[3.4 Övningar|Övningar]]
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[[3.4 Övningar|Exercises]]
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<div class="inforuta" style="width:580px;">
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'''Råd för inläsning'''
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'''Study advice'''
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'''Grund- och slutprov'''
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Efter att du har läst texten och arbetat med övningarna ska du göra grund- och slutprovet för att bli godkänd på detta avsnitt. Du hittar länken till proven i din student lounge.
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'''The basic and final tests'''
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After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.
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'''Tänk på att:'''
 
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Du kan behöva lägga ner mycket tid på logaritmer.
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'''Keep in mind that:'''
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Logaritmer brukar behandlas översiktligt i gymnasiet. Därför brukar många högskolestudenter stöta på problem när det gäller att räkna med logaritmer.
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You may need to spend much time studying logarithms.
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Logarithms usually are dealt with summarily in high school. Therefore, many college students tend to encounter problems when it comes to calculations with logarithms.
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Revision as of 16:57, 14 July 2008

       Teori          Övningar      

Innehåll:

  • Logarithmic Equations
  • Exponential equations
  • Spurious roots

Learning outcomes::

After this section, you will have learned to:

  • Solve equations that contain logarithm or exponential expressions and which can be reduced to first or second order equations.
  • Deal with spurious roots, and know when they arise.

Basic Equations

Equations where logarithms appear can vary a lot. Here are some examples where the solution is given almost immediately by the definition of a logarithm, that is.

\displaystyle \begin{align*} 
   10^x = y\quad&\Leftrightarrow\quad x = \lg y\\
   e^x = y\quad&\Leftrightarrow\quad x = \ln y\\
 \end{align*}

(We consider only 10-logarithms or natural logarithms.)

Example 1

Lös ekvationerna

  1. \displaystyle 10^x = 537\quad has a solution \displaystyle x = \lg 537.
  2. \displaystyle 10^{5x} = 537\quad gives \displaystyle 5x = \lg 537, i.e. \displaystyle x=\frac{1}{5} \lg 537.
  3. \displaystyle \frac{3}{e^x} = 5 \quad Multiplication of both sides with \displaystyle e^x and division by 5 gives \displaystyle \tfrac{3}{5}=e^x , which means that \displaystyle x=\ln\tfrac{3}{5}.
  4. \displaystyle \lg x = 3 \quad The definition gives directly \displaystyle x=10^3 = 1000.
  5. \displaystyle \lg(2x-4) = 2 \quad From the definition we have \displaystyle 2x-4 = 10^2 = 100 and it follows that \displaystyle x = 52.

Example 2

  1. Solve the equation \displaystyle \,(\sqrt{10}\,)^x = 25.

    Since \displaystyle \sqrt{10} = 10^{1/2} the left-hand side is equal to \displaystyle (\sqrt{10}\,)^x = (10^{1/2})^x = 10^{x/2} and the equation becomes
    \displaystyle 10^{x/2} = 25\,\mbox{.}
    This equation has a solution \displaystyle \frac{x}{2} = \lg 25, ie. \displaystyle x = 2 \lg 25.
  2. Solve the equation \displaystyle \,\frac{3 \ln 2x}{2} + 1 = \frac{1}{2}.

    Multiply both sides by 2 and then subtracting 2 from both sides
    \displaystyle 3 \ln 2x = -1\,\mbox{.}

    Divide both sides by 3

    \displaystyle \ln 2x = -\frac{1}{3}\,\mbox{.}

    Now, the definition directly gives \displaystyle 2x = e^{-1/3}, which means that

    \displaystyle x = {\textstyle\frac{1}{2}} e^{-1/3} = \frac{1}{2e^{1/3}}\,\mbox{.}

In many practical applications of exponential growth or decline there appear equations of the type

\displaystyle a^x = b\,\mbox{,}

where \displaystyle a and \displaystyle b are positive numbers. These equations are best solved by taking the logarithm of both sides

\displaystyle \lg a^x = \lg b

and use the law of logarithms for powers

\displaystyle x \cdot \lg a = \lg b

which gives the solution \displaystyle \ x = \displaystyle \frac{\lg b}{\lg a}.

Example 3

  1. Solve the equation \displaystyle \,3^x = 20.

    Take logarithms of both sides
    \displaystyle \lg 3^x = \lg 20\,\mbox{.}

    The left-hand side can be written as \displaystyle \lg 3^x = x \cdot \lg 3 giving

    \displaystyle x = \displaystyle \frac{\lg 20}{\lg 3} \quad ({}\approx 2{,}727)\,\mbox{.}
  2. Solve the equation \displaystyle \ 5000 \cdot 1{,}05^x = 10\,000.

    Divide both sides by 5000
    \displaystyle 1{,}05^x = \displaystyle \frac{ 10\,000}{5\,000} = 2\,\mbox{.}

    This equation can be solved by taking the lg logarithm of both sides of and rewriting the left-hand side as \displaystyle \lg 1{,}05^x = x\cdot\lg 1{,}05,

    \displaystyle x = \frac{\lg 2}{\lg 1{,}05} \quad ({}\approx 14{,}2)\,\mbox{.}

Example 4

  1. Solve the equation \displaystyle \ 2^x \cdot 3^x = 5.

    The left-hand side can be rewritten using the laws of powers giving \displaystyle 2^x\cdot 3^x=(2 \cdot 3)^x and the equation becomes
    \displaystyle 6^x = 5\,\mbox{.}

    This equation is solved in the usual way by taking logarithms giving

    \displaystyle x = \frac{\lg 5}{\lg 6}\quad ({}\approx 0{,}898)\,\mbox{.}
  2. Solve the equation \displaystyle \ 5^{2x + 1} = 3^{5x}.

    Take logarithms of both sides and use the laws of logarithms \displaystyle \lg a^b = b \cdot \lg a
    \displaystyle \eqalign{(2x+1)\lg 5 &= 5x \cdot \lg 3\,\mbox{,}\cr 2x \cdot \lg 5 + \lg 5 &= 5x \cdot \lg 3\,\mbox{.}\cr}

    Collect \displaystyle x to one side

    \displaystyle \eqalign{\lg 5 &= 5x \cdot \lg 3 -2x \cdot \lg 5\,\mbox{,}\cr \lg 5 &= x\,(5 \lg 3 -2 \lg 5)\,\mbox{.}\cr}

    The solution is

    \displaystyle x = \frac{\lg 5}{5 \lg 3 -2 \lg 5}\,\mbox{.}


Some more complicated equations

Equations containing exponential or logarithmic expressions can sometimes be treated as first order or second order equations by considering "\displaystyle \ln x" or "\displaystyle e^x" as the unknown variable.

Example 5

Solve the equation \displaystyle \,\frac{6e^x}{3e^x+1}=\frac{5}{e^{-x}+2}.

Multiply both sides by \displaystyle 3e^x+1 och \displaystyle e^{-x}+2 to eliminate the denominators

\displaystyle 6e^x(e^{-x}+2) = 5(3e^x+1)\,\mbox{.}

Note that since \displaystyle e^x and \displaystyle e^{-x} are always positive regardless of the value of \displaystyle x in this latest step we have multiply the equation by factors \displaystyle 3e^x+1 and \displaystyle e^{-x} +2both of which are different from zero, so this step cannot introduce new (spurious) roots of the equation.

Simplify both sides of the equation

\displaystyle 6+12e^x = 15e^x+5\,\mbox{,}

where we used \displaystyle e^{-x} \cdot e^x = e^{-x + x} = e^0 = 1. If we treat \displaystyle e^x as the unknown variable, the equation is essentially a first order equation which has a solution

\displaystyle e^x=\frac{1}{3}\,\mbox{.}

Taking logarithms then gives the answer

\displaystyle x=\ln\frac{1}{3}= \ln 3^{-1} = -1 \cdot \ln 3 = -\ln 3\,\mbox{.}

Example 6

Solve the equation \displaystyle \,\frac{1}{\ln x} + \ln\frac{1}{x} = 1.

The term \displaystyle \ln\frac{1}{x} can be written as \displaystyle \ln\frac{1}{x} = \ln x^{-1} = -1 \cdot \ln x = - \ln x and then the equation becomes

\displaystyle \frac{1}{\ln x} - \ln x = 1\,\mbox{,}

where we can consider \displaystyle \ln x as a new unknown. We multiply both sides with \displaystyle \ln x (which is different from zero when \displaystyle x \neq 1) gives us a quadratic equation in \displaystyle \ln x

\displaystyle 1 - (\ln x)^2 = \ln x\,\mbox{,}
\displaystyle (\ln x)^2 + \ln x - 1 = 0\,\mbox{.}

Completing the square on the left-hand side

\displaystyle \begin{align*} 
   \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 - \frac{5}{4}\\
 \end{align*}

We continue by taking the root giving

\displaystyle 
 \ln x = -\frac{1}{2} \pm \frac{\sqrt{5}}{2} \,\mbox{.}

This means that the equation has two solutions

\displaystyle 
 x= e^{(-1 + \sqrt{5})/2}
 \quad \mbox{och} \quad
 x= e^{-(1+\sqrt{5})/2}\,\mbox{.}


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 \displaystyle e^{(\ldots)} can only have positive values. The risk is otherwise that you get spurious roots.

Example 7

Solve the equation \displaystyle \,\ln(4x^2 -2x) = \ln (1-2x).

For the equation to be satisfied the arguments \displaystyle 4x^2-2x and \displaystyle 1-2x must be equal,

\displaystyle 4x^2 - 2x = 1 - 2x\,, \displaystyle (*)

and also positive. We solve the equation \displaystyle (*) by moving of all the terms to one side

\displaystyle 4x^2 - 1= 0

and take the root. This gives that

\displaystyle 
 \textstyle x= -\frac{1}{2}
 \quad\mbox{och}\quad
 x = \frac{1}{2} \; \mbox{.}

We now check if both sides of \displaystyle (*) are positive

  • If \displaystyle x= -\tfrac{1}{2} then both are sides are equal to \displaystyle 4x^2 - 2x = 1-2x = 1-2 \cdot \bigl(-\tfrac{1}{2}\bigr) = 1+1 = 2 > 0.
  • If \displaystyle x= \tfrac{1}{2} then both are sides are equal to \displaystyle 4x^2 - 2x = 1-2x = 1-2 \cdot \tfrac{1}{2} = 1-1 = 0 \not > 0.

So the logarithmic equation has only one solution \displaystyle x= -\frac{1}{2}.

Example 8

Solve the equation \displaystyle \,e^{2x} - e^{x} = \frac{1}{2}.

The first term can be written as \displaystyle e^{2x} = (e^x)^2. The whole equation is a quadratic with \displaystyle e^xas the unknown

\displaystyle (e^x)^2 - e^x = \tfrac{1}{2}\,\mbox{.}

The equation can be a little easier to manage if we write \displaystyle t instead of \displaystyle e^x,

\displaystyle t^2 -t = \tfrac{1}{2}\,\mbox{.}

Complete the square for the left-hand side.

\displaystyle \begin{align*} 
   \textstyle \bigl(t-\frac{1}{2}\bigr)^2 - \bigl(\frac{1}{2}\bigr)^2
     &= \frac{1}{2}\,\mbox{,}\\
   \bigl(t-\frac{1}{2}\bigr)^2
     &= \frac{3}{4}\,\mbox{,}\\
 \end{align*}

which gives solutions

\displaystyle 
 t=\frac{1}{2} - \frac{\sqrt{3}}{2}
 \quad\mbox{och}\quad
 t=\frac{1}{2} + \frac{\sqrt{3}}{2} \, \mbox{.}

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

\displaystyle 
 x = \ln \Bigl(\,\frac{1}{2}+\frac{\sqrt3}{2}\,\Bigr)

as the only solution to the equation.


Exercises

Study advice

The basic and final tests

After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.


Keep in mind that:

You may need to spend much 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.

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