Lösung 3.4:2b

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If we write the equation as
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<center> [[Image:3_4_2b-1(2).gif]] </center>
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<center> [[Image:3_4_2b-2(2).gif]] </center>
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<math>\left( e^{x} \right)^{2}+e^{x}=4</math>
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we see that
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<math>x</math>
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appears only in the combination
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<math>e^{x}</math>
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and it is therefore appropriate to treat
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<math>e^{x}</math>
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as a new unknown in the equation and then, when we have obtained the value of
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<math>e^{x}</math>, we can calculate the corresponding value of
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<math>x</math>
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by simply taking the logarithm.
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For clarity, we set
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<math>t=e^{x}</math>, so that the equation can be written as
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<math>t^{2}+t=4</math>
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and we solve this second-degree equation by completing the square,
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<math>t^{2}+t=\left( t+\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}=\left( t+\frac{1}{2} \right)^{2}-\frac{1}{4}</math>
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which gives
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<math>\left( t+\frac{1}{2} \right)^{2}-\frac{1}{4}=4\quad \Leftrightarrow \quad t=-\frac{1}{2}\pm \frac{\sqrt{17}}{2}</math>
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These two roots give us two possible values for
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<math>e^{x}</math>,
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<math>e^{x}=-\frac{1}{2}-\frac{\sqrt{17}}{2}</math>
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or
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<math>e^{x}=-\frac{1}{2}+\frac{\sqrt{17}}{2}</math>
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In the first case, the right-hand side is negative and because "
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<math>e</math>
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raised to anything" can never be negative, there is no
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<math>x</math>
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that can satisfy this equality. The other case, on the other hand, has a positive right-hand side (because
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<math>\sqrt{17}>1</math>
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) and we can take the logarithm of both sides to obtain
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<math>x=\ln \left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)</math>
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NOTE: It is a little tricky to check the answer to the original equation, so we can be satisfied with substituting
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<math>t=\frac{\sqrt{17}}{2}-\frac{1}{2}</math>
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into the equation
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<math>t^{\text{2}}+t=\text{4}</math>,
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LHS
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<math>=</math>
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<math>\begin{align}
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& =\left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)^{2}+\left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)=\frac{17}{4}-2\centerdot \frac{1}{2}\centerdot \frac{\sqrt{17}}{2}+\frac{1}{4}+\frac{\sqrt{17}}{2}-\frac{1}{2} \\
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& =\frac{17}{4}+\frac{1}{4}-\frac{1}{2}=\frac{17+1-2}{4}=\frac{16}{4}=4= \\
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\end{align}</math>
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<math>=</math>
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RHS.

Version vom 11:44, 6. Okt. 2008

If we write the equation as


\displaystyle \left( e^{x} \right)^{2}+e^{x}=4


we see that \displaystyle x appears only in the combination \displaystyle e^{x} and it is therefore appropriate to treat \displaystyle e^{x} as a new unknown in the equation and then, when we have obtained the value of \displaystyle e^{x}, we can calculate the corresponding value of \displaystyle x by simply taking the logarithm.

For clarity, we set \displaystyle t=e^{x}, so that the equation can be written as


\displaystyle t^{2}+t=4


and we solve this second-degree equation by completing the square,


\displaystyle t^{2}+t=\left( t+\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}=\left( t+\frac{1}{2} \right)^{2}-\frac{1}{4}


which gives


\displaystyle \left( t+\frac{1}{2} \right)^{2}-\frac{1}{4}=4\quad \Leftrightarrow \quad t=-\frac{1}{2}\pm \frac{\sqrt{17}}{2}


These two roots give us two possible values for \displaystyle e^{x},


\displaystyle e^{x}=-\frac{1}{2}-\frac{\sqrt{17}}{2} or \displaystyle e^{x}=-\frac{1}{2}+\frac{\sqrt{17}}{2}


In the first case, the right-hand side is negative and because " \displaystyle e raised to anything" can never be negative, there is no \displaystyle x that can satisfy this equality. The other case, on the other hand, has a positive right-hand side (because \displaystyle \sqrt{17}>1 ) and we can take the logarithm of both sides to obtain


\displaystyle x=\ln \left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)


NOTE: It is a little tricky to check the answer to the original equation, so we can be satisfied with substituting \displaystyle t=\frac{\sqrt{17}}{2}-\frac{1}{2} into the equation \displaystyle t^{\text{2}}+t=\text{4},

LHS \displaystyle =


\displaystyle \begin{align} & =\left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)^{2}+\left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)=\frac{17}{4}-2\centerdot \frac{1}{2}\centerdot \frac{\sqrt{17}}{2}+\frac{1}{4}+\frac{\sqrt{17}}{2}-\frac{1}{2} \\ & =\frac{17}{4}+\frac{1}{4}-\frac{1}{2}=\frac{17+1-2}{4}=\frac{16}{4}=4= \\ \end{align}


\displaystyle = RHS.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.4:2b“