Lösung 3.4:3b

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The expressions
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<center> [[Image:3_4_3b-1(2).gif]] </center>
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<math>\text{ln}\left( x^{\text{2}}+\text{3}x \right)\text{ }</math>
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and
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<math>\text{ln}\left( \text{3}x^{\text{2}}-\text{2}x \right)\text{ }</math>
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<center> [[Image:3_4_3b-2(2).gif]] </center>
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are equal only if their arguments are equal, i.e.
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<math>\left( x^{\text{2}}+\text{3}x \right)=\left( \text{3}x^{\text{2}}-\text{2}x \right)\text{ }</math>
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however, we have to be careful! If we obtain a value for
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<math>x</math>
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which makes the arguments equal but negative or zero, then it will not correspond to a genuine solution because ln is not defined for negative arguments. At the end of the exercise, we must therefore check that
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<math>x^{\text{2}}+\text{3}x\text{ }</math>
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and
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<math>\text{3}x^{\text{2}}-\text{2}x\text{ }</math>
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really are positive for those solutions that we have calculated.
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If we move all the terms over to one side in the equation for the arguments, we get the second-degree equation
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<math>2x^{2}-5x=0</math>
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and we see that both terms contain x, which we can take out as a factor:
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<math>x\left( 2x-5 \right)=0</math>
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From this factorized expression, we read off that the solutions are
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<math>x=0\text{ }</math>
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and
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<math>x={5}/{2}\;</math>.
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A final check shows that when
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<math>x=0\text{ }</math>
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then
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<math>x^{\text{2}}+\text{3}x=\text{ 3}x^{\text{2}}-\text{2}x\text{ }=0</math>, so
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<math>x=0\text{ }</math>
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is not a solution. On the other hand, when
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<math>x={5}/{2}\;</math>
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then
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<math>x^{\text{2}}+\text{3}x=\text{ 3}x^{\text{2}}-\text{2}x=\text{55}/\text{4}>0</math>, so
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<math>x={5}/{2}\;</math>
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is a solution.

Version vom 11:52, 26. Sep. 2008

The expressions \displaystyle \text{ln}\left( x^{\text{2}}+\text{3}x \right)\text{ } and \displaystyle \text{ln}\left( \text{3}x^{\text{2}}-\text{2}x \right)\text{ } are equal only if their arguments are equal, i.e.


\displaystyle \left( x^{\text{2}}+\text{3}x \right)=\left( \text{3}x^{\text{2}}-\text{2}x \right)\text{ }


however, we have to be careful! If we obtain a value for \displaystyle x which makes the arguments equal but negative or zero, then it will not correspond to a genuine solution because ln is not defined for negative arguments. At the end of the exercise, we must therefore check that \displaystyle x^{\text{2}}+\text{3}x\text{ } and \displaystyle \text{3}x^{\text{2}}-\text{2}x\text{ } really are positive for those solutions that we have calculated.

If we move all the terms over to one side in the equation for the arguments, we get the second-degree equation


\displaystyle 2x^{2}-5x=0


and we see that both terms contain x, which we can take out as a factor:


\displaystyle x\left( 2x-5 \right)=0


From this factorized expression, we read off that the solutions are \displaystyle x=0\text{ } and \displaystyle x={5}/{2}\;.

A final check shows that when \displaystyle x=0\text{ } then \displaystyle x^{\text{2}}+\text{3}x=\text{ 3}x^{\text{2}}-\text{2}x\text{ }=0, so \displaystyle x=0\text{ } is not a solution. On the other hand, when \displaystyle x={5}/{2}\; then \displaystyle x^{\text{2}}+\text{3}x=\text{ 3}x^{\text{2}}-\text{2}x=\text{55}/\text{4}>0, so \displaystyle x={5}/{2}\; is a solution.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.4:3b“