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Lösung 3.4:3b

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
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K (Lösning 3.4:3b moved to Solution 3.4:3b: Robot: moved page)
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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 lnx2+3x   and ln3x2−2x   are equal only if their arguments are equal, i.e.


x2+3x=3x2−2x  


however, we have to be careful! If we obtain a value for 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 x2+3x and 3x2−2x 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


2x2−5x=0


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


x2x−5=0 


From this factorized expression, we read off that the solutions are x=0 and x=52.

A final check shows that when x=0 then x2+3x= 3x2−2x =0, so x=0 is not a solution. On the other hand, when x=52 then x2+3x= 3x2−2x=5540, so x=52 is a solution.

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