Lösung 3.4:3b

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The expressions
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The expressions <math>\ln\bigl(x^2+3x\bigr)</math> and <math>\ln\bigl(3x^2-2x \bigr)</math> are equal only if their arguments are equal, i.e.
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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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are equal only if their arguments are equal, i.e.
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{{Displayed math||<math>x^2 + 3x = 3x^2 - 2x\,\textrm{.}</math>}}
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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 ''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 <math>x^2 + 3x</math> and <math>3x^2 - 2x</math> really are positive for those solutions that we have calculated.
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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
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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{{Displayed math||<math>2x^2-5x=0</math>}}
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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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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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{{Displayed math||<math>x(2x-5) = 0\,\textrm{.}</math>}}
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From this factorized expression, we read off that the solutions are
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From this factorized expression, we read off that the solutions are <math>x=0</math>
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<math>x=0\text{ }</math>
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and <math>x=5/2\,</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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A final check shows that when <math>x=0</math> then <math>x^2 + 3x = 3x^2 - 2x = 0</math>, so <math>x=0</math> is not a solution. On the other hand, when <math>x=5/2</math> then <math>x^2 + 3x = 3x^2 - 2x = 55/4 > 0</math>, so <math>x=5/2</math> is a solution.
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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.
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Version vom 14:25, 2. Okt. 2008

The expressions \displaystyle \ln\bigl(x^2+3x\bigr) and \displaystyle \ln\bigl(3x^2-2x \bigr) are equal only if their arguments are equal, i.e.

Vorlage:Displayed math

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 \displaystyle x^2 + 3x and \displaystyle 3x^2 - 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

Vorlage:Displayed math

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

Vorlage:Displayed math

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

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