Lösung 3.4:3b
Aus Online Mathematik Brückenkurs 1
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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. | 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. | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x^2 + 3x = 3x^2 - 2x\,\textrm{.}</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 <math>x^2 + 3x</math> and <math>3x^2 - 2x</math> really are positive for those solutions that we have calculated. | 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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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 | ||
| - | {{ | + | {{Abgesetzte Formel||<math>2x^2-5x=0</math>}} |
and we see that both terms contain ''x'', which we can take out as a factor, | and we see that both terms contain ''x'', which we can take out as a factor, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x(2x-5) = 0\,\textrm{.}</math>}} |
From this factorized expression, we read off that the solutions are <math>x=0</math> | From this factorized expression, we read off that the solutions are <math>x=0</math> | ||
Version vom 08:46, 22. 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.
| \displaystyle x^2 + 3x = 3x^2 - 2x\,\textrm{.} |
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
| \displaystyle 2x^2-5x=0 |
and we see that both terms contain x, which we can take out as a factor,
| \displaystyle x(2x-5) = 0\,\textrm{.} |
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.
