Aus Online Mathematik Brückenkurs 1

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.