Aus Online Mathematik Brückenkurs 1

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.