Processing Math: Done

From Förberedande kurs i matematik 1

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,

x(2x−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.