It's correct (if i haven't gone too fast).
Now, solve for x.
You know that y=3=e^x and y=2=e^x are solutions
(e^x)^2 - 5e^x + 6 = 0
how should I start?
I tried setting e^x = y
then -> y^2 - 5y + 6 = 0
solving this I get the roots 3 and 2
Am I doing this wrong?
UH ok =) I just found out this -> I get answers y=3 and y=2 .. I take e^x = 3 and e^x = 2
then I take ln 3 and ln 2 and I get the correct answer.
Just as a note.
What are the common situations for using the method of calling e^x = y ?
the common situations you will be able to identify using experience, but pretty much anywhere where you see the problem can be simplified into a form you are more familiar with (like in this case, it simplified to a quadratic equation) this technique can be used. it is mostly for visualization purposes though, you could have just as well solved the problem without it.