e^(2x)-6e^x+5=0 has two solutions, what are they?
I tried this:
1. moved +5 to the other side so there wouldnt be a ln 0 for the next step
2. applied ln to both sides [2x-6x=ln(-5)]
3. combined like terms = -4x=ln(-5)
4. isolated x = x=ln(-5)/-4
obviously this is wrong
the only thing i can think of that would yield two answers is the quadratic formula, but i don't know if thats the right route or how to set that up
edit still stuck here. any tips on where to start?
edit2: Ok. I just looked at this problem and realized that e^0 would work for one answer. I assume the other I can work out now.
edit3: still stuck. tried everything i know