1. The problem statement, all variables and given/known data Solve the equation for b. db/dx = (e^2x)(e^2b) -- therefore: db / (e^2b) = (e^2x)dx Also note that b(0) = 8. 2. Relevant equations 3. The attempt at a solution Following a list of steps my teacher gave me to solve these: 1) Integrate both sides. -0.5 e^(-2b) = 0.5 e^(2x) + C 2) Isolate b. e^(-2b) = -e^(2x) + C -2b = ln(-e^(2x) + C) And here we run in to the problem - you can't take ln of a negative value! If that e^(2x) was positive I would finish the problem like this: 2x + C = -2b x is 0 and b is 8 so C has to be -16. And my final answer would be b = -x + 8. But, I didn't think you could do this if there's a negative inside ln?