How do I solve e^(ax) = Ce^(bx), when a does not equal b?

Question

I know the answer has something to do with natural log, and that a has to be connected to b and c, but I don't know what.

Answer 1

take ln of both sides

ln [e^(ax)] = ln[ C e^(bx) ]
ax lne = lnC + bx lne, using the rules of logs, and recall that lne = 1

ax = lnC + bx
ax - bx = lnC

so when a = b

ax - ax = lnC
0 = ln C
C = 1

when a ≠ b
ax - bx = lnC
x(a-b) = lnC
x = lnC/(a-b)