taking natural log ... ax = bx + ln(C)
ax - bx = ln(C) ... x (a - b) = ln(C)
x = [ln(C)] / (a - b)
e^ax = Ce^bx, where a ≠ b
how do i solve for x?
3 answers
e^ax / e^bx = C
e^((a-b)x) = C
(a-b)x = lnC
x = lnC/(a-b)
e^((a-b)x) = C
(a-b)x = lnC
x = lnC/(a-b)
thank you!