Asked by lizzy

e^ax = Ce^bx, where a ≠ b

how do i solve for x?
Answered by R_scott
taking natural log ... ax = bx + ln(C)

ax - bx = ln(C) ... x (a - b) = ln(C)

x = [ln(C)] / (a - b)

Answered by oobleck
e^ax / e^bx = C
e^((a-b)x) = C
(a-b)x = lnC
x = lnC/(a-b)
Answered by lizzy
thank you!
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