Using dN/dt=kN(P-N) How long will it take for 99% of a pop. to become ill if 1 person was ill in the beginning

Thanks everyone!Using dN/dt=kN(P-N) How long will it take for 99% of a pop. to become ill if 1 person was ill in the beginning
[Ln(N) - Ln(P - N)]/P - [Ln(1) - Ln(P - 1)]/P = kt


[Ln(N/(P - N))]/P + [Ln(P - 1)]/P = kt


Ln(N(P - 1)/(P - N)) = kPt


Ln(0.99P(P - 1)/(P - 0.99P)) = kPt


t = (1/k)(1/P)Ln(0.99(P - 1)/(1 - 0.99))


t = (1/kP)Ln(99(P - 1))


or


t = Ln{[99(P - 1)]^(1/kP)}

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