Differential Equation: Bacterial culture?

The population of a bacterial culture increases at a rat that is proportional to the number of bacteria present at any time t days ----> can translated as : dM/dt = k * M . . . .where M is the population of bacteria

dM/dt = k * M
dM / M = k * dt ---> start integrating both sides

ln( M ) = k * t + C
e^( ln( M ) ) = e^( k * t + C )
M(t) = e^( k * t ) * e^( C )
M(t) = e^( k * t ) * A0 ---> calling e^( C ) as A0 as another constant

M(t) = e^( k * t ) * A0 ------> M(8) = the initial condition which is M(8) = 2A0
M(8) = e^( k * 8 ) * A0
2A0 = e^( 8k ) * A0
2 = e^( 8k )
ln(2) = ln( e^( 8k ) )
ln(2) = 8k
ln(2)/8 = k

M(t) = e^( ln(2)/8 * t ) * A0 -----> M(t) = 3A0 . . . .we are looking for the time
3A0 = e^( ln(2)/8 * t ) * A0
3 = e^( ln(2)/8 * t )
ln( 3 ) = ln( e^( ln(2)/8 * t ) )
ln( 3 ) = ln(2)/8 * t
8 * ln( 3 ) / ln(2) = t

t ≈ 13 days

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dM/dt = 2M/8

dM/M = 2/8dt

ln|M| = (1/4)t + c

M = Ce^{(1/4)t}

where C is the initial population at time t = 0.

(3/2)M = Ce^{(1/4)t3}

where t3 is the time that the population is 3 times the initial population.

M = C^e{(1/4)t2}

where t2 is the time that the population is 3 times the initial population.

Dividing the first equation by the second:

3/2 = e^{(1/4)(t3-t2)}

(1/4)(t3-t2) = ln(3/2)

(t3-t2) = (4)ln(3/2)

t3-t2 = ln(81/8)

t3-t2 ≈ 2.315 additional days

t3 ≈ 10.315 days