exponential growth?

Let dP / dt = kP , where P denotes the population of bacteria
Then dP / P = k dt
∫ (1/P) dp = ∫ k dt + c1
ln P = kt + c1
P = e^( c1 + kt ) = c * e^(kt)

P0 = P(0) = c , where P0 denotes the initial number, that is , P( t = 0 )
Hence P = P0 * e^(kt)

Since the number doubles in 1 week,
P(1) = P0 * e^(k*1) = 2 * P0
e^k = 2 , so we have
P = P0 * e^(kt) = P0 * (e^k)^t = P0 * 2^t

P(2) = P0 * 2^2 = 4 * P0

Ans: After 2 weeks, the number will be 4 times of the initial number.