How many bacteria would there be at the end of 10 hours?

Question

The number of bacteria increased exponentially. At first there were 400. Three hours later there were 1600.

Answer 1

Let B(t) be the number of bacteria after t hours. Then, according to the statement, B(t) = k* e^(at), where k and a are constants. According to the given conditions,

B(0) = k e^0 = k = 400
B(3) = ke^(3a) = 1600. It follows e^(3a) = 1600/400 = 4, so that 3a = ln(4) => a = ln(4)/3

So, our equation can be written as B(t) = 400 e^(ln(4)/3 a) = 400 4^(a/3)

At the end of 10 hours, there will be

B(10) = 400 4^(100/3) bacteria

Answer 2

The general form of an exponential growth is x(t) =x(0)e^(kt)
x(0) is the number at time, t=0. so the equation becomes
x(t) = 400e^(kt). At t = 3, there are 1600 so,
x(3) = 1600 = 400e^(k3) = 400e^(3k).
or e^(3k) = 4, then Taking the natural
logarithm of both sides (base e), we have
3k = ln(4) or k = (1/3) * 1.386 = 0.462
so the equation become x(t) = 400e^(0.462t)
then at 10 hours we have
x(10) = 400e^(0.462*10) = 400e^(4.62)
x(10) = 400 * 36.461 = 14,584.4