The bacteria in a certain culture double every 7.3 hours. the culture has 6,500 bacteria at the start.
how many bacteria will the culture contain after 5 hours?

Y = y(x) is the amount of bacteria after t hours.

Since we know the change is y WRT t is proportional to the amount of y:
dy/dt = my (m is some constant)

Solve for y using "Separation of Variables":
dy/y = mdt
Integrate both sides.
ln|y| = mt + C
|y| = e^(mt+C) = e^(mt) * e^C
y = ce^(mt) (where c = (+-)e^C)

Let's solve for the constant c by making use of the "initial values". We know at t=0 that y=6500.

6500 = ce^(m*0)
6500 = c

So now:
y = 6500e^(mt)
but we still need to solve for m somehow. Let's use the fact that the amount, y, doubles every Δt=7.3 hours.

At t=0:
y = 6500

So at t=7.3:
13000 = 6500e^(7.3m)
2 = e^(7.3m)
ln(2) = 7.3m
ln(2) / 7.3 = m

Substitute m:
y = 6500e^(ln(2)/7.3 * t)

We are interested in the amount of bacteria, y, after 5 hours (t=5):
y = 6500e^(ln(2)/7.3 * 5)
y = 10449.6 (approximately)

The correct answer is $90