Answer:
The population is of 500 after 10.22 hours.
Step-by-step explanation:
The rate of change of the population of a certain organism is proportional to the population at time t, in hours.
This means that the population can be modeled by the following differential equation:

In which r is the growth rate.
Solving by separation of variables, then integrating both sides, we have that:



Applying the exponential to both sides:

In which K is the initial population.
At time t = 0 hours, the population is 300.
This means that K = 300. So

At time t = 24 hours, the population is 1000.
This means that P(24) = 1000. We use this to find the growth rate. So








So

At what time t is the population 500?
This is t for which P(t) = 500. So








The population is of 500 after 10.22 hours.