Question

the population of A bacteria counts 500 individuals initially doubling each day. The population of B bacteria counts 50 individuals at the start and triples every day.

How long will it take for the population of B to exceed that of A?
what will then be the number of individuals in each population?

if the population of A doubles twice as slowly, how much less time will it take the population of B to overtake?

Answer 1

so you want t such that
500*2^t < 50*3^t
20 < (3/2)^t
log20 < t log(3/2)
log20/log(3/2) < t
5.678 < t
If you only check at the end of each day, then after 6 days,
A=32000 and B=36450

Now adjust for slower doubling. Post your work if you get stuck.