A bacteria doubles its population in 8 hours. At this rate how many hours would it take the population of the bacteria to triple?

Question

A bacteria doubles its population in 8 hours.  At this rate how many hours would it take the population of the bacteria to triple?

Eli Assouline · Accepted Answer

Time taken for 3 times the population to have grown is:

12 hours 40 minutes and say 47 seconds

Let the rate of growth be constant and of value #y%# Let time in hours be #t# Let count of bacteria at any time #t# be #b_t# So initial count of bacteria would be #b_o# Given that count of bacteria after 8 hours (#b_8#) is such that #b_8 =2b_o# Required to determine unknown time #x" for "3b_o->t_x# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Building the model for initial condition")# After an hour, the rise in bacteria is #b_o(1+y/100)^1# The rise in bacteria following two hours is #b_o(1+y/100)^2# The rise in bacteria following two hours is #b_o(1+y/100)^3# Eight hours later, we have: #color(blue)(b_o(1+y/100)^8 = 2b_o)#....................Equation(1) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Determine the value of "y)# Using equation(1) divide both sides by #b_o# #(1+y/100)^8=2# Establishing roots #1+y/100 = root(8)(2)# #color(blue)(y=100(root(8)(2)-1))# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Determine the time it takes for "3b_o)# Equation (1) gives us: #b_o(1+y/100)^x = 3b_o# #=> (1+y/100)^x=3# But #y= 100(root(8)(2)-1)# giving #(1+root(8)(2)-1)^x=3# #(root(8)(2))^x=3# Acquiring logs (I choose to utilize log to base 10) #xlog(root(8)(2))=log(3)# But #root(8)(2) = 2^(1/8)# so we have #log(2^(1/8)) = 1/8log(2)# #x/8log(2)=log(3)# #x=(8log(3))/(log(2))# Using my calculator, I get: #x=12.6797000058# Is it possible that there was some calculation error, let's say: #x=12.6797# hours exactly #x=#12 hours 40 minutes and say 47 seconds

Eli Assouline · Answer

undefined It would take 16 hours for the population of the bacteria to triple.