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

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#
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#color(blue)("Building the model for initial condition")#

The increase in bacteria after 1 hour is

#b_o(1+y/100)^1#

The increase in bacteria after 2 hour is

#b_o(1+y/100)^2#

The increase in bacteria after 2 hour is

#b_o(1+y/100)^3#

So after 8 hours we have:

#color(blue)(b_o(1+y/100)^8 = 2b_o)#....................Equation(1)

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#color(blue)("Determine the value of "y)#

Using equation(1) divide both sides by #b_o#

#(1+y/100)^8=2#

Taking roots

#1+y/100 = root(8)(2)#

#color(blue)(y=100(root(8)(2)-1))#
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#color(blue)("Determine the time it takes for "3b_o)#

Using Equation(1) we have:

#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#

Taking logs ( I elect to use 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))#

My calculator gives:

#x=12.6797000058#

Is there could be a degree of error in the calculation lets say:

#x=12.6797# hours exactly

#x=#12 hours 40 minutes and say 47 seconds