For this case we have a function of the form:
[tex]y = A * (b) ^ t
[/tex]
Where,
A: it is the initial amount of bacteria
b: growth rate
t: it's time
By the time the bacteria are double we have that y = 2A
Substituting values:
[tex]2A = A * (1,083) ^ t
[/tex]
From here, we clear t:
[tex](1,083) ^ t = 2
log1.083 ((1.083) ^ t) = 2
t = log1.083 (2)
t = 8.7 hours[/tex]
Answer:
it takes for the size of the sample to double about:
[tex]t = 8.7 hours[/tex]
