Explanation:
so, what is the question ?
all you are giving us here are some facts about what's going on. but what is the goal ? what do we need to find out ?
is it when the bacteria cultures have the same population ?
I assume that for my following answer :
"once every 2 hours" can be expressed by "x/2". with x being hours. so, instead of "x", which would mean "once every hour", we use "x/2".
"doubles (once) every 2 hours" is then
2×2×2×...×2 doubles on how many units of 2 hours we have.
this is
2^(x/2)
this is the factor we need to multiply with everything we want to behave that way (to double in size every 2 hours).
in our case here : 50 bacteria.
so, Colby's bacteria culture function is
c(x) = 50 × 2^(x/2)
similar for jaquan:
j(x) = 80 × 2^(x/3)
as his culture doubles every 3 hours.
now for the assumed question, when will both cultures have the same population ?
it is for that value of x that makes both function to deliver the same functional result.
50 × 2^(x/2) = 80 × 2^(x/3)
5 × 2^(x/2) = 8 × 2^(x/3)
now we use the log2 function on both sides and get
log2(5 × 2^(x/2)) = log2(8 × 2^(x/3))
remember the laws of logarithm. particularly
log(a×b) = log(a) + log(b)
and
loga(b) = logc(b)/logc(a)
log2(5) + log2(2^(x/2) = log2(8) + log2(2^(x/3)
log2(5) + x/2 = log2(8) + x/3 = 3 + x/3
3×log2(5) + 3x/2 = 9 + x
2×3×log2(5) + 3x = 2×9 + 2x
6×log2(5) + 3x = 18 + 2x
x = 18 - 6×log2(5) = 18 - 6×ln(5)/ln(2) = 4.068431431... hours
so, after a little bit more than 4 hours both bacteria cultures reach the exact same number of population : 204.8
which is, of course, only theoretical, as there cannot be "part bacteria" in the cultures.
now, we could calculate the x for when c(x) = 204, and the x for when c(x) = 205.
and then the same for j(x) and then see if or what overlap between these two x intervals there is.
during this overlap period both cultures will have the same (whole number) amount of bacteria.
but I think this is a topic for another question, as I am not even sure that I addressed the true original question.