colby and jaquan are growing bacteria in an experiment in a laboratory. colby starts with 50 bacteria in his culture and the number of bacteria doubles every 2 hours. jaquan start with 80 of a different type of bacteria that doubles every 3 hours

Step-by-step 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.

1) c 2143.6 cm

2) c 113.0 m

3) a 33.5 ft

4)   d 2 cm

5) c 20 mm

6) 180 in

7) d 120 ft

5.5

step-by-step explanation: