Csdrhrt

A certain type of bacteria doubles every $2$ hours. If a researcher counts $800$ bacteria at 11am on Monday, find the number of bacteria at each of the following:

a) 9am Monday

b) 11am Tuesday

I know that the formula is $y=800(2)^{t/2}$

I am confused as to what to do for part c), d), e), and f). Please help!

Sure, your original formula is correct. But we have to remember that we are given the initial population at an initial time.

$$P(t) = 800cdot2^{(t/2)}$$

Where $P$ is the population of the bacteria, and the variable $t$ represents the time after $11 text{AM}$. $800$ of course represents the initial population at $t_0 = 11 text{AM}$, and the $2$ and the $t/2$ mean that the population doubles, and it does so every two hours.

a) 1 PM, Monday; This is 2 hours after 11 AM, and that means we need to find $P(2)$.
$$P(2)=800cdot2^{2/2} = 1600$$

b) 3 PM Monday, is 4 hours after 11 AM. $P(4) = 800cdot2^{(4/2)} = 3200$

c) 4 PM Monday, is 5 hours after 11 AM. $P(5) = 800cdot2^{(5/2)} = 3200sqrt2 approx 4525$

d) 5:30 PM Monday, is 6.5 hours after. $P(6.5) = 800cdot2^{(6.5/2)} = 3200cdot(2)^{3/4} approx 5382$

et cetera, and you get the idea.

You've go it! Just plug the numbers in, including the fractional and-or negative ones!

The times that are not multiples of 2hr later give fractional numbers to be plugged into the formula: this is fine, as $$2^frac{1}{2}=√2$$$$2^frac{3}{2}=2√2$$$$2^frac{5}{2}=4√2$$ etc, etc; and the times of the readings made before the reference time yield negative numbers for the argument, & $$2^{-1}=frac{1}{2}$$$$2^{-2}=frac{1}{4}$$$$2^{-3}=frac{1}{8}$$$$2^{-frac{1}{2}}=frac{1}{√2}$$$$2^{-frac{3}{2}}=frac{1}{2√2}$$$$2^{-frac{5}{2}}=frac{1}{4√2}$$ etc etc etc.

And the denominator need not be 2: if it's 3, then cube roots are yielt; if 4, quartic; if 5, quintic; etc etc: nor must it be a rational number!

To put it more technically the function $operatorname{2uparrow}$ is a function of a real variable valid in the range (-∞, ∞) ... and, moreover, the entire complex plane. I know it is at first defined as iterated multiplication by 2; but it is one of the great feats of mathematics, which we take for granted now (I've been discussing this kind of thing elsewhere on here just recently), that the definition can be (and is) broadened in scope such that absolutely any input be a valid argument, and yet the function be the same function as that defined by iterated multiplication at arguments from the set of natural numbers. However ... when we are dealing with it as a function of absolutely any number, we would normally conceive of it as $operatorname{exp((ln2)×)}$.