Question 399·Hard·Math

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In a laboratory experiment, a bacterial culture grows by 50% every 6 hours. The culture begins with $N_0$ bacteria at 8:00 a.m. on Day 1. After exactly 24 hours, an antiseptic treatment instantly destroys 30% of the bacteria that are present at that moment. The culture is then returned to the incubator, where the same 24-hour "grow-then-treat" cycle is repeated once more.

Which of the following expressions gives the number of bacteria in the culture immediately after the second antiseptic treatment, in terms of $N_0$ ?

Repeated exponential change is the cue: the culture grows by the same percent every six hours, then loses a percent at the end of each day. Count growth intervals and treatments separately, and multiply one factor for each event. The tempting trap is using the number of 24-hour cycles as the growth exponent.

Hints

Hint 1
An interval is the stretch of time over which a stated change applies. Growth happens every six hours, not once per 24-hour cycle. How many growth intervals fit into one cycle?
Hint 2
A growth factor tells you what to multiply the previous amount by. A percent increase keeps all the bacteria already present and adds more, so include the original $100\%$ when you make the factor.
Hint 3
A decay factor tells you what remains after a decrease. Subtract the fraction destroyed from the whole, then use that factor once per treatment. How many treatments take place?

Step-by-step

Count each kind of event
Step 1
Count growth intervals in one cycle
No Desmos needed. The choices ask you to count changes, not calculate a population. Growth repeats every $6$ hours, so divide the 24-hour cycle by $6$ : $\frac{24}{6}=4$ . The culture grows four times before a treatment.
Step 2
Find the growth factor
A growth factor is the number that multiplies the current population. A $50\%$ increase keeps the original $100\%$ and adds $50\%$ , so each growth interval multiplies by: $1+0.50=1.50$ . Don't use $0.50$ ; that's only the amount added, not the new total.
Step 3
Find the treatment factor
The treatment destroys $30\%$ of the bacteria present at that moment. It leaves the other $70\%$ , so its decay factor is: $1-0.30=0.70$ . Multiply the population at treatment time by $0.70$ , not by $0.30$ .
Step 4
Write one complete cycle
Start with $N_0$ . The first cycle has four growth intervals followed by one treatment, so the population immediately after that treatment is: $N_0(1.5)^4(0.7)$ . The treatment factor acts on the population after its growth.
Step 5
Repeat the entire cycle
The second cycle starts with what the first left behind. Repeat both multipliers, not the starting population $N_0$ . Square the one-cycle multiplier: $N_0\left[(1.5)^4(0.7)\right]^2$ . Apply the power to each factor: $N_0(1.5)^{4\cdot2}(0.7)^2$ . Multiply the growth exponents: $N_0(1.5)^8(0.7)^2$ . This is the number of bacteria immediately after the second treatment. Choice C.

Question 399·Hard·Math

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