Question

A scientist places 17 mg of bacteria in a culture for an experiment and he finds that the mass of the bacteria triples every day. The mass of the bacteria in the culture on any given day is what percent of the mass of bacteria in the culture exactly one day prior? Each day that passes, the mass of bacteria in the culture changes by what percent? What is the mass of the bacteria in the culture 3 days after the start of the experiment?

Answer 1

Answer: Hello!

usually the growth of a population is represented with an expoenential growth. And we also know that the mass triples every day,

this is $P(t) = A*3^{t - 1}$

where t is time in this case days, A is the initial population and r is the rate of grouth.

then in this case the initial population is 17mg, then A = 17mg.

so the function is : $P(t) = 17mg*3^{t - 1}$

first question: The mass of the bacteria in the culture on any given day is what percent of the mass of bacteria in the culture exactly one day prior?

Not need to do math here, you know that each day the mass of bactery triples, then for a given day, the mass is 200% bigger than the prior day (if the first day is 17, the second is 3*17 = 17 + 2*17, this is 100% + 200% of the population in the prior day, then the increment is of 200%).

then the third day you have 3*3*17 = 9*17 = 17 + 8*17, so here you have an 800% increment with respect of the first day.

the fourth day you have an increment of 3*3*3*17 = 27*17 = 17 + 26*17, so here you have an increment of 2600% with respect at the initial population, and so on. So there is a pattern:

the population that increments every day is $(3^{t} - 1)*100$ percent, where again, t is days.

What is the mass of the bacteria in the culture 3 days after the start of the experiment?

Here we need to calculate $P(3) = 17mg*3^{3-1} = 17mg*9 = 153mg$