The population of bacteria triples each day on a petri dish. If it takes 20 days for the population of bacteria to fill the entire dish, how many days will it take bacteria to fill half of the petri dish?
My doubt 1: if we don't have a initial population can be solve this problem? Like $A(d)$= initial population times $3^{20}$ days but as I don't have this I am stuck.
My doubt 2: Also, can we assume a number for population at day 20 and move our way backwards?
On
The population of bacteria triples each day on a petri dish.
Let $\{ a_n \}$ the population of bacteria on the n-th day. We have $$a_{n+1} = 3 \cdot a_n$$ Furthermore, we have $$a_n = 3^{n-1} \cdot a_1$$
If it takes 20 days for the population of bacteria to fills the entire dish.
Let $S$ the total population of bacteria when the entire dish is filled. We have $$a_{20} \geqslant S > a_{19}$$ or $$3^{19} \cdot a_1 \geqslant S > 3^{18} \cdot a_1$$
How many days it will take bacteria to fill half of the petri dish?
Just divide the inequity by 2, we have $$\frac{3^{19}}{2} a_1 \geqslant \frac{S}{2} > \frac{3^{18}}{2} a_1$$
Let $3^{n-1} = \frac{3^{18}}{2}$ and $3^{m-1} = \frac{3^{19}}{2}$, we have $n = 19 - log_3 2 \approx 18.369$ and $m = 20 - log_3 2 \approx 19.369$.
It's obvious that on the 19th day, the dish is half-filled.
Let the population $N_t$ of bacteria after time (in days) $t$ with initial population $N_0$ be given by: $$N_t = N_0 e^{\lambda t}$$ where $\lambda$ is a constant. It is given $N_1 = 3N_0 = N_0 e^{\lambda}$. From this you can rearrange terms to get $\lambda = \ln (3)$. This means: $$N_t = N_0 e^{t \ln(3)} = N_03^t.$$
Let $M$ be the population that covers the whole petridish. It is given that $M = N_03^{20} \implies M/N_0 = 3^{20}$. The population that covers half the petridish should $M/2 = N_03^{T}$ where $T$ is the required time. $M/2N_0 = 3^{20}/2 = 3^{\log_3 (0.5)+ 20} = 3^T \implies T = \log_3(0.5) + 20 = 20 - \log_3 (2).$
I think this answers your questions.