Interpolation of bacteria growth 
Sandy decided to do an experiment with a sandwich she left in her locker. She found it $3$ days after she had put it there, took it to the lab, and counted $41$ bacteria. $2$ days later she counted $153$ bacteria. She estimates that it will take $400$ bacteria to completely cover her sandwich.
How many days would it take this to happen, starting from the day she first put her sandwich in her locker?

I put it in a table so it can be easier to understand:

*

*$3$ days $= 41$ bacteria

*$5$ ($2$ days later) days = $153$ bacteria

I tried to identify what the problem was, but I am lost. Please help!
 A: We assume that the bacteria population keeps doubling after every constant time period,  $T_{\text{double}}$ which is known as the doubling time. Since the the bacteria population keeps doubling after some constant time period, it undergoes exponential growth. Let $P(t)$ be the population of the bacteria after time $t$, and let $P'(t)=\frac{d}{dt}P(t)$. From the fact that the bacteria grows exponentially, we know that $P(t)=kP'(t)$, where k is some constant. Furthermore, $P(t)=Ce^{kt}$, where C some constant. Given that $P(3)=41$ and $P(5)=153$, we are supposed to  find the time, $T$ that satisfies $P(T)=400$. 
In order to solve the problem, just plug in values of  $P(3)=41$ and $P(5)=153$ into the equation $P(t)=Ce^{kt}$. This gives us the equations convert the equations $P(3)=41$ and $P(5)=153$ into $Ce^{3k}=41$ and $Ce^{5k}=153$. Use these equations to find the values of $C$ and $k$. Then, solve the equation $$Ce^{kT}=400 \leadsto T=\frac{1}{k}\ln\frac{400}{C}$$
Just as a side note, the double time is given by the equation $T_{\text{double}}=\frac{\ln 2}{k}$
