Exponential Growth

Question

A colony if bacteria in a petri dish grows exponentially. At noon, there 1000 bacteria cells in the dish. At 8 pm, there 3000 cells in the dish. How long does it take the bacteria to double its population ?

from the above we know that it triples every 8 hours, based on that triple rate I don't know how to get the double.

Answer 1

if the population function is $P(t)$ $$P(t)=P_0e^{kt}$$ $P_0=1000$,let $t=0$ start at noon . then,$$ P(t)=1000e^{kt} $$ find $k$ with the second condition. $$3000=1000e^{8k}$$ $$k=\frac{1}{8}\ln 3$$ Then find $t$ for $P(t)=2P_0$

Answer 2

You are given that the population is $1000\cdot 3^{\frac t8}$, where $t$ is measured in hours since noon, so the equation to solve is $2000=1000\cdot 3^{\frac t8}$ Now take the logs of both sides and isolate the $t$

Answer 3

It is an exponential grow in the form $p(t) = p_0e^{at}$.

For $t=0$ (noon), then $p(0)=p_0e^{0a}=p_0$, then $p_0$ is the initial population of $1000$.

For $t=8$ (8 pm) then $p(8)=1000e^{8a}=3000$, replace and you have $e^{8a}=3$ or $a=\frac18\ln(3)$, so the formula becomes:

$$p(t)=1000e^{\frac18\ln 3\cdot t}=1000\cdot3^{t/8}$$

Now, we want $t$ such as $p(t)=2000$, then \begin{align} p(t) = 1000\cdot e^{\frac18\ln3\cdot t} &= 2000\\ e^{\frac18\ln3\cdot t} &= 2\\ \frac18\ln3\cdot t &= \ln2 \\ t &= 8\frac{\ln2}{\ln3} = 8\log_32=8\frac1{\log_23} \end{align}

Answer 4

The defining property of exponential growth is that, when you take the log of all dependent values of the function, you end up with a straight line function. In this case, the slope is $ \frac{\ln 3000-\ln 1000}{8-0} =\frac{\ln 3}{8} $ And you want $t$ such that $ \frac{\ln 2000 - \ln 1000 }{t-0} = \frac{\ln 3}{8} $ That is, $t= \frac{\ln 2}{\ln 3}8$ hours, which makes rough sense because you have a tripling after 8 hours--so it better double after fewer hours than 8.