A colony of bacteria grows at a rate proportional to its size. Initially, there are 500 bacteria present. The colony doubles in size every 3 hours. How many bacteria are present after 14 hours?

I understand that the formulaa A=Ce^kt must be used but I am unsure about how to find each variable aand how it relates to finding the answer?

Question

A colony of bacteria grows at a rate proportional to its size. Initially, there are 500 bacteria present. The colony doubles in size every 3 hours. How many bacteria are present after 14 hours?

I understand that the formulaa A=Ce^kt must be used but I am unsure about how to find each variable aand how it relates to finding the answer?

anonymous · Answer

$$500\times 3^{\frac{14}{3}}$$ then a calculator

anonymous · Answer

you can do it your way if  you like but it is much more cumbersome because first you have to find k and then you have to compute

anonymous · Answer

want to do it that way?

anonymous · Answer

Haha id much rather do it your way as it seems a lot simpler. However, our professor is requiring us to work it out painfully using the above method

anonymous · Answer

yes it is much much simpler and more accurate too

anonymous · Answer

ok C is easy, C = 500

anonymous · Answer

to find k note that in 3 hours it triples, so replace t by 3 and A by 1500 and solve 
$$1500=500e^{3k}$$
$$3=e^{3k}$$
$$3k=\ln(3)$$
$$k=\frac{\ln(3)}{3}$$

anonymous · Answer

then you have your k, i guess you could use a calculator to find the decimal 
k =  0.3362

anonymous · Answer

now your formula is 
$$500e^{.3362t}$$ and then replace t by 5 to get your answer

anonymous · Answer

awesome. Im trying to work it out and use your help as an answer key. Thank you very much

anonymous · Answer

yw