Question

A laconic exponential growth question

Answer 1

Why is the number of bacterium= (bacteria initial)*e^kt
Shouldn't it be 2^kt?

What is the proof for this?

Answer 2

either equation works

Answer 3

well if you are given the doubling time condition, then you can use
y = initial * 2^t/(doubling time)

Answer 4

, we generally dont use y = initial * 2^(kt) , thats a bit odd . usually we use 2^t when we have some extra information

Answer 5

http://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/
e is the growth constant, magically appearing in almost all exponential growth problems. The website provides excellent explanation why e appears. If you want to read it.

Answer 6

right, e has very nice properties., easy to integrate and differentiate. but technically you can use any base greater than 1 , if you have exponential growth

Answer 7

for instance, if i am given the condition, there are 50 bacteria at time zero. and they double every 10 minutes, that is the equation
y = 50 * 2^( t / 10) .
now if you use y = 50 e^(kt) you have to first find k

Answer 8

but the benefit of using 'e' really comes out in calculus, differential equations, etc

Answer 9

Thanks

Answer 10

but technically for exponential growht, you can use any base greater than 1 , so you can use
y = initial * 3^(kt) , but that seems a bit odd

Answer 11

the most common bases are 2 (doubling , growth) , 1/2 (halving , half-life)
and e (for continuous problems )