exponential functions

**doctorgk** · March 25th 2008, 07:56 AM

okay, this is a weird problem...thanks to anyone who can solve it

A certain strain of bacteria is living in a can. They are good at reproducing and are capable of doubling their population every 60 seconds. Unfortunately, they only live for 1 hour. The first bacterium invaded the can at 10:00 PM and at 10:57 PM one bacterium shouts, "We've been alive for 57 minutes and the can is only one-eighth full. Party."
--A--How much longer will the party go on?

As 11:00 approaches, an expedition is formed to locate a new can to live. A returning bacterium announces, "Hooray, we have located 3 more empty cans. Thats three times more space than we've ever had in our entire existence.
--B--How much more time has the expedition provided?

**earboth** · March 25th 2008, 08:36 AM

Originally Posted by doctorgk

okay, this is a weird problem...thanks to anyone who can solve it

A certain strain of bacteria is living in a can. They are good at reproducing and are capable of doubling their population every 60 seconds. Unfortunately, they only live for 1 hour. The first bacterium invaded the can at 10:00 PM and at 10:57 PM one bacterium shouts, "We've been alive for 57 minutes and the can is only one-eighth full. Party."
--A--How much longer will the party go on?

...

If the time is measured in minutes the amount of bacteria can be calculated by:

$a(t) = 1\cdot 2^t$

After 60 minutes the first bacterium will die. There are missing 3 minutes to this moment. Since 2³ = 8 the population of bacteria will increase 8 times and will fill the can completely!

... and the party will become a funeral

**earboth** · March 25th 2008, 08:43 AM

Originally Posted by doctorgk

...

As 11:00 approaches, an expedition is formed to locate a new can to live. A returning bacterium announces, "Hooray, we have located 3 more empty cans. Thats three times more space than we've ever had in our entire existence.
--B--How much more time has the expedition provided?

At 11:00 the can is filled completely minus 1 bacterium.

If the amount of bacteria is measured in cans the number of cans can be calculated by:

$c(t) = 2^t$

There are 3 cans available:

$3 = 2^t~\iff~ \ln(3)=t \cdot \ln(2)~\iff~ t =\frac{\ln(3)}{\ln(2)} ~ \approx ~1.5849625\ min =$ $1 min 35 s$

**iknowone** · March 25th 2008, 08:55 AM

Ok so the can is able to hold $(8)(2^{57}) = 2^{60}$ bacteria since at minute n there are $2^n$ bacteria and after 57 minutes the can is 1/8 full. Provided that a dead bacteria disintigrates into the ether upon death and doesn't take up any room in the can then at 11:00 there will be $2^{60} - 1$ bacteria in the can, (the first guy died). In the next minute however the population will double and only 1 more bacterium dies (the 1 created at 10:01). Since $2(2^{60} -1) - 1 >> 2^{60}$ the party will end at 11:01 (4 minutes after the first said "Party").

Given 3 times the space, the group is able to support $(3)2^{60}$ bacteria. $2(2^{60} -1) - 1 < (3)2^{60} < 2(2(2^{60} -1) - 1) - 2$
-the least value in this inequality is the population at 11:01, the greatest value is the population at 11:02, so it bought them 1 minute.

What do you think? I guess I am assuming the doubling happens instantly at the end of the minute and immediately after the bacteria 1 hour old die. (more of a mind game interpretation than a population growth).

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