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math help

C

crazeinc

Member
I'm embarassed to ask this, but I could use some help. Given a count of 400 after 2 hours, and a count of 25600 after 6 hours, how do I find the rate? (it's constant)
 
Oh if it helps ... sqrt of 400 = 20 and sqrt of 25600 is 160 (since 16^2 = 256)

i dunno if it helps

i didnt read the whole problem
 
It doesn't start at 0 at time 0 does it? Because then it's not a constant rate... are you sure it's constant?
 
I worded it incorrectly. The rate is somewhere around 2.85, but I need to figure out how to find it. The size is not zero at time 0.
 
Originally posted by: crazeinc
I worded it incorrectly. The rate is somewhere around 2.85, but I need to figure out how to find it. The size is not zero at time 0.
Click to expand...

Could you try to word it correctly then?
 
Originally posted by: YOyoYOhowsDAjello
Originally posted by: crazeinc
I worded it incorrectly. The rate is somewhere around 2.85, but I need to figure out how to find it. The size is not zero at time 0.
Click to expand...

Could you try to word it correctly then?
Click to expand...

A bacteria culture grows with a constant relative growth rate. The count was 400 after 2 hours and 25,600 after 6 hours. What's the initial population?
 
The answer is 50. I've been using this formula for other problems dealing with growth rates but I'm stuck: y(t)=y(original)e^(kt) k= constant
 
Originally posted by: crazeinc

A bacteria culture grows with a constant relative growth rate. The count was 400 after 2 hours and 25,600 after 6 hours. What's the initial population?
Click to expand...

Ok heres some info:

400*64 = 400*(2^6) = 25600
That means the bacteria duplicated 6 times in 4 hours
In 2 hours there are 3 duplications

400/(2^3) = 400/8 = 50
 
well i'm still kinda stuck on this...part b asked to find an expression for the population after t hours. The answer is y(t)=500e^((t*ln 8)/2) but I'm lost where the ln 8/2 came from
 
Originally posted by: crazeinc
well i'm still kinda stuck on this...part b asked to find an expression for the population after t hours. The answer is y(t)=500e^((t*ln 8)/2) but I'm lost where the ln 8/2 came from
Click to expand...
I think you have to solve for k for the exp. growth equation. If you solve for k, you get: k = [ln(P2/P1)]/(t2-t1) = (ln 64)/4 = [ln (8^2)]/4 = (ln 8)/2 = 1.0397. e^1.0397 = 2.828.

edit: I don't know where the 500 comes from, is that a typo and should be 50?
 
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