Question 487797: Please help!!!
I have this problem to solve with a detailed answer please
A colony of bacteria can be modeled by N(t)= 1000e ^0.0014t, where N is measured in bacteria per millilter and t is minutes, evaluate N(0) and interpret results, estimate how long it takes for N to double
Thank you please I need to have this answered for my final exam today
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! colony of bacteria are modeled by n(t) = 1000 * e^(0.00145 * t).
n is measured in bacteria per milliliter.
t is in minutes.
evaluate n(0) and interpret results.
estimate how long it takes for n to double.
n(0) would be equal to 1000 * e^(0.00145 * 0) which would be equal to 1000 * e^0 which would be equal to 1000 * 1 which would be equal to 1000 bacteria per milliliter.
this is because e^0 = 1.
actually, anything to the 0 power is equal to 1 except 0.
0^0 cannot be determined.
that's not your problem though.
in your problem, e^0 = 1 and the formula of:
n(0) = 1000 * e^(0.00145 * t) becomes:
n(0) = 1000 * e^(0.00145 * 0) which becomes:
n(0) = 1000 * e^0 which becomes:
n(0) = 1000 * 1 which becomes:
n(0) = 1000
since n is expressed in number of bacteria per milliliter then:
n(0) = 1000 bacteria per milliliter.
your formula of:
n(0) = 1000 * e^(0.00145 * 0) becomes:
1000 = 1000 * e^(0.00145 * 0)
divide both sides of this equation by 1000 and you get:
1 = e^(0.00145*0)
this becomes:
1 = e^0 which becomes:
1 = 1
this confirms the calculation was good.
in order for the bacteria to double, it would have to go from 1000 to 2000.
you have n(t) = 1000 * e^(0.00145 * t)
you want to know the value of t when n(t) is equal to 2000.
your formula becomes:
2000 = 1000 * e^(0.00145 * t)
divide both sides of this equation by 1000 to get:
2 = e^(0.00145 * t)
take the natural log of both sides of this equation to get:
ln(2) = ln(e^(0.00145 * t))
since, in general, log(a^x) = x * log(a), then your equation of:
ln(2) = ln(e^(0.00145 * t)) becomes:
ln(2) = 0.00145 * t * ln(e)
since ln(e) is equal to 1, then your this equation becomes:
ln(2) = 0.00145 * t
divide both sides of this equation by 0.00145 and you get:
ln(2) / 0.00145 = t
use your calculator to solve for t to get:
t = 478.l0325383
if we did this right, it should take 478.10325383 minutes for the bacteria to double.
we test this by substituting that value for t in the equation of:
2000 = 1000 * e^(0.00145 * t) to get:
2000 = 1000 * e^(0.00145 * 478.10325383) which becomes:
2000 = 1000 * e^(.693147181) which becomes:
2000 = 2000
the formula works, so the answer is that:
the bacteria will double in 478.10325383 minutes.
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