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In just 5 weeks the population goes from 100 to 1000. What is the doubling time of this population? As in, how long does the population take to double?

i tried 1000=100(2)(t/5), where t is time in weeks
10=2(t/5)
log210=t/5
log(10)/log(2)=t/5
5*3.322=t
t=16.61

16.6 is quite a bit more than 5, which was the number of weeks it took to grow by 10x, so I knew that obviously wasn't correct. Can anyone help me figure out how this is solved or what formula is used?

Aleguan  Feb 1, 2018
 #1
avatar+86889 
+1

We  have this function :

 

P(t)  =  100(e)kt

 

And when t  = 5, we have that

 

1000 =  100(e)k5            divide both sides by  100

 

10  =  e5k        take the ln of both sides

 

ln 10  =  ln e5k     and we can write

 

ln 10  =  5k  ln e                  {  ln e   =  1...so we can ignore this }

 

ln 10 =  5k        divide both sides by 5

 

ln 10 / 5  =  k

 

The function is

 

P(t)  = 100(e) ln10 * t / 5

 

 

So.....we want to find the doubling time....and we have that

 

200  = 100(e)ln10 * t / 5       divide both sides by  100

 

2  =  e ln 10 * t / 5      take the ln  again

 

ln 2  = ln e ln 10 *  / 5      amd we can write

 

ln 2  =  [  (ln10) / 5 ] * t          

 

5 ln 2  =  ln 10 * t

 

5 ln 2 / ln 10   =  t  ≈     1.5 weeks

 

 

cool cool cool

CPhill  Feb 1, 2018
edited by CPhill  Feb 1, 2018
 #2
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Let the number of doublings=n, then:

 

2^n =1,000/100

2^n =10 take the log of both sides

n =3.3219 doublings in 5 weeks..

5/3.3219 =1.505 - weeks - period of 1 doubling, so that:

2^3.3219 =10x - population growth in 5 weeks.

Guest Feb 1, 2018

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