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Question

Find the half-lives of the substances

1) einsteinium-253, which decays at a rate of 3.406% per day

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2) You place $800 in an account that earns 4% annual interest, compounded annually. How long will it be until you have $2000?

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3) A population grows from 11000 to 13000 in three years. Assuming the growth is exponential, find the:

a) annual growth rate

b) continuous growth rate

 

Answer 1

Find the half-lives of the substances
1) einsteinium-253, which decays at a rate of 3.406% per day

1/2 = 1*e^(-0.03406 h), where h is the half-life

ln(1/2) = -0.03406 h

h = -ln(1/2)/0.03406 ≈ 20.35076866001014 days


2) You place $800 in an account that earns 4% annual interest, compounded annually. How long will it be until you have $2000?

2000 = 800(1.04)^t

(1.04)^t = 5/2

t log(1.04) = log(5/2)

t = log(5/2)/log(1.04) ≈ 23.36241894157182 yrs

 

→ 24 years because you have to wait until the interest is added that year.


3) A population grows from 11000 to 13000 in three years. Assuming the growth is exponential, find the:

a) annual growth rate
b) continuous growth rate

13000 = 11000 e^(3k)

3k = ln(13/11)
k = ln(13/11)/3 = continuous growth rate ≈ 5.568469488772 %

P = Ae^((ln(13/11)/3)t)

r = e^(ln(13/11)/3) - 1 ≈ 5.726427034643 % = annual growth rate

Answer 2

Sarah,

       1.  Let A₀ be the amount of Es at t=0 then A(t)=A₀e^-kt where his the decay rate in days^-1  so the half life is the time for an initial amount to decay to 1/2 it's original amount. You can rearrange the above and solve for t_1/2 =- Ln(1/2)/k, k=.03406 day^-1 so t_1/2=20.35 days. 

 

       2. Solve 2000=800(1.04)ⁿ for n,  hint divide both sides by 800 and then take logs of both sides.

 

       3. a.13000=11000(1+i)³ solve for i, i is the annual growth rate

           b. 13000=11000e^3k solve for k, k is the continuous growth rate.

           k and i are related through (1+i)=e^k and for "small" growth rates are approximately equal.

Regards

 

Answer 3

a)

A = 800 ( 1 + 0.04) ^n

 

2000 = 800 ( 1.04) ^n

 

5/2 = ( 1.04 ) ^n

 

n( log 1.04) = log ( 10/4)

 

    n = log( 5/2) / log (1.04)

   

   n = 23.36

 

 or

 

2000 = 800 e^0.04n=

 

5/2 = e^0.04n

 

0.04 n = ln( 5/2)

 

n = 22.91

 

   Both methods are equally valid , and the answer will be slightly different.

 

 

S =S₀ e^{- 0.03406t}   

 

1/2 = e^{- 0.0346t}

 

ln( 1/2 ) = - 0.0346t

 

t = 20.02 days

 

 

2)

 

In order to get the same answer:

   

 ( 1.04 ) ^ n = e ^x

 

 

3 )

    13,000 = 11,000 e^3r 

 

    13/11 = e^3r

 

    3r = ln( 13/11)

 

     r = ln( 13/11) /3 = 0.056

 

     r = 5.6 %

 

 

 

   n ln1.04 = x

 

     2000 = 800 e^ ( nln1.04)

 

   ln  (5/2)  =n ln ( 1.04)

 

  n = ln( 5/2) / ln( 1.04) = 23.36

  

     But this method is not promoted by majority of text books, and any of the 1st and 2nd method with good enough approximation, slightly different is acceptable.