how to write an exponential function

Question

Andrew measures the amount of a very unstable substance to be 100 moles.  The half-life of  this substance is 3 days (after 3 days, half is gone).

 

How will I write an exponenential function that models this situation where, y, is the of substance and , x, is time in days.

 

 

How will I complete this chart:

 

        Days                            Amount

 

  -3(3 days prior)                      ??

_____________________________

       0                                      ??

_____________________________

       6                                      ??

_____________________________

 

 

Thank you

Answer 1

You can use the universal formula for exponential growth or decay.

y = 100(0.5)^x/3

y(-3) = 200 moles

y(0) = 100 moles

y(6) = 25 moles

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Attn:

Universal formula for exponential growth or decay

N(t) = N₀[N₁/N₀]^t/t₁, where t₁ is the time for N to reach N₁ from N₀.

Answer 2

The classical expression for radioactive decay is

 

y_t = y₀e^(-k)(x)

 

where y_t is the amount left after time x has passed, y₀ is the initial amount, and k is a constant that depends on the specific isotope, the units in which time is expressed, etc.

 

Notice that in the problem, they do not give us a value for k, but they did give us a way of calculating it from the half-life data.

 

At half-life (in this case 3 days) y_t/y₀ = 1/2

 

y_t/y₀ = 1/2 = e(-k)(x) = e^(-3k)

 

so that

 

-3k = ln(1/2)

 

k = [ln(2)/3] = ≈ 0.231

 

So that

 

y_t = y₀e^(-0.231)(x)

 

Remember, since we calculated k based on days, our equation will only work if subsequent value for x are also expressed in days.  Similarly, y must always be expressed in moles.

 

After six days, for example

 

y_t = (100 moles)e^{(-0.231)(6 days)}

 

y_t = 25.007 moles

 

Days          Amount

  -3                100

    0                 50

    6                 25.007

 

 

 

Answer 3

This type of question is asked before questions requiring an equation. It's meant to help the student get a handle on "half-life".

 

On day 0 Andrew will have the measured amount of 100 moles.

 

On day -3 there was 2 * 100 = 200 moles.

 

On day 3, 100 * 1/2 = 50 moles.

 

On day 6, 50 * 1/2 = 25 moles.

 

The equation can then be easily written as y = 100 * (1/2)^(x/3) because every 3rd day the amount is multiplied by another 1/2.

 

Let k = ln((1/2)^(1/3)) = -(ln 2)/3,

then y = 100 * e^(-x*(ln 2)/3).

This equation can be used to provide an "exact" answer for any particular value of x.

Answer 4

 N = N0 ( 1 - r ) ^t   the rate at which decays - t is in time, r is a rate of decay.

 

   100 ( 1-r ) ^3 = 50 / Half life of 100 moles after the decay

 

    ( 1-r ) ^3 = 1/2

 

   3 log ( 1-r)  = log 1/2 = - 0.3

 

             log ( 1-r ) = -0.3/ 3 = - 0.1

 

                1- r = 10 ^ (-0.1)= 0.79

           r = 1- 0.79 = 0.21

         

       100( 1- 0.79)^ t             t is in days,

  

         N =N0 ( 0.21^t)