# I think I have this problem almost solved but I am stuck in some areas. Andrew measures the amount of a very unstable substance to be 100 moles. The half-life of this substance is 3 days (after 3...

I think I have this problem almost solved but I am stuck in some areas.

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).

1. Write an exponential function that models this situation where, y, is the

amount of substance and ,x, is time in days. I have y = 100(.5) ^x/3

2. Complete this table: " Days " "Amount"

_____ ________

-3(3 days earlier) 200

0 100

3 50

6 25

9 12.50

12 6.25

3. If the table is correct, how is this information then shown on a graph

to graph the above equation? (Is my equation correct?)

4. Then how would I show the calculation and consider the trend of

the graph if I calculated the expected amount of substance if Andrew

had taken his measurement 9 days earlier?

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To calculate the expected amount of substance if Andrew had taken his measurement 9 days earlier, substitute x=-9 in the original equation i.e `y=100*(1/2)^(x/3).`

**So, the equation would look like this**:

`y=100*(1/2)^(-9/3)`

Now simplify to get the expected amount of substance i.e:

`y=100*(1/2)^(-3)=100*8=800` moles.

An exponential decay process can be described by the formula `N(t)=N_0(1/2)^(t/t_(1/2))` where `N_0` is the initial quantity of the substance that will decay, `N(t)` is the quantity that still remains and has not yet decayed after a time `t` , and `t_(1/2)` is the half-life of the decaying quantity.

The half life of the substance is 3 days and the initial amount is 100 moles.

Considering, y, is the amount of substance and ,x, is time in days, **the exponential function that models the situation is `y=100*(1/2)^(x/3)` .**

To determine the amount of substance at any moment of time, after it was initially measured or before it, substitute the value of x in the formula.

For instance 3 days earlier, the amount of substance would have been

`y(-3)=100*(1/2)^-(3/3)=100*(1/2)^-1=100*2=200`

` `

Similarly,

`y(0)=100`

`y(3)=50`

`y(6)=25`

`y(9)=12.5`

`y(12)=6.25`

So, the above table is correct. The values at the other instances of time can be estimated in a similar way.

The graph of the amount of substance versus time elapsed is: