Question

A chemist measures 8 grams of a substance that

has a half-life of 6 hours. Which equation models

the weight, y, of the substance after 15 hours?

Answer 1

`y(t) =8 (b)^t` Where b is the decay rate for this case. Using the condition given we have:

`4 = 8 b^6`

`(1)/(2)= b^6`

`b = ((1)/(2))^(1/6)`

And our model would be given by:

`y(t) = 8 ((1)/(2))^(t/6)`

And replacing the value of t=15 we got:

`y(15) = 8 ((1)/(2))^(15/6) = 1.414 grams`

Explanation:

For this case since the half life is 6 hours we have the following condition:

`y(6) = (1)/(2)A_o`

Where
`A_o=4` is the initial amount

Our model for this case is given by this expression:

`y(t) =8 (b)^t` Where b is the decay rate for this case. Using the condition given we have:

`4 = 8 b^6`

And solving for b we got:

`(1)/(2)= b^6`

And solving for b we got:

`b = ((1)/(2))^(1/6)`

And our model would be given by:

`y(t) = 8 ((1)/(2))^(t/6)`

And replacing the value of t=15 we got:

`y(15) = 8 ((1)/(2))^(15/6) = 1.414 grams`