| t |
Actual |
Predicted |
| 0 |
500 |
500 |
| 0.5 |
420 |
420 |
| 1 |
360 |
353 |
| 1.5 |
290 |
296 |
| 2 |
245 |
249 |
| 2.5 |
216 |
209 |
We can apply y=ae^(kt).
When t=0, y=a, so a=500. y=500e^(kt).
Now put some values for y and t:
y=420, t=0.5: 420=500e^(0.5k), (420/500)=e^(0.5k)
ln(0.84)=0.5k, k=2ln(0.84)=-0.3487 approx.
This would imply y=500e^(-0.3487t).
Now put t=2, when we expect y=245. But let’s see what the model gives: y=500e^(-0.3487×2)=249 approx. This is about 2% in error.
In the table above the third column shows the prediction according to the model. The predicted values are in error by around 2%.
The predicted half life is when y=250: half-life=-ln(2)/(2ln(0.84))=2 approximately. From the given table, 245 is the actual value when t=2, while the predicted value is 249—close to 250.
If we had used t=2 instead of t=0.50, k=-0.3567. So the predicted half-life would be 1.94, that is, approximately 2, as before. The average of the two k values is -0.3523. This value would give a half-life of 1.96, approximately 2.