Decay and Logarithm Problem: Natural vs Compounding

[noobforlife]

Here's the problem:

Bismuth-210 is an isotope that decays by about 13% each day. What is the half-life of Bismuth-210?

I wanted to solve this using Euler's number. But, the answer seemed far off from what the book suggested. Am I doing something wrong?

THE BOOK'S SOLUTION

.5a = a(.87)^t
.5 = (.87)^t
log(.5) = t * log(.87)
(log .5)/(log .87) = t
4.977 days = t, approximately

That all makes sense to me and how they got the answer.

MY APPROACH

.5a = ae^-.13t
.5 = e^-.13t
ln(.5) = -.13t
(ln .5)/(-.13) = t
5.33 = t approximately

A 7% spread in answers seemed wrong to me on this. Should I not be using continuous decay? I think I should since it decays continuously. So, am I doing the calculation incorrectly?

Thanks!

 

[MarkFL]

If you are going to use \(e\), then you should write:

[MATH]A(t)=A_0e^{-kt}[/MATH]
Now, we are told:

[MATH]A(1)=0.87A_0[/MATH]
Hence:

[MATH]A(1)=A_0e^{-k}=0.87A_0[/MATH]
[MATH]e^{-k}=0.87[/MATH]
[MATH]-k=\ln(0.87)[/MATH]
Thus:

[MATH]A(t)=A_0e^{t\ln(0.87)}=A_0e^{\ln(0.87^t)}=A_0(0.87)^t[/MATH]
It's quite a bit easier to just recognise that 87% remains at the end of each day relative to the end of the previous day, and simply begin with:

[MATH]A(t)=A_0(0.87)^t[/MATH]
Either way, the solution given by your book then follows.

 

[HallsofIvy]

Yes, a 7% difference in answer is a bit much- and it is "your approach" that is wrong! You rounded incorrectly.

In "your approach" you have \(\displaystyle e^{-0.13t}\). Where did you get that "-0.13"? \(\displaystyle 0.87^t= e^{ln(0.87^t)}= e^{t ln(0.87)}\). And ln(0.87)= -0.13926 which, to two decimal places is -0.14, not -0.13. with \(\displaystyle .5a= ae^{-0.14t}\), \(\displaystyle ln(0.5)= -0.14t\) so \(\displaystyle t= \frac{-0.69}{-0.14}= 4.95\) which is much closer.

 

[Dr.Peterson]

Here's the problem:

Bismuth-210 is an isotope that decays by about 13% each day. What is the half-life of Bismuth-210?

I wanted to solve this using Euler's number. But, the answer seemed far off from what the book suggested. Am I doing something wrong?

THE BOOK'S SOLUTION

.5a = a(.87)^t
.5 = (.87)^t
log(.5) = t * log(.87)
(log .5)/(log .87) = t
4.977 days = t, approximately

That all makes sense to me and how they got the answer.

MY APPROACH

.5a = ae^-.13t
.5 = e^-.13t
ln(.5) = -.13t
(ln .5)/(-.13) = t
5.33 = t approximately

A 7% spread in answers seemed wrong to me on this. Should I not be using continuous decay? I think I should since it decays continuously. So, am I doing the calculation incorrectly?

Thanks!

It appears that you are assuming (wrongly) that the k in ae^kt, which is the rate in continuous compounding, is the same as the r in a(1-r)^t, which represents annual compounding. Your problem is not about compound interest, but is equivalent - as long as you identify the correct rate. That is, both formulas do the same thing, but in terms of different ways to measure growth or decay rate.

Consider the case of compound interest. Using continuous compounding at, say 10%, after one year $1 grows to 1*e^-.10*1 = $1.105, which amounts to 10.5% annual interest (off by 5% in this case). If you are told that the annual interest rate is 10%, then calculating 10% continuous interest would give the wrong result. That is effectively what you are doing. In your problem you are, in effect, told that you earn -13% "interest" each year -- not as a continuous rate. So that should be your r, not your k.

 

[noobforlife]

Thank you all so much, both for the verbal and mathematical approaches. Once you realize that the rate will be different with continuous decay, it's obvious that I need to solve for the correct k.

Duh!