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cbear12345 · Aug 21, 2013, 05:45 AM

On my assignment I am really struggling with this question...

Bones A and B are x and y thousand years old respectively. A Geiger Counter confirms that Bone A contains 3 times as much C-14 as Bone B. What can you say about the ages of Bones A and B. Justify your answer

The equation used is

a=15.3 x 0.886^t

Thank you in advanced

ebaines · Aug 21, 2013, 06:00 AM

You can set up an equation for the ratio of C14 for bone A and bone B:

$ a_A = 15.3(0.866)^{t_A}, \ a_B = 15.3 (0.866)^{t_B} \\ \\ \\ \frac {a_A}{a_B} = 3 = \frac {15.3(0.866)^{t_A}}{15.3 (0.866)^{t_B}} $

Now you can manipulate this using logarithms to get an expression for $t_A - t_B$ . Post back with what you find and we'll help if you get stuck.

cbear12345 · Aug 21, 2013, 06:26 AM

I am not sure why you have the 3 awkwardly placed in that last line, like I know its times 3 but why is it there? Also do I use lo law and bring the ta and tb to the front?
Thank you so much for your help

ebaines · Aug 21, 2013, 06:47 AM

The 3 isn't awkward - it's the ratio of C14 in Bone A (which I called " $a_A$ ") to the amount in Bone B. Thus $a_A/a_B = 3$ .

cbear12345 · Aug 21, 2013, 06:51 AM

Oh okay, I understand
So then where do I go from there?

ebaines · Aug 21, 2013, 06:57 AM

Hint: take the log of both sides of the equation.

cbear12345 · Aug 21, 2013, 07:03 AM

I think I have it...
I got the answer
Ta/Tb=14.893

Is that correct?

ebaines · Aug 21, 2013, 07:12 AM

Originally Posted by cbear12345

I think I have it....
I got the answer
Ta/Tb=14.893

Is that correct?

No, not correct. Please show us the steps you took to get this and we can see where you went wrong.

cbear12345 · Aug 21, 2013, 07:14 AM

ln3=ln15.3(0.886)^ta / ln15.3(0.886)^tb
ln3=ta13.5558/tb13.5558
ln3 x 13.5558= ta/tb

hope that makes sense

ebaines · Aug 21, 2013, 07:20 AM

First step is wrong. It's not true that

$ \ln(\frac x y) = \frac {\ln(x)}{\ln(y)} $

Instead note that the 15.3 coefficients cancel out, and use the identity:

$ \ln(\frac x y) = \ln(x) - \ln(y) $

cbear12345 · Aug 21, 2013, 07:33 AM

ahhh true
so is it

ln3=ta-tb

ebaines · Aug 21, 2013, 07:45 AM

Getting closer, but you left out the 0.866 factor...