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

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.

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

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.

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

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

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

Is that correct?

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.

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

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)

ahhh true
so is it

ln3=ta-tb

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