Re: Radioactive Logarithim

If it takes 200 years for half of the isotope to decay, then we may state:

$\displaystyle \frac{1}{2}m_0=m_0\cdot10^{-200k}$

$\displaystyle \frac{1}{2}=10^{-200k}$

$\displaystyle \log(2)=200k$

$\displaystyle k=\frac{1}{200}\log(2)$

$\displaystyle M=m_0\cdot10^{-\frac{t}{200}\log(2)}=m_0\cdot2^{-\frac{t}{200}}$

$\displaystyle m_0=M\cdot2^{\frac{t}{200}}$

Re: Radioactive Logarithim

Well then, I was a little off.

But thank you so much, that makes more sense and helps me with the next part aswell :D