May 20th, 2014, 05:58 PM  #1 
Banned Camp Joined: May 2014 From: london Posts: 21 Thanks: 0  logarithms
There are two primary equations used when dealing with standard radioactive decay. 1.) N = N0e^kt 2.) K = ln2/r N represents the final amount remaining N0 represents the initial amount e is the exponential constant k is the constant exponent for that sample t represents the amount of time that passed r is the halflife of that sample ln stands for natural logarithm the halflife of radium226 is 1600 years. Suppose you have a 22mg sample. After how long will only 18mg of the sample remain? Last edited by skipjack; May 21st, 2014 at 01:21 PM. 
May 20th, 2014, 11:34 PM  #2 
Senior Member Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry 
If I understood it correctly $\displaystyle n=n_0 \cdot e^{kt } \\ k= \frac{\ln 2}{r}$ Therefore $\displaystyle k= \frac{\ln 2}{1600} \\ 18=22 \cdot e^{kt} \\ kt =\ln {(9/11)}$ $\displaystyle t=1600 \cdot \log_{2} (9/11)$ 
May 21st, 2014, 01:25 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,845 Thanks: 1567 
I can't be certain which is expected, but a value obtained by use of a calculator would often be acceptable for this type of question.

May 21st, 2014, 05:52 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,274 Thanks: 2435 Math Focus: Mainly analysis and algebra 
I think you are supposed to understand how the answer is arrived at rather than simply typing some stuff into a calculator. It's possible to train monkeys to push buttons, after all.

May 22nd, 2014, 01:11 AM  #5 
Senior Member Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry 
$\displaystyle \log_2 (9/11) \approx 0.289475$, therefor $\displaystyle t = 1600 \cdot (0.289475) = 463.16$

May 22nd, 2014, 06:59 AM  #6 
Senior Member Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry  