How do I find an equation for the amount of Thorium$-234$ at any time $t$? The isotope of Thorium-$234$ has a half-life of $24.5$ days. Find an equation which gives the amount of Thorium-$234$ at any time $t$ for a sample with initial size $2$ kg. At what time does the sample weight $0.75$ kg? > $$\ y = Ce^{-kt} \ $$ $$\ y = 2e^{-k(24.5)} \ $$ $$\ \ln{y} = \ln{\Big(6e^{-k(24.5)}\Big)} \ $$ $$\ \ln{y} = -6k(24.5) \ $$ I know to solve for $k$. However, if $y$ is unknown, then I can only get so far. Must I use $0.75$ kg as $y$ to find $k$? The above approach is incorrect, I am almost positive. Is the following the correct approach: > $$\ T_{1/2} = \frac{\ln{2}}{k} \ $$ $$\ k = \frac{\ln2}{T_{1/2}} = \frac{\ln{2}}{24.5} \approx 0.028 \ $$ Giving me > $$\ y = 2e^{-0.028t} \ $$ $$\ 0.75 = 2e^{-0.028t} \ $$ $$\ t = \frac{\ln(0.375)}{-0.028} \approx 35.0 \ $$ I want to clear myself of doubt of whether this is correct. You are all appreciated! Thanks, Michael

Yes, your second approach is correct. In the first approach, I'm not sure exactly how you got the $6$ to be the constant, and you would not be using half-life but rather $e$ as the exponential base. Not quite what you want. Also, when taking the natural logarithm of both sides, the $6$ disappeared. But yes, the approach you took the second time is correct, but I would suggest using the exact value instead of $-0.28$ and use $\frac{\ln(2)}{24.5}$ to avoid round-off.