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 May 1st, 2017, 07:46 PM #1 Senior Member   Joined: Apr 2008 Posts: 193 Thanks: 3 What did I do wrong in my solution? A certain radioactive substance has a half-life period of 35 days. How long will it take a 2 gram sample to decay to 0.05 grams? my solution dy/dt = ky where y is the amount of the substance Solving it gives lny=kt+c ------ (1) When t=0 days, y= 2 grams. So, plugging them into (1) above gives c=ln2 lny = kt + ln2 When t=35 grams, y=1 day. Next, I use the values to find k and get k=-ln2/35 So, lny= - (ln2)t/35 +ln2 ------ (2) When y = 0.05 grams, I plug it into (2). t=186 days. This is wrong. The correct answer is 167 days. I cannot get the answer. Can someone help me? Thanks.
 May 1st, 2017, 08:43 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,200 Thanks: 1155 I find $k=-\dfrac{\ln(2)}{35}$ just as you do and I get 186.267 days for 2g to decay to 0.05g. I think your answer key is mistaken. Thanks from topsquark
 June 28th, 2017, 10:01 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 Given that this substance "has a half life of 35 days", we can write $\displaystyle y(t)= A(1/2)^{t/35}$ where A is the original amount and t is in days. The original amount was 2 grams so $\displaystyle y(t)= 2(1/2)^{t/35}a$. To determine when it is reduced to 0.05 grams, solve $\displaystyle 2(1/2)^{t/35}= 0.05$. Divide both sides by 2: $\displaystyle (1/2)^{t/35}= 0.025$. Take the logarithm, base 10, $\displaystyle (t/35)log(1/2)= -(t/35)log(2)= log(0.025)=-1.602$. Since log(2)= 0.3010, t/35= 1.602/.3010= 5.322 and t= 35(5.322)= 186.2 days. Thanks from topsquark

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