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A 15-g sample of radioactive iodine decays in such a way that the mass remaining after t days is given by 

m(t) = 15e 0.071t,

 where 

m(t)

 is measured in grams. After how many days is there only 1 g remaining?

 Jul 9, 2014

Best Answer 

 #1
avatar+99279 
+5

m(t) = 15e −0.071t

 

$$\begin{array}{rll}
m(t)&=&15e^{-0.071t}\\\\
1&=&15e^{-0.071t}\\\\
\frac{1}{15}&=&e^{-0.071t}\\\\
ln^{\frac{1}{15}}&=&ln{e^{-0.071t}}\\\\
ln^{\frac{1}{15}}&=&-0.071t\\\\
\frac{ln^{\frac{1}{15}}}{-0.071}&=&t\\\\
t&=&\frac{ln^{\frac{1}{15}}}{-0.071}\\\\


\end{array}$$

 

$${\frac{{ln}{\left({\frac{{\mathtt{1}}}{{\mathtt{15}}}}\right)}}{-{\mathtt{0.071}}}} = {\mathtt{38.141\: \!552\: \!128\: \!200\: \!140\: \!8}}$$

 

So in 38 days there will be just over 1g left.

In 39 days there will be less than 39 grams left.

 Jul 13, 2014
 #1
avatar+99279 
+5
Best Answer

m(t) = 15e −0.071t

 

$$\begin{array}{rll}
m(t)&=&15e^{-0.071t}\\\\
1&=&15e^{-0.071t}\\\\
\frac{1}{15}&=&e^{-0.071t}\\\\
ln^{\frac{1}{15}}&=&ln{e^{-0.071t}}\\\\
ln^{\frac{1}{15}}&=&-0.071t\\\\
\frac{ln^{\frac{1}{15}}}{-0.071}&=&t\\\\
t&=&\frac{ln^{\frac{1}{15}}}{-0.071}\\\\


\end{array}$$

 

$${\frac{{ln}{\left({\frac{{\mathtt{1}}}{{\mathtt{15}}}}\right)}}{-{\mathtt{0.071}}}} = {\mathtt{38.141\: \!552\: \!128\: \!200\: \!140\: \!8}}$$

 

So in 38 days there will be just over 1g left.

In 39 days there will be less than 39 grams left.

Melody Jul 13, 2014

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