We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
1708
1
avatar

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+101054 
+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+101054 
+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

8 Online Users

avatar