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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

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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?

Guest Jul 9, 2014

#1
+91053
+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.

Melody  Jul 13, 2014
Sort:

#1
+91053
+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.

Melody  Jul 13, 2014

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