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**+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

#1**+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