APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

	APPLICATIONS OF EXPONENTIAL
	AND
	LOGARITHMIC FUNCTIONS

DECAY WORD PROBLEMS:

To solve an exponential or logarithmic word problem, convert the narrative to an equation and solve the equation.

Example 2: Hospitals utilize the radioactive substance iodine-131 in the diagnosis of conditions of the thyroid gland. The half-life of iodine-131 is eight days.

a: Determine the decay constant b.
b: If a hospital acquires 2 g of iodine-131, how much of this sample will remain after 20 days ?
c: How long will it be until only 0.01 g remains?

Solution and Explanation:

First, what does it mean to say that the half-life of iodine-131 is eight days ?

It means that after eight days only half of the original amount remains. After another eight days one-half of that amount remains. Another way of saying this is that after 16 days only $\displaystyle \frac{1}{2}$ of $\displaystyle \frac{1}{2}$ or $\displaystyle \frac{1}{4}$ of the original amount remains. Make a table showing the relationship between the number of days that have passes and the remaining amount of the iodine-131.

$\begin{eqnarray*}&& \\ 0 \mbox{ days } &:&{\quad }2{ g} \\ && \\ 8 \mbox{ day... ...{\quad }\displaystyle \frac{1}{2}\left( 1{ g}\right) =0.50 { g} \end{eqnarray*}$

$\begin{eqnarray*}24 \mbox{ days } &:&{\quad }\displaystyle \frac{1}{2}\left( 0.5... ...d }\displaystyle \frac{1}{2}\left( 0.125{ g}\right) =0.0625{ g} \end{eqnarray*}$

Let's us form the equation with base e:

$\begin{eqnarray*}&& \\ f\left( t\right) &=&a\cdot e^{bt} \\ && \\ && \end{eqnarray*}$

At time 0, the hospital had 2 g. We can say the same thing with the equation

$\begin{eqnarray*}&& \\ f\left( 0\right) &=&2. \\ && \\ && \end{eqnarray*}$

The equation is now

$\begin{eqnarray*}&& \\ f\left( t\right) &=&2\cdot e^{bt}. \\ && \\ && \end{eqnarray*}$

After 8 days, there is only 1 g left. Another way to say this is

$\begin{eqnarray*}&& \\ f\left( 8\right) &=&2\cdot e^{b\left( 8\right) } \\ && \\ f(8) &=&1 \end{eqnarray*}$

$\begin{eqnarray*}1 &=&2\cdot e^{b\left( 8\right) } \\ && \\ \displaystyle \frac{1}{2} &=&e^{b\left( 8\right) } \\ && \\ && \end{eqnarray*}$

Take the natural logarithm of both sides of the equation what you started The model now can be written as

$\begin{eqnarray*}&&\\ \displaystyle \frac{1}{2}&=&e^{b\left( 8\right) } \\ &&\... ...le \frac{1}{2}\right) &=&\ln \left( e^{b\left( 8\right) }\right) \end{eqnarray*}$

$\begin{eqnarray*}\ln \left( \displaystyle \frac{1}{2}\right) &=&8b\cdot \ln \lef... ... &&\\ \ln \left( \displaystyle \frac{1}{2}\right) &=&8b\cdot 1 \end{eqnarray*}$

$\begin{eqnarray*}b&=&\displaystyle \frac{\ln \left( \displaystyle \frac{1}{2}\ri... ...\ &&\\ f\left( t\right)&=&2\cdot e^{-0.086643\left( t\right) } \end{eqnarray*}$

The decay constant is $b=-0.086643.\bigskip\bigskip\bigskip$

How much of the iodine-131 will remain after 20 days? Just replace t in the equation with 20.

$\begin{eqnarray*}&& \\ f\left( t\right) &=&2\cdot e^{-0.086643\left( t\right) }... ... \\ f\left( 20\right) &=&2\cdot e^{-0.086643\left( 20\right) } \end{eqnarray*}$

$\begin{eqnarray*}&=&2\cdot \left( 0.176778\right) \\ && \\ &=&0.353556{ } \\ && \\ && \\ && \end{eqnarray*}$

After 20 days, there will be 0.353556 g left.

How long will it be until only 0.01 g remains? Replace $f\left( t\right)$ with 0.01 g and solve for t.

$\begin{eqnarray*}&& \\ f\left( t\right) &=&2\cdot e^{-0.086643\left( t\right) }... ...c{0.01}{2} &=&e^{-0.086643\left( t\right) } \\ && \\ && \\ && \end{eqnarray*}$

Take the natural logarithm of both sides of the equation.

$\begin{eqnarray*}&& \\ \ln \left( \displaystyle \frac{0.01}{2}\right) &=&\ln \l... ...ystyle \frac{0.01}{2}\right) &=&-0.086643\left( t\right) \cdot 1 \end{eqnarray*}$

$\begin{eqnarray*}t &=&\displaystyle \frac{\ln \left( \displaystyle \frac{0.01}{2... ...t) }{-0.086643} \\ && \\ t &\approx &61.151130{ \mbox{ days }} \end{eqnarray*}$

It would take a little over 61 days for the sample to be reduced to 0.01 g. To be more specific, it would take

$\begin{eqnarray*}&& \\ && \\ 61.151130 \mbox{ days } &=&61{ }\mbox{ days }+0.1... ... \mbox { day }} &=&61{ }\mbox{ days }+3.627120{ }\mbox{ hours } \end{eqnarray*}$

$\begin{eqnarray*}61{ }\mbox{ days }+3.627120{ }\mbox{ hours } &=&61{ }\mbox{ day... ...\mbox{ hours }{ }37.6272{ }\mbox{ minutes } \\ && \\ && \\ && \end{eqnarray*}$

It would take 61 days, 3 hours, and about 37 minutes for the 2 g sample to be reduced to 0.01 g.

Let's us check this answer

$\begin{eqnarray*}&& \\ f\left( 61.151130\right) &=&2\cdot e^{-0.086643\left( 61.151130\right) } \\ && \\ &=&0.01 \\ && \\ && \end{eqnarray*}$

If you would like to work another example, click on example

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