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 3: The radioactive isotope sodium-24 is used as a tracer to measure the rate of flow in an artery or vein. The half-life of sodium-24 is 14.9 hours. Suppose that a hospital buys a 40-g sample of sodium-24 and will reorder when the sample is reduced to 3 g.

a: How much of the sample will remain after 50 hours?.
b: How often before the hospital has to reorder sodium-24?
c: How much of the sample will remain after 1 year?

Solution and Explanation:

First, what does it mean to say that the half-life of sodium-24 is 14.9 hours? It means that after 14.9 hours only half of the original amount remains. After another 14.9 hours one-half of that one-half amount remains. Another way of saying that is that after 29.8 hours 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 hours that have passes and the amount of sodium-24 remaining.

$\begin{eqnarray*}&& \\ && \\ 0\mbox{ hours } &:&{\quad 40 g} \\ && \\ 14.9\m... ...&{\quad }\displaystyle \frac{1}{2}\left( 20{ g}\right) =10 { g} \end{eqnarray*}$

$\begin{eqnarray*}44.7\mbox{ hours } &:&{\quad }\displaystyle \frac{1}{2}\left( 1... ...\quad }\displaystyle \frac{1}{2}\left( 2.5{ g}\right) =1.25 { g} \end{eqnarray*}$

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 40 g. We can say the same thing with the equation

$\begin{eqnarray*}&& \\ f\left( 0\right) &=&40. \\ && \\ && \end{eqnarray*}$

The equation is now

$\begin{eqnarray*}&& \\ f\left( t\right) &=&40\cdot e^{bt}. \\ && \\ && \end{eqnarray*}$

After 14.9 hours,there is only $\displaystyle \frac{1}{2}$ of 40=20 g left. Another way of say this is

$\begin{eqnarray*}&& \\ f\left( 14.9\right) &=&20 \\ && \\ f(14.9) &=&40\cdot e^{b\left( 14.9\right) } \end{eqnarray*}$

$\begin{eqnarray*}\displaystyle \frac{20}{40} &=&e^{b\left( 14.9\right) } \\ && \\ \displaystyle \frac{1}{2} &=&e^{14.9b} \\ && \\ && \end{eqnarray*}$

Take the natural log of both sides of the equation.

$\begin{eqnarray*}&& \\ \ln \left( \displaystyle \frac{1}{2}\right) &=&\ln \left... ...&\\ \ln \left( \displaystyle \frac{1}{2}\right) &=&14.9b\cdot 1 \end{eqnarray*}$

$\begin{eqnarray*}b &=&\displaystyle \frac{\ln \left( \displaystyle \frac{1}{2}\right) }{14.9} \\ && \\ b &\approx &-0.046520 \end{eqnarray*}$

The equation can be written as

$\begin{eqnarray*}f\left( t\right) &=&40\cdot e^{-0.046520t} \end{eqnarray*}$

The decay constant is $b=-0.046520.\bigskip\bigskip$

How much of the sodium-24 will remain after 50 hours? Just replace t in the formula with 50.

$\begin{eqnarray*}f\left( 50\right) &=&40\cdot e^{-0.046520\left( 50\right) } \\ ... ...\\ &=&40\cdot \left( 0.097686\right) \\ && \\ &=&3.907428{ g} \end{eqnarray*}$

How often before the hospital has to reorder sodium-24? or How long will it before the sample is reduced to 3 g? Replace $f\left( t\right)$ with 3 g and solve for t.

$\begin{eqnarray*}f\left( t\right) &=&40\cdot e^{-0.046520t} \\ && \\ 3 &=&40\c... ...046520t} \\ && \\ \displaystyle \frac{3}{40} &=&e^{-0.046520t} \end{eqnarray*}$

Take the natural logarithm of both sides of the equation.

$\begin{eqnarray*}\displaystyle \frac{3}{40} &=&e^{-0.046520t} \\ && \\ \ln \le... ...laystyle \frac{3}{40}\right) &=&\ln \left( e^{-0.046520t}\right) \end{eqnarray*}$

$\begin{eqnarray*}\ln \left( \displaystyle \frac{3}{40}\right) &=&-0.046520t\cdot... ...ln \left( \displaystyle \frac{3}{40}\right) &=&-0.046520t\cdot 1 \end{eqnarray*}$

$\begin{eqnarray*}t &=&\displaystyle \frac{\ln \left( \displaystyle \frac{3}{40}\right) }{-0.046520} \\ && \\ t &\approx &55.680722\mbox{ hours } \end{eqnarray*}$

How much of the sample will remain after 1 year?

First convert 1 year to hours.

$\begin{eqnarray*}1\mbox{ year }\cdot \displaystyle \frac{365\mbox{ days }}{1\mbo... ...le \frac{24\mbox{ hours }}{1\mbox{ day }} &=&8,760\mbox{ hours } \end{eqnarray*}$

Substitute 8,760 for t in the equation $f\left( t\right) =40\cdot e^{-0.046520t}$ .

$\begin{eqnarray*}f\left( 8,760\right) &=&40\cdot e^{-0.046520\left( 8,760\right) } \\ && \\ &=&4.17308604588\times 10^{-176} \end{eqnarray*}$

This is an extremely small number.

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