We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation [latex]N\left(t\right)=80{b}^{t}[/latex] to find b:

[latex]\begin{array}{c}N\left(t\right)\hfill & =80{b}^{t}\hfill & \hfill \\ 180\hfill & =80{b}^{6}\hfill & \text{Substitute using point }\left(6, 180\right).\hfill \\ \frac{9}{4}\hfill & ={b}^{6}\hfill & \text{Divide and write in lowest terms}.\hfill \\ b\hfill & ={\left(\frac{9}{4}\right)}^{\frac{1}{6}}\hfill & \text{Isolate }b\text{ using properties of exponents}.\hfill \\ b\hfill & \approx 1.1447 & \text{Round to 4 decimal places}.\hfill \end{array}[/latex]

NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.

The exponential model for the population of deer is [latex]N\left(t\right)=80{\left(1.1447\right)}^{t}[/latex]. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)

We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below passes through the initial points given in the problem, [latex]\left(0,\text{ 8}0\right)[/latex] and [latex]\left(\text{6},\text{ 18}0\right)[/latex]. We can also see that the domain for the function is [latex]\left[0,\infty \right)[/latex], and the range for the function is [latex]\left[80,\infty \right)[/latex].

Because we don’t have the initial value, we substitute both points into an equation of the form [latex]f\left(x\right)=a{b}^{x}[/latex], and then solve the system for a and b.

• Substituting [latex]\left(-2,6\right)[/latex] gives [latex]6=a{b}^{-2}[/latex]
• Substituting [latex]\left(2,1\right)[/latex] gives [latex]1=a{b}^{2}[/latex]

Use the first equation to solve for a in terms of b:

[latex]\begin{array}{l}6=ab^{-2}\\\frac{6}{b^{-2}}=a\,\,\,\,\,\,\,\,\text{Divide.}\\a=6b^{2}\,\,\,\,\,\,\,\,\text{Use properties of exponents to rewrite the denominator.}\end{array}[/latex]

Substitute a in the second equation, and solve for b:

[latex]\begin{array}{l}1=ab^{2}\\1=6b^{2}b^{2}=6b^{4}\,\,\,\,\,\text{Substitute }a.\\b=\left(\frac{1}{6}\right)^{\frac{1}{4}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Use properties of exponents to isolate }b.\\b\approx0.6389\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Round 4 decimal places.}\end{array}[/latex]

Use the value of b in the first equation to solve for the value of a:

[latex]a=6b^{2}\approx6\left(0.6389\right)^{2}\approx2.4492[/latex]

Thus, the equation is [latex]f\left(x\right)=2.4492{\left(0.6389\right)}^{x}[/latex].

We can graph our model to check our work. Notice that the graph below passes through the initial points given in the problem, [latex]\left(-2,\text{ 6}\right)[/latex] and [latex]\left(2,\text{ 1}\right)[/latex]. The graph is an example of an exponential decay function.

We can choose the y-intercept of the graph, [latex]\left(0,3\right)[/latex], as our first point. This gives us the initial value, [latex]a=3[/latex]. Next, choose a point on the curve some distance away from [latex]\left(0,3\right)[/latex] that has integer coordinates. One such point is [latex]\left(2,12\right)[/latex].

[latex]\begin{array}{c}y=a{b}^{x}& \text{Write the general form of an exponential equation}. \\ y=3{b}^{x} & \text{Substitute the initial value 3 for }a. \\ 12=3{b}^{2} & \text{Substitute in 12 for }y\text{ and 2 for }x. \\ 4={b}^{2} & \text{Divide by 3}. \\ b=\pm 2 & \text{Take the square root}.\end{array}[/latex]

Because we restrict ourselves to positive values of b, we will use b = 2. Substitute a and b into the standard form to yield the equation [latex]f\left(x\right)=3{\left(2\right)}^{x}[/latex].

Since the account is growing in value, this is a continuous compounding problem with growth rate r = 0.10. The initial investment was $1,000, so P = 1000. We use the continuous compounding formula to find the value after t = 1 year:

[latex]\begin{array}{c}A\left(t\right)\hfill & =P{e}^{rt}\hfill & \text{Use the continuous compounding formula}.\hfill \\ \hfill & =1000{\left(e\right)}^{0.1} & \text{Substitute known values for }P, r,\text{ and }t.\hfill \\ \hfill & \approx 1105.17\hfill & \text{Use a calculator to approximate}.\hfill \end{array}[/latex]

The account is worth $1,105.17 after one year.

Since the substance is decaying, the rate, 17.3%, is negative. So, r = –0.173. The initial amount of radon-222 was 100 mg, so a = 100. We use the continuous decay formula to find the value after t = 3 days:

[latex]\begin{array}{c}A\left(t\right)\hfill & =a{e}^{rt}\hfill & \text{Use the continuous growth formula}.\hfill \\ \hfill & =100{e}^{-0.173\left(3\right)} & \text{Substitute known values for }a, r,\text{ and }t.\hfill \\ \hfill & \approx 59.5115\hfill & \text{Use a calculator to approximate}.\hfill \end{array}[/latex]

So 59.5115 mg of radon-222 will remain.