6.1 Exponential Functions - College Algebra | OpenStax (2024)

Follow the order of operations. Be sure to pay attention to the parentheses.

$$\begin{array}{lll}f\left(x\right)\hfill & =5{\left(3\right)}^{x+1}\hfill & \hfill \\ f\left(2\right)\hfill & =5{\left(3\right)}^{2+1}\hfill & \text{Substitute}x=2.\hfill \\ \hfill & =5{\left(3\right)}^3\hfill & \text{Addtheexponents}.\hfill \\ \hfill & =5\left(27\right)\hfill & \text{Simplifythepower}\text{.}\hfill \\ \hfill & =135\hfill & \text{Multiply}\text{.}\hfill \end{array}$$

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 $N(t)=80b^t$ to find $b:$

$$\begin{array}{lll}N(t)\hfill & =80b^t\hfill & \hfill \\ 180\hfill & =80b^6\hfill & \text{Substituteusingpoint}(6,180).\hfill \\ \frac{9}{4}\hfill & =b^6\hfill & \text{Divideandwriteinlowestterms}.\hfill \\ b\hfill & ={\left(\frac{9}{4}\right)}^{\frac{1}{6}}\hfill & \text{Isolate}b\text{usingpropertiesofexponents}.\hfill \\ b\hfill & \approx 1.1447\begin{array}{cccc}& & & \end{array}\hfill & \text{Roundto4decimalplaces}.\hfill \end{array}$$

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 $N(t)=80{\left(1.1447\right)}^t.$ (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 in Figure 3 passes through the initial points given in the problem, $\left(0,80\right)$ and $\left(\text{6},180\right).$ We can also see that the domain for the function is $[0,\infty ),$ and the range for the function is $[80,\infty ).$

Figure 3 Graph showing the population of deer over time, $N(t)=80{\left(1.1447\right)}^t,$ $t$ years after 2006

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

• Substituting $\left(-2,6\right)$ gives $6=a{b}^{-2}$
• Substituting $\left(2,1\right)$ gives $1=a{b}^2$

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

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

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

Thus, the equation is $f(x)=2.4492{(0.6389)}^x.$

We can graph our model to check our work. Notice that the graph in Figure 4 passes through the initial points given in the problem, $\left(-2,\text{6}\right)$ and $\left(2,\text{1}\right).$ The graph is an example of an exponential decay function.

Figure 4 The graph of $f(x)=2.4492{(0.6389)}^x$ models exponential decay.

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

$$\begin{array}{ll}y=a{b}^x\hfill & \text{Writethegeneralformofanexponentialequation}.\hfill \\ y=3b^x\hfill & \text{Substitutetheinitialvalue3for}a.\hfill \\ 12=3b^2\hfill & \text{Substitutein12for}y\text{and2for}x.\hfill \\ 4=b^2\hfill & \text{Divideby3}.\hfill \\ b=\pm 2\begin{array}{cccc}& & & \end{array}\hfill & \text{Takethesquareroot}.\hfill \end{array}$$

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 $f(x)=3{(2)}^x.$

Follow the guidelines above. First press [STAT], [EDIT], [1: Edit…], and clear the lists L1 and L2. Next, in the L1 column, enter the x-coordinates, 2 and 5. Do the same in the L2 column for the y-coordinates, 24.8 and 198.4.

Now press [STAT], [CALC], [0: ExpReg] and press [ENTER]. The values $a=6.2$ and $b=2$ will be displayed. The exponential equation is $y=6.2\cdot 2^x.$

Because we are starting with $3,000, $P=3000.$ Our interest rate is 3%, so $r=\mathrm{0.03.}$ Because we are compounding quarterly, we are compounding 4 times per year, so $n=4.$ We want to know the value of the account in 10 years, so we are looking for $A\left(10\right),$ the value when $t=10.$

$$\begin{array}{lll}A(t)\hfill & =P{\left(1+\frac{r}{n}\right)}^{nt}\hfill & \text{Usethecompoundinterestformula}.\hfill \\ A(10)\hfill & =3000{\left(1+\frac{0.03}{4}\right)}^{\mathrm{4\cdot 10}}\begin{array}{cccc}& & & \end{array}\hfill & \text{Substituteusinggivenvalues}.\hfill \\ \hfill & \approx \text{\$}4045.05\hfill & \text{Roundtotwodecimalplaces}.\hfill \end{array}$$

The account will be worth about $4,045.05 in 10 years.

The nominal interest rate is 6%, so $r=\mathrm{0.06.}$ Interest is compounded twice a year, so $k=2.$

We want to find the initial investment, $P,$ needed so that the value of the account will be worth $40,000 in $18$ years. Substitute the given values into the compound interest formula, and solve for $P.$

$$\begin{array}{lll}A(t)\hfill & =P{\left(1+\frac{r}{n}\right)}^{nt}\hfill & \text{Usethecompoundinterestformula}.\hfill \\ \mathrm{40,000}\hfill & =P{\left(1+\frac{0.06}{2}\right)}^{2(18)}\begin{array}{cccc}& & & \end{array}\hfill & \text{Substituteusinggivenvalues}A\text{,}r,n\text{,and}t.\hfill \\ \mathrm{40,000}\hfill & =P{(1.03)}^{36}\hfill & \text{Simplify}.\hfill \\ \frac{\mathrm{40,000}}{{(1.03)}^{36}}\hfill & =P\hfill & \text{Isolate}P.\hfill \\ P\hfill & \approx \text{\$}13,801\hfill & \text{Divideandroundtothenearestdollar}.\hfill \end{array}$$

Lily will need to invest $13,801 to have $40,000 in 18 years.

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

$$\begin{array}{lll}A(t)\hfill & =P{e}^{rt}\hfill & \text{Usethecontinuouscompoundingformula}.\hfill \\ \hfill & =1000{(e)}^{0.1}\begin{array}{cccc}& & & \end{array}\hfill & \text{Substituteknownvaluesfor}P,r,\text{and}t.\hfill \\ \hfill & \approx 1105.17\hfill & \text{Useacalculatortoapproximate}.\hfill \end{array}$$

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

Since the substance is decaying, the rate, $17.3\%$ , is negative. So, $r=-\mathrm{0.173.}$ 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:

$$\begin{array}{lll}A(t)\hfill & =a{e}^{rt}\hfill & \text{Usethecontinuousgrowthformula}.\hfill \\ \hfill & =100e^{-0.173(3)}\begin{array}{cccc}& & & \end{array}\hfill & \text{Substituteknownvaluesfor}a,r,\text{and}t.\hfill \\ \hfill & \approx 59.5115\hfill & \text{Useacalculatortoapproximate}.\hfill \end{array}$$

So 59.5115 mg of radon-222 will remain.