Intermediate Algebra: Functions and Graphs

We would like to find values of \(a\) and \(b\) so that the given function values satisfy \(f(x) = ab^x\) . By substituting the function values into the formula, we can write two equations.

This is a system of equations in the two unknowns, \(a\) and \(b\text{,}\) but it is not a linear system. We can solve the system by the method of elimination, but instead of adding the equations, we will divide one of the equations by the other.

Note that by dividing the two equations, we eliminated \(a\text{,}\) and we can now solve for \(b\text{.}\)

Next we substitute \(b = 3\) into either of the two equations and solve for \(a\text{.}\)

Thus, \(a = 0.5\) and \(b = 3\text{,}\) and the function is \(f(x) = 0.5(3^x)\text{.}\)

We want to find a function \(C(t) = ab^t\) for the number of cedi to the dollar, where \(t = 0\) in 1986. We have two function values, \(C(7) = 720\text{,}\) and \(C(10) = 1620\text{,}\) and with these values we can write two equations.

We divide the second equation by the first to find

Now we can solve this last equation for \(b\) to get \(b = \sqrt[3]{2.25}\approx 1.31\text{.}\) Finally, we substitute \(b = 1.31\) into the first equation to find \(a\text{.}\)

Thus, \(C(t) = 108.75(1.31)^t\) , and the annual inflation rate was 31%.

1. The population of Egypt is growing according to the formula

   \begin{equation*} P(t) = 74(1.02)^t \end{equation*}

   where \(t\) is in years and \(P(t)\) is in millions. We would like to know when the population will reach 148 million (twice 74 million), so we solve the equation

   \begin{align*} 74(1.02)^t \amp= 148\amp\amp \blert{\text{Divide both sides by 74.}}\\ 1.02^t \amp= 2\amp\amp \blert{\text{Take the log of both sides.}}\\ t \log 1.02 \amp= \log 2\amp\amp \blert{\text{Divide both sides by log 1.02.}}\\ t \amp= \frac{\log 2}{\log 1.02} \approx 35 \text{ years} \end{align*}

   It will take the population about 35 years to double.

2. Twice 148 million is 296 million, so we solve the equation

   \begin{align*} 148(1.02)^t \amp= 296\amp\amp \blert{\text{Divide both sides by 148.}}\\ 1.02^t \amp= 2\amp\amp \blert{\text{Take the log of both sides.}}\\ t \log 1.02 \amp= \log 2\amp\amp \blert{\text{Divide both sides by log 1.02.}}\\ t \amp= \frac{\log 2}{\log 1.02} \approx 35 \text{ years} \end{align*}

   It will take the population about 35 years to double again.

3. A graph of \(P(t) = 74(1.02)^t\) is shown below. Note that the population doubles every 35 years.

   

1. The current population of the United States is not given, so we represent it by \(P_0\text{.}\) With \(t\) expressed in years, the formula is then
   \begin{equation*} P(t) = P_0 2^{t/115.87} \end{equation*}
2. We write \(2^{t/115.87}\) in the form \(\left(2^{1/115.87}\right)^t\) to see that the growth factor is \(b = 2^{1/115.87}\text{,}\) or 1.006. For exponential growth, \(b = 1 + r\text{,}\) so \(r = 0.006\text{,}\) or 0.6%.

1. To find the half-life, we calculate the time elapsed when only half the original amount, or 100 mg, is left.

   \begin{align*} 200(0.73)^t \amp = 100\amp\amp \blert{\text{Divide both sides by 200.}}\\ 0.73^t \amp = 0.5\amp\amp \blert{\text{Take the log of both sides.}}\\ t\log 0.73 \amp = \log 0.5\amp\amp \blert{\text{Divide both sides by log 0.73.}}\\ t \amp= \frac{\log 0.5}{\log 0.73} = 2.2 \end{align*}

   The half-life is 2.2 hours.

2. After 2.2 hours, 100 mg of ibuprofen is left in the body. After another 2.2 hours, half of that amount, or 50 mg, is left. Thus, 50 mg remain after 4.4 hours.

3. We locate multiples of 2.2 hours on the horizontal axis. After each interval of 2.2 hours, the amount of ibuprofen is reduced to half its previous value. The graph is shown below.

   

   \(t\)	\(0\)	\(2.2\)	\(4.4\)	\(6.6\)	\(8.8\)
   \(Q(t)\)	\(200\)	\(100\)	\(50\)	\(25\)	\(12.5\)

1. We let \(Q(t)\) denote the amount of cobalt-60 left after \(t\) years, and let \(Q_0\) denote the initial amount. Every 5.27 years, \(Q(t)\) is reduced by a factor of 0.5, so
   \begin{equation*} Q(t) = Q_0 (0.5)^{t/5.27} \end{equation*}
2. We rewrite the decay law in the form \(Q(t) = Q_0 (1 - r )^t\) as follows:
   \begin{equation*} Q(t) = Q_0 (0.5)^{t/5.27}=Q_0 \left((0.5)^{1/5.27}\right)^t = Q_0 (0.8768)^t \end{equation*}
   Thus, \(1 - r = 0.8768\text{,}\) so \(r = 0.1232\text{,}\) or 12.32%.