Intermediate Algebra: Functions and Graphs

First, we evaluate each function for several convenient values, as shown in the table.

Then we plot the points for each function and connect them with smooth curves. Note that \(a=1\) for both functions, so their graphs have the same \(y\)-intercept, \((0,1)\text{.}\)

For positive \(x\)-values, \(g(x)\) is always larger than \(f(x)\text{,}\) and is increasing more rapidly. In the figure, we can see that \(g(x) = 4^x\) climbs more rapidly than \(f(x) = 3^x\text{.}\) However, for negative \(x\)-values, \(g(x)\) is smaller than \(f(x)\text{.}\)

\(x\)	\(f(x)\)	\(g(x)\)
\(-2\)	\(\dfrac{1}{9}\)	\(\dfrac{1}{16}\)
\(-1\)	\(\dfrac{1}{3}\)	\(\dfrac{1}{4}\)
\(0\)	\(1\)	\(1\)
\(1\)	\(3\)	\(4\)
\(2\)	\(9\)	\(16\)

We can tell that \(f\) is the exponential function because its values increase by a factor of 3 for each unit increase in \(x\text{.}\) The base of the function is \(b=3\text{.}\) We also see from the table that \(a=f(0)=2\text{,}\) so \(f(x)=2(3^x)\text{.}\) Thus, \(h\) must be the power function, \(h(x)=kx^3\text{.}\) To find \(k\text{,}\) we notice that \(h(1)=k\text{,}\) so \(k=2\text{,}\) and \(h(x)=2x^3\text{.}\)

1. Using the fact that \(9 = 3^2\text{,}\) we write each side of the equation as a power of \(3\text{:}\)
   \begin{align*} 3^{x-2} \amp = \left(3^2\right)^3 \amp \amp \blert{\text{Simplify the right side.}}\\ 3^{x-2} \amp = 3^6 \end{align*}
   Now we equate the exponents to obtain
   \begin{align*} x - 2 \amp = 6 \end{align*}
   The solution is \(x=8\text{.}\)
2. We write each factor as a power of \(3\text{.}\)
   \begin{align*} 3^3 \cdot 3^{-2x} \amp= \left(3^2\right)^{x+1}\amp\amp\blert{27=3^3~\text{and}~9=3^2.} \end{align*}
   We use the laws of exponents to simplify each side:
   \begin{equation*} 3^{3-2x} = 3^{2x+2} \end{equation*}
   Now we equate the exponents to obtain
   \begin{align*} 3 - 2x \amp = 2x + 2\\ -4x =\amp -1 \end{align*}
   The solution is \(x = \dfrac{1}{4}\text{.}\)

1. We let \(P\) represent the number of fleas present after \(t\) days. The original population of 10 fleas doubles not every day, but every 5 days. A table of values for \(P\) would look like this.

   \(t\)	\(P(t)\)	\(\hphantom{0000}\)
   \(0\)	\(10\)	\(\hphantom{0000}\)
   \(5\)	\(20\)	\(P(\alert{5}) = 10 \cdot 2^{\alert{1}} = 20\)
   \(10\)	\(40\)	\(P(\alert{10}) = 10 \cdot 2^{\alert{2}} = 40\)
   \(15\)	\(80\)	\(P(\alert{15}) = 10 \cdot 2^{\alert{3}} = 80\)
   \(20\)	\(160\)	\(P(\alert{20}) = 10 \cdot 2^{\alert{4}} = 160\)

   Notice that the original population is multiplied by another factor of 2 every 5 days. We must divide \(t\) by 5 to see how many times the population doubles. The formula for \(P\) is thus

   \begin{equation*} P(t) = 10 \cdot 2^{t/5} \end{equation*}
2. We set \(P = \alert{10,240}\) and solve for \(t\text{:}\)
   \begin{align*} \alert{10,240} \amp = 10\cdot 2^{t/5}\amp\amp \blert{\text{Divide both sides by 10.}}\\ 1024 \amp = 2^{t/5} \amp\amp \blert{\text {Write 1024 as a power of 2.}}\\ 2^{10} \amp = 2^{t/5} \end{align*}
   We equate the exponents to get \(10 = \dfrac{t}{5}\text{,}\) or \(t = 50\text{.}\) The population will grow to 10,240 fleas in 50 days.

We enter \(Y_1 = 2\) ^ X and use the standard graphing window (ZOOM 6) to obtain the graph shown in figure (a). We are looking for a point on this graph with \(y\)-coordinate \(5\text{.}\)

Using the TRACE feature, we see that the \(y\)-coordinates are too small when \(x \lt 2.1\) and too large when \(x \gt 2.4\text{.}\) The solution we want lies somewhere between \(x = 2.1\) and \(x = 2.4\text{,}\) but this approximation is not accurate enough.

To improve our approximation, we will use the intersect feature. Set \(Y_2 = 5\) and press GRAPH. The \(x\)-coordinate of the intersection point of the two graphs is the solution of the equation \(2^x = 5\) Activating the intersect command results in figure (b), and we see that, to the nearest hundredth, the solution is \(2.32\text{.}\)

We can verify that our estimate is reasonable by substituting into the equation:

We enter 2 ^ 2.32 ENTER to get \(4.993322196\text{.}\) This number is not equal to \(5\text{,}\) but it is close, so we believe that \(x = 2.32\) is a reasonable approximation to the solution of the equation \(2^x = 5\text{.}\)