Definition 12.1.1.
An exponential function generally has the following form:
\begin{equation*}
f(x)=a\cdot b^x = a(b)^x = a(b^x)
\end{equation*}
Section 12.1 Equations of Exponential Functions
Basically, an exponential function is a function with the variable in the exponent. Our usual, garden-variety exponential function looks like the one above, but don’t forget that we could have some transformations involved.
Definition 12.1.2.
The initial value of an exponential function is the \(y\)-value of the \(y\)-intercept. When we have our exponential function written as just \(f(x)=a\cdot b^x\text{,}\) the initial value turns out to be just \(a\text{.}\)
Example 12.1.3.
Suppose we want to find the initial value of \(f(x)=3\cdot1.07^x\text{.}\) Going by the official definition, we need to find the \(y\)-intercept, which you should remember is where \(x=0\text{.}\)
\begin{equation*}
y=3\cdot1.07^0=3\cdot 1 = 3
\end{equation*}
Indeed, we do get that the initial vlaue is just \(3\text{.}\)
Definition 12.1.4.
The growth/decay rate of an exponential function determines how fast it is growing or decaying. When we have our exponential function written as just \(f(x)=a\cdot b^x\text{,}\) we have \(b=1+\text{rate(as a decimal)}\text{.}\) We always write our rate as a percentage.
-
It is a growth rate if it is positive (in other words, if \(b\gt 1\))
-
It is a decay rate if it is negative (in other words, if \(b\lt 1\))
Example 12.1.5.
Suppose we want to find the growth/decay rate of \(f(x)=3\cdot1.07^x\text{.}\) We see that \(b=1.07\text{,}\) which we know is 1 plus the rate (as a decimal). Subtracting 1, we get that the rate is \(0.07\text{,}\) except that we have to convert it to a percentage. We do that by multiplying by 100 (which is the same thing as moving the decimal to the right 2 places). Since it’s positive, it’s a growth rate. So, our final answer is that the growth rate is 7%.
Checkpoint 12.1.6.
Find the initial value and the growth/decay rate of \(f(x)=-7(1.4)^x\text{.}\)
Solution.
To find the initial value, we can either plug in \(x=0\) to get the \(y\)-intercept, or just remember the fact we learned above. Either way, the initial value is \(-7\text{.}\)
To find the growth/decay rate, we see that \(b=1.4\text{,}\) which we know is 1 plus the rate (as a decimal). Subtracting 1, we get that the rate is \(0.4\text{,}\) except that we have to convert it to a percentage. We do that by multiplying by 100 (which is the same thing as moving the decimal to the right 2 places). Since it’s positive, it’s a growth rate. So, our final answer is that the growth rate is 40%.
Checkpoint 12.1.7.
Find the initial value and the growth/decay rate of \(f(x)=\frac{1}{5}(0.59)^x\text{.}\)
Solution.
To find the initial value, we can either plug in \(x=0\) to get the \(y\)-intercept, or just remember the fact we learned above. Either way, the initial value is \(\frac{1}{2}\text{.}\)
To find the growth/decay rate, we see that \(b=0.59\text{,}\) which we know is 1 plus the rate (as a decimal). Subtracting 1, we get that the rate is \(-0.41\text{,}\) except that we have to convert it to a percentage. We do that by multiplying by 100 (which is the same thing as moving the decimal to the right 2 places). Since it’s positive, it’s a growth rate. So, our final answer is that the growth rate is -41%.
Sometimes, we want to use what we just learned about to compare different functions or situations.
Example 12.1.8.
Suppose the population of five towns are modeled by the equations below
| Town | Population |
|---|---|
| Town A | \(134(1.05)^t\) |
| Town B | \(657(0.93)^t\) |
| Town C | \(628(1.213)^t\) |
| Town D | \(721(0.54)^t\) |
| Town E | \(212(0.31)^t\) |
We want to answer the following questions about these five towns:
-
Which town had the largest initial population?
-
Which towns are shrinking?
-
Which town is shrinking the fastest?
Since the first question involves the initial value and the other two questions involve the growth/decay rate, let’s figure out those values for all of the towns and add that information to the table:
| Town | Population | Initial Value | Growth/Decay Rate |
|---|---|---|---|
| Town A | \(134(1.05)^t\) | \(134\) | \((1.05-1)*100=(0.05)*100=5\) |
| Town B | \(657(0.93)^t\) | \(657\) | \((0.93-1)*100=(-0.07)*100=-7\) |
| Town C | \(628(1.213)^t\) | \(628\) | \((1.213-1)*100=(0.213)*100=21.3\) |
| Town D | \(721(0.54)^t\) | \(721\) | \((0.54-1)*100=(-0.46)*100=-46\) |
| Town E | \(212(0.31)^t\) | \(212\) | \((0.31-1)*100=(-0.69)*100=-69\) |
Now we have all of the information we need to answer the questions.
-
Which town had the largest initial population? This is asking for the largest initial value. Looking at the column of initial values we computed, we see that Town D has the largest initial population.
-
Which towns are shrinking? This is asking about which towns are decaying, which means they have a negative rate. Looking at the column of rates we computed above, we see that Towns B, D, and E are all shrinking.
-
Which town is shrinking the fastest? This is looking for the rate that is the most negative. Looking at the column of rates we computed above, we see that Town E is shrinking the fastest.
Checkpoint 12.1.11.
Suppose the amount of money in five bank accounts is modeled by the equations below
| Account | Money |
|---|---|
| Account A | \(5673(1.32)^t\) |
| Account B | \(4445(1.93)^t\) |
| Account C | \(7134(1.314)^t\) |
| Account D | \(3516(2.54)^t\) |
| Account E | \(2126(1.31)^t\) |
Answer the following questions about these five accounts:
-
Which account started with the most money?
-
Which account has the best (highest) interest rate, and what is its rate?
Answer.
Solution.
Since the first question involves the initial value and the other two questions involve the growth/decay rate, let’s figure out those values for all of the accounts and add that information to the table:
| Account | Money | Initial Value | Growth/Decay Rate |
|---|---|---|---|
| Account A | \(5673(1.32)^t\) | \(5673\) | \((1.32-1)*100=(0.32)*100=32\) |
| Account B | \(4445(1.93)^t\) | \(4445\) | \((1.93-1)*100=(0.93)*100=93\) |
| Account C | \(7134(1.314)^t\) | \(7134\) | \((1.314-1)*100=(0.314)*100=31.4\) |
| Account D | \(3516(2.54)^t\) | \(3516\) | \((2.54-1)*100=(1.54)*100=154\) |
| Account E | \(2126(1.31)^t\) | \(2126\) | \((1.31-1)*100=(0.31)*100=31\) |
Now we have all of the information we need to answer the questions.
-
Which account started with the most money? This is asking for the largest initial value. Looking at the column of initial values we computed, we see that Account C has the largest initial value.
-
Which account has the best (highest) interest rate, and what is its rate? This is asking about which account has the highest growth rate. Looking at the column of rates we computed above, we see that Account D has the highest interest rate with a rate of 154%.
Now that we know how to pull out the pieces of an exponential function when we give you the formula, let’s go backwards: write the formula based on the pieces that we give you.
The key will be the template that we saw above: \(f(x)=a\cdot b^x\text{.}\) When we did linear functions, our template is \(f(x)=mx+b\text{.}\) For quadratic functions, it was \(f(x)=A(x-r_1)(x-r_2)\text{.}\) For polynomials, it was similar to quadratics but we could have more roots and we had to deal with some exponents. But when the question asks for an exponential function, our template is \(f(x)=a\cdot b^x\text{,}\) and then we need to fill in \(a\) and \(b\text{.}\)
Example 12.1.14.
Suppose we want to write the equation of the exponential function with initial value \(19\) and growth rate 108%. Since it’s an exponential function, we know the template we are working with is
\begin{equation*}
f(x)=a\cdot b^x
\end{equation*}
We already know that the initial value is \(19\text{,}\) so we can fill that in right away:
\begin{equation*}
f(x)=19\cdot b^x
\end{equation*}
Now we just have to use the growth rate of 108% to find \(b\text{.}\) First, we need to convert our rate to a decimal by dividing by 100 (which is the same thing as moving the decimal place to the left 2). So, we get \(1.08\text{.}\) We still need to add 1 to get \(b\text{,}\) so we have \(b=2.08\text{.}\) So, our final answer is
\begin{equation*}
f(x)=19(2.08)^x
\end{equation*}
Checkpoint 12.1.15.
Find the equation of the exponential function with initital value \(-2\) and decay rate -32%.
Solution.
Since it’s an exponential function, we know the template we are working with is
\begin{equation*}
f(x)=a\cdot b^x
\end{equation*}
We already know that the initial value is \(-2\text{,}\) so we can fill that in right away:
\begin{equation*}
f(x)=-2\cdot b^x
\end{equation*}
Now we just have to use the decay rate of -32% to find \(b\text{.}\) First, we need to convert our rate to a decimal by dividing by 100 (which is the same thing as moving the decimal place to the left 2). So, we get \(-0.32\text{.}\) We still need to add 1 to get \(b\text{,}\) so we have \(b=0.68\text{.}\) So, our final answer is
\begin{equation*}
f(x)=-2(0.68)^x
\end{equation*}
Example 12.1.16.
Suppose we want to find the equation of the exponential function with initital value \(2\) and goes through the point \((3,10)\text{.}\) We are still working with an exponential function, so we use the same template.
\begin{equation*}
f(x)=a\cdot b^x
\end{equation*}
Since we know the initial value is \(2\text{,}\) we can fill that in right away.
\begin{equation*}
f(x)=2\cdot b^x
\end{equation*}
Now, in the previous example, we were given a rate to be able to find \(b\text{.}\) In this case, we don’t know the rate, but we do have a point on the graph. So, we will use a strategy we have seen lots of times before: plug in the values for \(x\) and \(y\) and solve for the thing we don’t know.
\begin{align*}
10\amp=2\cdot b^3\\
5\amp= b^3\\
\sqrt[3]{5}\amp=b
\end{align*}
So, we have \(b=\sqrt[3]{5}\text{,}\) which we can also write as \(5^{1/3}\text{.}\) Filling this in, we have our final answer.
\begin{equation*}
f(x)=2\left(\sqrt[3]{5}\right)^x
\end{equation*}
Checkpoint 12.1.17.
Find the equation of the exponential function with growth rate 50% and goes through the point \((2,18)\text{.}\)
Solution.
Since we are still working with an exponential function, we use the same template.
\begin{equation*}
f(x)=a\cdot b^x
\end{equation*}
This time, we are given the growth rate, so we can compute \(b=1+0.5=1.5\text{.}\)
\begin{equation*}
f(x)=a\cdot 1.5^x
\end{equation*}
\begin{align*}
18\amp= a\cdot 1.5^2\\
18\amp= a\cdot 2.25\\
\frac{18}{2.25}\amp= a\\
8\amp=a
\end{align*}
So, we have our final answer
\begin{equation*}
f(x)=8\cdot 1.5^x
\end{equation*}
