Can solving an exponential function predict the future? Can exponential functions be a fast track to wealth?
As a matter of fact, yes! Successful entrepreneurs all over the world know the value of modeling population and economic growth and decay. By constructing an accurate exponential equation to model the past and present growth of a particular area of the country, state, or even town, the ideal location for a new business to thrive can be predicted.
Some things cannot be modeled of course, such as natural disasters and the like, but often a few well educated guesses make the difference between stunning success and dismal failure in the business world.
Watch This
Embedded Video:
 James Sousa: Solving Exponential Equations  Part 1 of 2
Guidance
In this lesson you will learn about exponential functions, a family of functions different from the other function families because the variable x is in the exponent. For example, the functions f(x) = 2^{x} and g(x) = 100(2)5^{x} are exponential functions.
Evaluating Exponential Functions
Consider the function f(x) = 2^{x}. When we input a value for x, we find the function value by raising 2 to the exponent of x. For example, if x = 3, we have f(3) = 2^{3} = 8.
If we choose larger values of x, we will get larger function values, as the function values will be larger powers of 2. For example, f(10) = 2^{10} = 1,024.
If we choose smaller values of x, we will quickly end up with fractions. For instance, if x = 0, we have f(0) = 2^{0} = 1. If x = 3, we have \begin{align*}f(3)=2^{3}=\left( \tfrac{1}{2} \right)^{3}=\tfrac{1}{8}\end{align*}
In general, if we have a function of the form f(x) = a^{x}, where a is a positive real number, the domain of the function is the set of all real numbers, and the range is limited to the set of positive real numbers. This restricted domain will result in a specific shape of the graph.
Solving exponential equations
Solving an exponential equation means determining the value of x for a given function value. The solution to the equation 2^{x} = 8 is the value of x that makes the equation a true statement, therefor x = 3, since 2^{3} = 8.
Example A
Solve for x: 3 (2^{x} ^{+ 1}) = 24.
Solution
We can solve this equation by writing both sides of the equation as a power of 2:





\begin{align*}3(2^{x + 1}) = 24\end{align*}
3(2x+1)=24

\begin{align*}3(2^{x + 1}) = 24\end{align*}








\begin{align*}\frac{3(2^{x + 1})} {3} = \frac{24} {3}\end{align*}
3(2x+1)3=243

\begin{align*}\frac{3(2^{x + 1})} {3} = \frac{24} {3}\end{align*}








\begin{align*}2^{x + 1} = 8\end{align*}
2x+1=8

\begin{align*}2^{x + 1} = 8\end{align*}








\begin{align*}2^{x + 1} = 2^3\end{align*}
2x+1=23

\begin{align*}2^{x + 1} = 2^3\end{align*}



To solve the equation now, recall a property of exponents: if b^{x} = b^{y}, then x = y. That is, if two powers of the same base are equal, the exponents must be equal. This property tells us how to solve:






\begin{align*}2^{x + 1} = 2^3\end{align*}
2x+1=23

\begin{align*}2^{x + 1} = 2^3\end{align*}









\begin{align*}\Rightarrow x + 1 = 3 \end{align*}
⇒x+1=3

\begin{align*}\Rightarrow x + 1 = 3 \end{align*}









\begin{align*} \,\! x = 2 \end{align*}
x=2

\begin{align*} \,\! x = 2 \end{align*}




Example B
For the function g(x) = 3^{x}, find g(2), g(4), g(0), g(2), g(4).
Solution:
\begin{align*}g(2) = 3^2 = 9\end{align*}
\begin{align*}g(4) = 3^4 = 81\end{align*}
\begin{align*}g(0) = 3^0 = 1\end{align*}
\begin{align*}g(2) = 3^{2} = \frac{1} {3^2} = \frac{1} {9}\end{align*}
\begin{align*}g(4) = 3^{4} = \frac{1} {3^4} = \frac{1} {81}\end{align*}
The values of the function g(x) = 3^{x} behave much like those of f(x) = 2^{x}: if we choose larger values, we get larger and larger function values. If x = 0, the function value is 1. And, if we choose smaller and smaller x values, the function values will be smaller and smaller fractions. Also, the range of g(x) is limited to positive values.
Example C
Use a graphing utility to solve each equation:
a. 2^{3}^{x} ^{ 1} = 7  b. 6^{4}^{x} = 2^{8}^{x} ^{ 5} 

Solution:
a. 2^{3}^{x} ^{ 1} = 7
 Graph the function y = 2^{3}^{x} ^{ 1} and find the point where the graph intersects the horizontal line y = 7. The solution is x ≈ 1.27
b. 6^{4}^{x} = 2^{8}^{x} ^{ 5}
 Graph the functions y = 6^{4}^{x} and y = 2^{8}^{x} ^{ 5} and find their intersection point.
The solution is approximately x ≈ 0.27. (Your graphing calculator should show 9 digits: 0.272630365.)
Vocabulary
An exponential function is a function where the input variable is found in the exponent, e.g.: \begin{align*}y = 3^{x}\end{align*}
An exponential function with outputs that increase is known as an exponential growth function.
An exponential function with outputs that decrease is known as an exponential decay function.
Guided Practice
1) Graph \begin{align*}f(x) = 2^x\end{align*}
2) Graph \begin{align*}g(x) = (\frac{1}{2})^x\end{align*}
3) Solve the equation 5^{6}^{x} ^{+ 10} = 25^{x} ^{ 1}
Solutions
1) First create a table showing values for x and f(x):

x f(x) (3) 1/8 (2) 1/4 (1) 1/2 0 1 1 2 2 4 3 8

Observe that \begin{align*}f\end{align*}
f is increasing and onetoone. Also, \begin{align*}f\end{align*}f is strictly positive, the range of \begin{align*}a^x\end{align*}ax is \begin{align*}(0, \infty)\end{align*}(0,∞) Finally, it is important to notice that \begin{align*}f(0) = 1\end{align*}f(0)=1 and \begin{align*}f(1) = 2\end{align*}f(1)=2
 The graph looks like this:
2) To graph \begin{align*}g(x) = (\frac{1}{2})^x\end{align*}
 At first, it looks like a completely new function, but consider:

\begin{align*}(\frac{1}{2})^x = \frac{1}{2^x} = 2^{x}\end{align*}
(12)x=12x=2−x 
Since \begin{align*}2^x\end{align*}
2x is the inverse of \begin{align*}2^{x}\end{align*}2−x (which could also be written as \begin{align*}g(x) = f(x)\end{align*}g(x)=f(−x) ) the graph of \begin{align*}g\end{align*}g is just the reflection of the graph of \begin{align*}f\end{align*}f across the yaxis!

The graph below shows \begin{align*}f(x) = 2^x\end{align*}
f(x)=2x in blue and \begin{align*}g(x) = (\frac{1}{2})^x\end{align*} in red:
3) Use the technique explained in the lesson:




 5^{6}^{x} ^{+ 10} = 25^{x} ^{ 1}







 5^{6}^{x} ^{+ 10} = (5^{2})^{x} ^{ 1}







 5^{6}^{x} ^{+ 10} = 5^{2}^{x} ^{ 2}






 \begin{align*}\Rightarrow\end{align*} 6x + 10 = 2x  2






 4x + 10 = 2







 4x = 12







 x = 3



Practice
 Describe an exponential function, and give an example.
 What is an exponential function with increasing outputs known as?
 What is an exponential function with outputs that decrease known as?
Evaluate each function at the given value.
 \begin{align*}f(x) = \frac{1}{3}\cdot6^x\end{align*} at x = 2
 \begin{align*}f(n) = 10\cdot2^n\end{align*} at n = 5
 \begin{align*}g(x) = \frac{1}{5}\cdot(\frac{1}{3})^x\end{align*} at x = 3
 Look at the graph below of the exponential function \begin{align*}g(x) = a^x\end{align*}. Find \begin{align*}a\end{align*}
Sketch the graph of each exponential function and explicitly evaluate each function for at least four values of x:
 \begin{align*}f(x) = \pi^x\end{align*}
 \begin{align*}f(x) = (\frac{2}{3})^{x}\end{align*}
 \begin{align*}f(x) = 2^{x+4} 2\end{align*}
 Find the average rate of change of \begin{align*}f(x) = 2^x\end{align*} from \begin{align*}x = 3\end{align*} to \begin{align*}x = 6\end{align*} Draw a graph that illustrates your answer
HINT for problems 12  14:
\begin{align*}A = P (1+\frac{r}{n})^{nt}\end{align*} where:
 "A" = ending Amount
 "P" = Principal (beginning amount)
 "r" = interest rate (as a decimal value  NOT as a percentage!)
 "n" = number of compoundings per period (commonly one year)
 "t" = time (the number of periods/years)
 If you invest $28,000 in an account that gets 10% annual interest compounded quarterly, how much would you have in 15 years?
 If you invested a penny on Jan 1, 1776 at 12% interest compounded daily, how much would you have on Jan 1, 2013 ?
 How much would you need to invest to get $20,000 in 3 years at an annual interest rate of 6.5% compounded monthly?
 The halflife of element 137m is 2.552 minutes. How much of an 10 gram sample would be left after 8 minutes?
 The half – life of carbon 14 is 5,730 years. How much of a 100 gram sample would be left after 25,000 years ?
 Sketch and use the graphs of \begin{align*}f(x) = (\frac{1}{3})^x\end{align*} and \begin{align*}p(x) = (\frac{1}{4})^x\end{align*} to graph the inequality \begin{align*}(\frac{1}{3})^x \leq g(x) \leq (\frac{1}{4})^x\end{align*}