Learning Objectives
- Describe the meaning of solving exponential equations
- Use the property of equality to solve exponential equations
- Determine the intersection point of two exponential functions
The Meaning of Solving Exponential Equations
Intersection point of functions
In chapter 3, we learned that the meaning of solving an equation is to find the intersection point(s) between two functions. The intersection point(s) between the graphs of any two functions [latex]f(x)[/latex] and [latex]g(x)[/latex] can be found algebraically by setting the two functions equal to each other:
[latex]f(x)=g(x)[/latex]
When the functions are equal, the value of [latex]x[/latex] is the same for both functions, as is the function value. In other words, [latex]f(x)=g(x)[/latex] means that the two functions have the same input [latex]x[/latex] as well as the same output (i.e [latex]f(x)=g(x)[/latex]). For example, solving the equation [latex]32\left(2^{x+3}\right)=64[/latex] means finding the intersection point between the two functions [latex]f(x)=32\left(2^{x+3}\right)[/latex] and [latex]g(x)=64[/latex] (figure 1).
FIgure 1. Intersection of [latex]y=f(x)[/latex] and [latex]y=g(x)[/latex]
Finding the [latex]x[/latex]-value given a function value
Another interpretation of the equation [latex](32)2^{x+3}=64[/latex] is finding the [latex]x[/latex] value when the function value of [latex]f(x)=(32)2^{x+3}[/latex] is 64 (figure 2).
Figure 2. Meaning of solving an equation
Graphically, this means finding the function value as a given [latex]y[/latex]-value on a graph, then moving vertically down to the [latex]x[/latex]-axis to determine the corresponding [latex]x[/latex]-value. In figure 2, to solve [latex](32)2^{x+3}=64[/latex], we graph the function [latex]f(x)=(32)2^{x+3}[/latex], look for 64 on the [latex]y[/latex]-axis then determine which [latex]x[/latex]-value has a function value at 64. In this case, [latex]x=-2[/latex].
Solving an equation in one variable
Algebraically, [latex](32)2^{x+3}=64[/latex] is an equation in one variable. When solving an equation in one variable, we find the value of the variable that satisfies the equation (e.g., the [latex]x[/latex]-value). There is no function value to report as the equation is in just one variable.
For example, when solving the equation, [latex]3^x=9[/latex], we find the value of [latex]x[/latex] that makes the equation true. The value of [latex]x[/latex] is 2 because [latex]3^2=9[/latex]. We need algebraic methods to solve equations, including the properties of equality covered in chapter 3.
In summary, when we set functions equal to each other, we are finding the intersection point between two functions (e.g., [latex]f(x)=3^x[/latex] and [latex]g(x)=9[/latex]). In this example, the intersection point between the graphs of the two functions [latex]f(x)[/latex] and [latex]g(x)[/latex] is (2, 9) because the [latex]y[/latex]-value is the function value [latex]f(2) = 3^2 = 9[/latex] and [latex]g(x)=9[/latex]. However, when we are solving an equation algebraically, there is no function in sight so we can just report the value of the variable (e.g., [latex]x[/latex]).
Solving Exponential Equations Using the Property of Equality
We may use the property of equality for solving an exponential equation if the exponential expressions on each side of the equation have the same base.
Property of Equality
For any real numbers [latex]a, \;x[/latex] and [latex]y[/latex],
If [latex]a^x = a^y[/latex], then [latex]x = y[/latex].
For example, to solve the equation [latex]3^x = 9[/latex], we can write [latex]9 = 3^2[/latex] so that [latex]3^x = 3^2[/latex]. The property of equality then tells us that [latex]x = 2[/latex] because with the same base, the exponents must be equal. Although it may at first appear that the exponential expressions in an equation do not have the same base, they can often be transformed to expressions with the same base.
Example 1
Solve the equation [latex]32\left(2^{x+3}\right)=64[/latex].
Solution
[latex]\begin{aligned}32\left(2^{x+3}\right)&=64&&\text{The base of the exponential is 2, so we try to write 32 as a power of 2}\\2^5\times 2^{x+3}&=2^6&&\text{Product rule}\\2^{x+3+5}&=2^6\\2^{x+8}&=2^6&&\text{Now each side has the same base}\\x+8&=6&&\text{The exponents must be equal}\\x&=-2\end{aligned}[/latex]
Example 2
Solve the equation [latex]32\left(2^{x+3}\right)=64[/latex].
Solution
Another, and more efficient way to solve the same equation as example 1, is to notice that each side of the equation is divisible by 32:
[latex]\begin{aligned}32\left(2^{x+3}\right)&=64&&\text{Divide each side by 32}\\2^{x+3}&=2&&\text{Same base so exponents must be equal}\\x+3&=1&&\text{Subtract 3 from both sides}\\x&=-2\end{aligned}[/latex]
TRY IT 1
Solve the equation [latex]15\left(3^{x+4}\right)=45[/latex].
Try It 2
Solve the equation [latex]\dfrac{1}{4}\left(6^{2x-1}\right)=9[/latex]
Example 3
Solve the equation [latex]27^{x+1}=81^{x-2}[/latex].
Solution
The bases of the exponential expressions are 27 and 81. Both of these can be written as a power of 3: [latex]27 = 3^3[/latex] and [latex]81 = 3^4[/latex].
[latex]\begin{aligned}27^{x+1}&=81^{x-2}\\{\left(3^3\right)}^{x+1}&={\left(3^4\right)}^{x-2}&&\text{Power to a power rule}\\3^{3(x+1)}&=3^{4(x-2)}&&\text{Same base so exponents must be equal}\\3(x+1)&=4(x-2)&&\text{Distribute}\\3x+3&=4x-8&&\text{Subtract }3x\text{ from both sides}\\3&=x-8&&\text{Add 8 to both sides}\\11&=x\end{aligned}[/latex]
Try It 3
Solve the equation [latex]2^{-x+5}=16^{x+1}[/latex].
Example 4
Determine the intersection point of the functions [latex]f(x)=5\left(2^{x-3}\right)[/latex] and [latex]g(x)=10\left(2^{-x+16}\right)[/latex].
Solution
The intersection point occurs at an [latex]x[/latex]-value where [latex]f(x)=g(x)[/latex].
Set [latex]f(x)=g(x)[/latex] and solve for [latex]x[/latex]:
[latex]\begin{aligned}5\left(2^{x-3}\right)&=10\left(2^{-x+16}\right)\\2^{x-3}&=2\left(2^{-x+16}\right)\\2^{x-3}&=2^{-x+17}\\x-3&=-x+17\\2x-3&=17\\2x&=20\\x&=10\end{aligned}[/latex]
Now that we know the value of [latex]x[/latex], we can find the corresponding value of [latex]y[/latex]:
[latex]\begin{aligned}y&=f(10)\\&=5\left(2^{10-3}\right)\\&=5\left(2^7\right)\\&=5(128)\\&=640\end{aligned}[/latex]
Consequently, the intersection point is [latex](10, 640)[/latex].
Note: We could also have used [latex]g(x)[/latex] to find the value of [latex]y[/latex]:
[latex]\begin{aligned}y&=g(10)\\&=10\left(2^{-10+16}\right)\\&=10\left(2^{6}\right)\\&=10\cdot 64\\&=640\end{aligned}[/latex]
This is a good check for the answer.
Try It 4
Determine the intersection point of the functions [latex]f(x)=6\left(3^x\right)[/latex] and [latex]g(x)=18\left(3^{-x+1}\right)[/latex].
