7.2 Relationship Between Linear and Exponential
In this section we will explore exponential and linear functions through the perspective of isomorphisms. Let \(a\) be a positive real number. Then we define the exponential function with base \(a\) as the function \[\exp_a: (\mathbb{R},+) \rightarrow (\mathbb{R}^+,\cdot)\] with \(\exp_a{(x)}= a^x\), where \(a^x\) is defined in Section 4.5.
Using the properties of exponents in Theorem 4.12 we have that \(\exp_a\) is an increasing function (i.e. \(x<y \Rightarrow \exp_a{(x)}<\exp_a{(y)}\)). This implies that \(\exp_a\) is an injection. Using properties of limits from analysis, one can prove that \(\exp_a\) is a surjection. From Section 4.5 we have that \[\exp_a{(x+y)} = a^{x+y} = a^x \cdot a^y = \exp_a{(x)}\cdot \exp_a{(y)}\] and so \(\exp_a\) is a homomorphis. And since \(\exp_a\) has all of these properties, it is an isomorphism.
We will now explore the isomorphism between \((\mathbb{R},+)\) and \((\mathbb{R}^+,\cdot)\).
\[\begin{array}{c|c} \mbox{Property in } (\mathbb{R},+) & \mbox{Property in } (\mathbb{R}^+,\cdot) \\ \hline m+m=2\cdot m & a\cdot a = a^2 \\ mk=m+m+m+\cdots +m, \: k \mbox{ times} & a^k=a \cdot a \cdot a\cdots a, \: k \mbox{ times} \\ a\cdot (b+c) = (a\cdot b) + (a\cdot c) & (b\cdot c)^a = (b^a) \cdot (c^a) \\ 0+m=m+0=m \mbox{ (additive identity)} & 1\cdot a=a\cdot 1= a \mbox{ (multiplicative identity)} \\ m+(-m)=0 \mbox{ (additive inverse)} & a\cdot \frac{1}{a}=1 \mbox{ (multiplicative inverse)} \\ \end{array}\]
One can see that repeated addition in \((\mathbb{R},+)\) corresponds to repeated multiplication in \((\mathbb{R}^+,\cdot)\). One can also see that \[\exp_a\left( m\cdot x + b\right) = \exp_a(b) \cdot \left(\exp_a(m)\right)^x\] and so we can see that linear expressions in \((\mathbb{R},+)\) correspond to exponential expressions in \((\mathbb{R}^+,\cdot)\). An implication of this is that the teaching of linear and exponential functions should be intertwined. At a minimum, teachers should use their horizon content knowledge when initially teaching students about exponential functions to directly related the properties of the functions to those of linear functions.
Related Content Standards
- (HSF.LE.1) Distinguish between situations that can be modeled with linear functions and with exponential functions.
- Prove that linear functions grow by equal differences over equal intervals, and that exponential functions grow by equal factors over equal intervals.
- Recognize situations in which one quantity changes at a constant rate per unit interval relative to another.
- Recognize situations in which a quantity grows or decays by a constant percent rate per unit interval relative to another.
In order to compare linear and exponential functions, we will study their properties side-by-side. Let \[f(x)=mx+b \quad \mbox{ and } \quad g(x)=b\cdot m^x\] be linear and exponential functions, respectively. The graphs of the functions are then given by the set of points \((x,y)\) such that \[y=mx+b \quad \mbox{ and } \quad y=b\cdot m^x.\] Setting \(x=0\) in each equation implies that \((0,b)\) is the \(y\)-intercept for both functions. Exploring the relationship of \(f(x)\) to \(f(x+1)\) and \(g(x)\) to \(g(x+1)\) we see that \[f(x+1)-f(x) = m \quad \mbox{ and } \quad \frac{g(x+1)}{g(x)} = m\] implying that in both cases the parameter \(m\) represents a constant rate of change. For the linear function \(m\) represents an additive rate of change, while in the exponential function \(m\) represents a multiplicative rate of change.
Related Content Standards
- (HSF.LE.5) Interpret the parameters in a linear or exponential function in terms of a context.
For the linear function, the function is increasing if and only if \(m\) is greater than the additive identity. With the exponential function, the values of \(m\) must be positive and the function is increasing if and only if \(m\) is greater than the multiplicative identity.
If we know that two points, \((x_1,y_1)\) and \((x_2,y_2)\), are on the graph of the functions, then we know that \[\begin{array}{ccc} y_2=m\cdot x_2 + b & & y_2 = b\cdot m^{x_2} \\ & \mbox{or} & \\ y_1=m\cdot x_1 + b & & y_1 = b\cdot m^{x_1}.\\ \end{array}\]
Using subtraction for the linear function and division for the exponential function we find that \[ (y_2-y_1) = m(x_2-x_1) \quad \mbox{ or } \quad \frac{y_2}{y_1} = m^{(x_2-x_1)}\] and so \[ m= \frac{y_2-y_1}{x_2-x_1} \quad \mbox{ or } \quad m= \left( \frac{y_2}{y_1} \right)^{\frac{1}{x_2-x_1}}.\] After determining the rate of change for the function one can now include that information in a point-`rate of change’ form of the function as \[ y= m(x-x_1)+y_1 \quad \mbox{ or } \quad y = y_1 \cdot m^{(x-x_1)}.\]