Question

The general exponential function has the form \(f(x)=a e^{b x},\) where \(a\) and \(b\) are constants and \(e\) is Euler's constant \((\approx\) 2.718). We want to find the equation of the exponential function that goes through the points (1,2) and (2,4) . (a) Show why we cannot simply subsitute in values for \(x\) and \(y\) in \(y=a e^{b x}\) and solve using the techniques we used for polynomials. (b) Show how the equality \(y=a e^{b x}\) leads us to the linear equation \(\ln y=\ln a+b x\) (c) Use the techniques we developed to solve for the unknowns In \(a\) and \(b\). (d) Knowing In \(a,\) find \(a\); find the exponential function \(f(x)=a e^{b x}\) that goes through the points (1,2) and (2,4)

Short answer

The exponential function is \(f(x) = e^{(\ln 2)x}\).

Step-by-step solution

Understanding the limitation of substitution method

The general form of the exponential function is \(y = a e^{bx}\). If we directly substitute the points (1, 2) and (2, 4) into this equation, we obtain two equations:1. \(2 = a e^{b}\)2. \(4 = a e^{2b}\).These are nonlinear equations due to the presence of \(e^{bx}\), meaning regular techniques like solving linear equations for polynomials won't work here.

Transforming to a linear form using natural logarithms

Given \(y = a e^{bx}\), taking the natural logarithm of both sides yields \(\ln y = \ln a + bx\). This equation is now linear in terms of \(\ln y\) and \(x\), enabling us to use linear techniques for solving.

Setting up the system of equations

Using the transformational equation \(\ln y = \ln a + bx\), create equations for the points (1,2) and (2,4):1. \(\ln 2 = \ln a + b\)2. \(\ln 4 = \ln a + 2b\).

Solving for \(b\)

Subtract equation (1) from equation (2) to eliminate \(\ln a\):\[(\ln 4 - \ln 2) = (\ln a + 2b) - (\ln a + b)\] This simplifies to \(b = \ln 2\) because \(\ln 4 - \ln 2 = \ln 4/2 = \ln 2\).

Solving for \(\ln a\)

Substitute \(b = \ln 2\) back into equation (1):\[\ln 2 = \ln a + \ln 2\]From which \(\ln a = 0\).

Finding \(a\)

Since \(\ln a = 0\), we find that \(a = e^0 = 1\).

Writing the final exponential equation

Substitute \(a = 1\) and \(b = \ln 2\) into the exponential equation:\[f(x) = 1 \cdot e^{(\ln 2) x} = e^{(\ln 2)x}\].

Key concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a fundamental concept when dealing with exponential functions. It is the logarithm to the base \( e \), where \( e \approx 2.718 \), also known as Euler's number. This logarithm function is particularly useful since it can transform an exponential form \( y = a e^{bx} \) into a linear form.
This transformation is achieved by taking the natural logarithm of both sides, leading to \( \ln y = \ln a + bx \). In this equation, \( \ln y \) corresponds to a "y" value, and it reveals a linear relationship in terms of \( \ln y \) and \( x \).

• \( \ln y \) is the dependent variable.
• \( x \) remains the independent variable.
• \( \ln a \) is akin to the y-intercept in linear equations.
• \( b \) acts as the slope in this transformed equation.

This linear relationship is crucial because it allows us to apply simple linear equation techniques to solve for unknown variables.

Linearization

Linearization is the process of converting a nonlinear equation into a linear one, making it easier to manipulate and solve. In the context of solving exponential functions like \( y = a e^{bx} \), linearization involves using the natural logarithm.
By applying \( \ln \) to both sides of the equation, the exponential relationship becomes \( \ln y = \ln a + bx \). This new equation is now easier to handle using linear techniques.

• Linearization simplifies the equation, revealing direct relationships between variables.
• It turns complex exponential forms into manageable linear equations.
• Linear techniques, such as subtraction or algebraic manipulation, can then be used.

This method significantly aids in finding the constants \( a \) and \( b \) when working with exponential functions that pass through specific points.

Equation Solving

Solving equations involving exponential functions requires clever manipulation of variables and applying appropriate mathematical techniques. With the exponential function \( y = a e^{bx} \), direct substitution of known points can lead to complex nonlinear equations. These are typically difficult to solve using ordinary algebraic methods.
Instead, as shown in the solution, taking the natural logarithm transforms the equation into \( \ln y = \ln a + bx \). This form is linear and manageable. By setting up simultaneous equations from known points, such as \( \ln 2 = \ln a + b \) and \( \ln 4 = \ln a + 2b \), we can strategically eliminate variables:

• Subtracting equations helps find the slope \( b \).
• Plugging the value of \( b \) back into one of the original linear forms reveals \( \ln a \).
• With \( \ln a \) known, we can calculate \( a \).

By these steps, we access values for \( a \) and \( b \), ultimately leading to the solution for the exponential function.

Constants

In the context of the exponential function \( f(x) = a e^{bx} \), constants \( a \) and \( b \) are essential parameters that characterize the function's behavior. These constants determine the scale and the rate of growth or decay of the function.
Finding these constants requires transforming the original nonlinear equation into a form where they can be isolated. Using the exercises and the derived linear equations, \( b \) is identified through relationships such as \( b = \ln 2 \).

• The constant \( b \) relates to the function's rate of growth, acting similarly to a slope in linear equations.
• The constant \( a \) is derived next, through \( \ln a = 0 \), which implies \( a = 1 \).

Recognizing and calculating these constants are fundamental in constructing an exponential equation that fits through given data points. These values provide a complete picture of the function's properties and behavior, allowing us to understand and use the function effectively in different mathematical contexts.