How To Find Exponential Function From Two Points

How to Find an Exponential Function from Two Points

Finding an exponential function that passes through two given points is a common problem in mathematics and its applications, ranging from population growth modeling to financial calculations. While seemingly straightforward, the process requires careful application of exponential function properties and logarithmic manipulation. This comprehensive guide will walk you through the process step-by-step, providing clear explanations and examples to solidify your understanding.

Understanding Exponential Functions

Before diving into the solution process, let's refresh our understanding of exponential functions. An exponential function is a function of the form:

f(x) = ab^x

where:

• a is the initial value (the y-intercept, the value of the function when x=0).
• b is the base, a constant representing the rate of growth or decay. If b > 1, the function represents exponential growth; if 0 < b < 1, it represents exponential decay.
• x is the independent variable.

Our goal is to determine the values of 'a' and 'b' given two points (x₁, y₁) and (x₂, y₂).

The Two-Point Method: A Step-by-Step Guide

The key to finding the exponential function lies in setting up a system of two equations using the given points and solving for 'a' and 'b'. Here's the detailed procedure:

Step 1: Substitute the points into the general equation

Substitute the coordinates of both points (x₁, y₁) and (x₂, y₂) into the general exponential function equation, f(x) = ab^x:

• y₁ = ab^x₁
• y₂ = ab^x₂

This creates a system of two equations with two unknowns (a and b).

Step 2: Solve for 'a' and 'b'

Solving this system of equations requires a clever approach using the properties of logarithms. Here's how:

1. Divide the equations: Divide the second equation by the first equation:

   (y₂/y₁) = (ab^x₂)/(ab^x₁)

2. Simplify using exponent rules: The 'a' terms cancel out, and using the rule for dividing exponents (b^m/bⁿ = b^m-n), we get:

   (y₂/y₁) = b^(x₂-x₁)

3. Solve for 'b': To isolate 'b', take the logarithm of both sides of the equation. Any base logarithm will work, but the natural logarithm (ln) is often preferred:

   ln(y₂/y₁) = ln(b^(x₂-x₁))

   Using the logarithm power rule (ln(xⁿ) = nln(x)), we simplify to:

   ln(y₂/y₁) = (x₂-x₁)ln(b)

   Finally, solve for 'b':

   ln(b) = ln(y₂/y₁) / (x₂-x₁)

   b = e^{[ln(y₂/y₁) / (x₂-x₁)]} or equivalently, b = (y₂/y₁)^{1/(x₂-x₁)}

4. Substitute 'b' back into one of the original equations to solve for 'a': Choose either of the original equations (y₁ = ab^x₁ or y₂ = ab^x₂) and substitute the value of 'b' you just calculated. Solve for 'a':

   For example, using the first equation:

   a = y₁ / b^x₁

Step 3: Write the exponential function

Once you have calculated the values of 'a' and 'b', substitute them back into the general exponential function equation:

f(x) = ab^x

This gives you the specific exponential function that passes through the two given points.

Examples: Putting the Method into Practice

Let's illustrate the process with a few examples:

Example 1: Find the exponential function that passes through the points (1, 2) and (3, 8).

1. Substitute the points:

   • 2 = ab¹
   • 8 = ab³

2. Divide the equations:

   • 8/2 = (ab³)/(ab¹)
   • 4 = b²

3. Solve for 'b':

   • b = √4 = 2

4. Solve for 'a': Using the first equation, 2 = a(2)¹, so a = 1.

5. Write the function: The exponential function is f(x) = 1(2)^x or simply f(x) = 2^x

Example 2: Find the exponential function that passes through the points (0, 5) and (2, 20).

1. Substitute the points:

   • 5 = ab⁰
   • 20 = ab²

2. Divide the equations:

   • 20/5 = (ab²)/(ab⁰)
   • 4 = b²

3. Solve for 'b':

   • b = 2

4. Solve for 'a': Using the first equation, 5 = a(2)⁰, so a = 5.

5. Write the function: The exponential function is f(x) = 5(2)^x

Example 3: Dealing with Decay Find the exponential function passing through (1, 100) and (2, 50).

1. Substitute points:

   • 100 = ab¹
   • 50 = ab²

2. Divide equations:

   • 50/100 = b
   • b = 0.5

3. Solve for a: Using the first equation, 100 = a(0.5)¹, so a = 200

4. Write the function: f(x) = 200(0.5)^x This represents exponential decay.

Handling Special Cases and Potential Errors

While the method described above is generally applicable, a few special cases warrant attention:

• Points with the same y-value: If both points have the same y-coordinate, the base 'b' will be 1, resulting in a constant function (not a true exponential function).
• Points with the same x-value: This situation is undefined for an exponential function because it implies a vertical line, not a function.
• Negative 'b' values: While the equations might yield a negative 'b', exponential functions generally assume positive bases. A negative base leads to complex numbers and isn't typically considered within the scope of standard exponential functions.
• Computational Errors: Be mindful of rounding errors during calculations. Use sufficient significant figures to maintain accuracy.

Applications of Finding Exponential Functions from Two Points

The ability to find an exponential function from two points is crucial in various fields:

• Population Growth: Modeling population growth based on census data.
• Financial Modeling: Calculating compound interest, investment growth, or loan amortization.
• Radioactive Decay: Determining the half-life of radioactive substances.
• Spread of Diseases: Simulating the spread of infectious diseases over time.
• Scientific Experiments: Analyzing data in experimental settings involving exponential changes.

Conclusion

Finding an exponential function from two points is a valuable mathematical skill with wide-ranging applications. By systematically applying the two-point method, carefully performing calculations, and understanding the underlying concepts of exponential functions, you can accurately model and analyze various real-world phenomena. Remember to carefully check your work and consider any special cases to ensure the accuracy and validity of your results. With practice, this process becomes intuitive and efficient.