Lecture Notes on Solving Equations Using the Lambert W Function

Introduction

• Focus on solving the equation involving finding X.
• Problem: Solve the equation (2^x + x = 5).
• Methods:
  • Sketch the graph of the function.
  • Use Desmos or a calculator to find the intersection point.

Graphical Approach

• Finding where (Y = 5) crosses the graph of (2^x + x).
• Graphing is a viable option but not always desired.

Introduction to the Lambert W Function

• The Lambert W function allows for algebraic solutions similar to the natural log function.
• Main idea: If (e^x = a), then (x = \ln(a)) can be easily solved.
• Example: (e^x = 5 \Rightarrow x = \ln(5)).

Properties of Natural Log

• (\ln(e^x) = x) (inverse properties).
• (\ln(5)) is a number approximately (1.609).

The Lambert W Function Explained

• The Lambert W function undoes the product of a number with its exponential expression:
  • If ( x e^x = a ) then ( x = W(a) ).
• This function is also referred to as the product log function.
• Requires transformation of equations to fit the format (x e^x).

Steps to Solve Using the Lambert W Function

1. Equation Manipulation: Start with the original equation (2^x + x = 5).
2. Rearrange to isolate exponential terms:
   • (2^x = 5 - x).
3. Divide both sides by (2^x):
   • (1 = \frac{5 - x}{2^x}).
4. Transform (2^x) to the natural base (e):
   • Rewrite (2 = e^{\ln(2)}), so (2^x = e^{x \ln(2)}).
5. Set up the argument for the Lambert W function:
   • Rearranging leads to (-x + 5 = 32\ln(2)).
6. Plug into the Lambert W function:
   • (W(32 \ln(2))).

Final Steps to Find X

• After applying the Lambert W function:
  • (-x + 5 \ln(2) = W(32 \ln(2))).
• Solve for (X):
  • (X = 5 - \frac{W(32 \ln(2))}{\ln(2)}).

Conclusion

• The approximate value for X using a product log calculator is (X \approx 1.716).
• Final answer can be kept as (X = 5 - \frac{W(32 \ln(2))}{\ln(2)}) or approximated numerically.

Key Takeaways

• Graphical methods vs algebraic methods for solving equations.
• Understanding the Lambert W function is crucial for solving complex exponential equations.
• Always continue learning; knowledge is essential for growth.