How to Solve Exponential and Polynomial Equations?

  • Thread starter Ben-CS
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In summary, to solve e^-x = x, one can use the Lambert W function, which is defined as the inverse of x*e^x. By using this function, we can rewrite the equation as x = W(e^-x), and then solve for x using a calculator or a numerical method. Another approach is to graph both sides of the equation and find the intersection point, which will give the solution for x. Additionally, we can use a series expansion to approximate the solution for x. Overall, the key to solving e^-x = x is to manipulate the equation into a form that can be solved using mathematical techniques or tools.
  • #1
Ben-CS
Problem 1:

e^-x = x

Solve for x.


Problem 2:

x^y - y^x = xy - x - y

Solve for y.
 
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  • #2
Neither of these have "elementary" solutions (solutions in terms of the "elementary" functions: polynomial, rational, trig, log, exponential).

The first can be solved numerically or they can be solved in terms of the "Lambert W function".
 
  • #3
Problem 1:

e^-x = x
e^x = 1/x
x e^x = 1
x = ProductLog(1), where ProductLog (a.k.a. Lambert's W-function) is defined as the inverse of f(W) = W e^W

Numerically, the answer is about 0.567143290409784

Problem 2:

I don't know. Sorry.
 

FAQ: How to Solve Exponential and Polynomial Equations?

What are the possible solutions to the equation e^-x = x?

The equation e^-x = x has two possible solutions, which can be found using numerical methods such as graphing or using a calculator. The first solution is x = 0, and the second solution is x ≈ 0.56714329.

How can I solve the equation e^-x = x without using numerical methods?

The equation e^-x = x can be solved using the Lambert W function, which is defined as the inverse of the function f(x) = xe^x. The solution to the equation is x = W(1), where W(1) is the principal branch of the Lambert W function evaluated at x = 1. This solution can also be written as x = -W(-1).

Can I use logarithms to solve the equation e^-x = x?

Yes, logarithms can be used to solve the equation e^-x = x. Taking the natural logarithm of both sides of the equation results in ln(e^-x) = ln(x), which simplifies to -x = ln(x). This can then be solved using algebraic methods to find the solution x ≈ 0.56714329.

Are there any other methods for solving the equation e^-x = x?

Yes, there are other methods for solving the equation e^-x = x, such as using the Newton-Raphson method or the secant method. These methods involve iteratively finding better approximations for the solution until a desired level of accuracy is reached.

Why does e^-x = x have two solutions instead of just one?

This is because the function e^-x and the function x intersect at two points, resulting in two values of x that satisfy the equation e^-x = x. These points of intersection can be seen in a graph of the two functions, and can also be found using numerical or analytical methods.

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