- #1
Ben-CS
Problem 1:
e^-x = x
Solve for x.
Problem 2:
x^y - y^x = xy - x - y
Solve for y.
e^-x = x
Solve for x.
Problem 2:
x^y - y^x = xy - x - y
Solve for y.
Last edited by a moderator:
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.
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).
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.
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.
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.