xe^(x)-1 =0
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I ran this through my homemade "Newton's Method" program and got
x = .5671433
for that value xe^x
had a value of 1.000000026 on my calculator
To find the value of x that satisfies the equation xe^(x) - 1 = 0, we can use algebraic manipulation and a numerical method called the Newton-Raphson method.
Step 1: Algebraic Manipulation
Add 1 to both sides of the equation to isolate the term with the exponential function:
xe^(x) = 1
Step 2: Apply the Natural Logarithm
Take the natural logarithm (ln) of both sides of the equation:
ln(xe^(x)) = ln(1)
Step 3: Use Logarithmic Properties
Using logarithmic properties, simplify the equation:
ln(x) + ln(e^(x)) = ln(1)
ln(x) + x = 0
Step 4: Rearrange the Equation
Rearrange the equation to isolate x:
ln(x) = -x
Step 5: Apply the Newton-Raphson Method
Now, to find the value of x, we can use the Newton-Raphson method which is an iterative procedure to approximate the root of an equation.
First, choose an initial guess, let's say x0 = 1.
The next step is to apply the following recursive formula:
x1 = x0 - (f(x0) / f'(x0))
Where f(x) = ln(x) + x and f'(x) is the derivative of f(x).
Derivative of f(x) = 1/x + 1.
Using x0 = 1, we can calculate x1 as follows:
x1 = 1 - (ln(1) + 1) / (1/1 + 1) = 0
Repeat this process until you get a value that satisfies the equation within the desired precision. In this case, the solution is x ≈ 0.5671.