To solve the equation $e^{-x} = 3 \log(x)$ using the fixed-point iteration method, we need to rewrite the equation in a form $x = g(x)$.

Step 1: Rearrange the Equation

The equation can be rearranged into a form suitable for fixed-point iteration. There are several ways to do this, but let's try one possible form: $x = -\ln\left(\frac{1}{3} e^{-x}\right)$ or $x = -\ln\left(\frac{e^{-x}}{3}\right) = -\ln(e^{-x}) + \ln(3) = x + \ln(3)$ However, this doesn't simplify well for iteration. Instead, let's try: $x = \frac{-\ln(x)}{3}$

So, let's use the iteration function: $x_{n+1} = \frac{-\ln(x_n)}{3}$

Step 2: Fixed-Point Iteration Process

1. Choose an initial guess: $x_0$. This value should be positive since the logarithm of a non-positive number is undefined.
2. Iterate using the equation $x_{n+1} = \frac{-\ln(x_n)}{3}$ until the values of $x_n$ converge (i.e., the difference between consecutive iterates is less than a predefined tolerance).

Example Iteration:

Let’s start with an initial guess, say $x_0 = 1.5$.

1. $x_1 = \frac{-\ln(1.5)}{3}$
2. Calculate $x_1$.
3. Continue iterating until the difference $|x_{n+1} - x_n|$ is sufficiently small.

Convergence Check:

You iterate until the change between two successive values is smaller than a chosen tolerance, say $10^{-5}$.

Tip:

The choice of the initial guess $x_0$ and the form of the function $g(x)$ can affect the convergence of the method.

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Would you like to see a detailed example of the iteration steps or the code to implement this in a programming language?

Related Questions:

1. What is the impact of different initial guesses on the convergence of fixed-point iteration?
2. How does the choice of the function $g(x)$ affect the convergence rate?
3. Can we use other methods like Newton-Raphson to solve this equation? How does it compare to fixed-point iteration?
4. What happens if the equation $e^{-x} = 3 \log(x)$ has multiple solutions?
5. How do we determine if the fixed-point iteration will converge for a given function?

Tip:

Always ensure that the function you are using for iteration $g(x)$ is such that $|g'(x)| < 1$ near the root to ensure convergence.