How do you use Newton's method to find the approximate solution to the equation #e^x+lnx=0#?

Question

William Abdalla · Accepted Answer

#x=0.26987# to 6dp

Let #f(x) = e^x + ln x# Then our aim is to solve #f(x)=0#

First let us look at the graphs:
graph{e^x + ln x [-5, 5, -20, 15]}

We can see there is one solution in the interval # 0 < x < 1 #.

We can find the solution numerically, using Newton-Rhapson method

# f(x) = e^x+lnx => f'(x) = e^x+1/x #, and using the Newton-Rhapson method we use the following iterative sequence

# { (x_0,=1), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :} #

Then using excel working to 6dp we can tabulate the iterations as follows:

And we conclude that the remaining solution is #x=0.26987# to 6dp

Ezra Adams · Answer

undefined To use Newton's method to find the approximate solution to the equation $e^x + \ln x = 0$, follow these steps:

1. Start with an initial guess for the solution, denoted as $x_0$.
2. Calculate the derivative of the function $f(x) = e^x + \ln x$, which is $f'(x) = e^x + \frac{1}{x}$.
3. Use the formula for Newton's method: 
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
4. Substitute the initial guess $x_0$ into the formula to find $x_1$.
5. Iterate the formula using the result from step 4 as the new guess, i.e., $x_1$ becomes the new $x_0$, and continue until you reach a desired level of accuracy or convergence.

By repeating these steps, you'll approach the solution to the equation $e^x + \ln x = 0$.