Question

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

Answer 1

`x=0.26987` to 6dp

Explanation:

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