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

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Answer 1 · 2017-01-02T11:58:08

$x = 0.26987$ $x=0.26987$ to 6dp

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

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 $0 < x < 1$ $0 < x < 1$ .

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:

enter image source here

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

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