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

Let

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

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

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

And we conclude that the remaining solution is

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

- Start with an initial guess for the solution, denoted as (x_0).
- Calculate the derivative of the function (f(x) = e^x + \ln x), which is (f'(x) = e^x + \frac{1}{x}).
- Use the formula for Newton's method: [x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}]
- Substitute the initial guess (x_0) into the formula to find (x_1).
- 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).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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