1. e^x+lnx=7

I don't know how to solve this

$\displaystyle e^x+lnx=7$

I know the answer is approx 1.8536 (Wolfram alpha)

and I know what the graph $\displaystyle y=e^x+lnx-7$ looks like (Desmos)

I have played with it but I have not had any success.

Thank you.

2. Re: e^x+lnx=7

Originally Posted by Melody2
$\displaystyle e^x+lnx=7$
I know the answer is approx 1.8536 (Wolfram alpha)
and I know what the graph $\displaystyle y=e^x+lnx-7$
There is no simple solution.
But you need to understand Newton Method.

3. Re: e^x+lnx=7

Thanks Plato,
I think that I do understand Newton's method.
I'll try it that way and see how I go.

4. Re: e^x+lnx=7

Newton's method, also known as the Newton-Raphson method, is probably the first method that would spring to mind. For this example though a simple fixed point iteration will suffice.

Rewrite the equation as

$\displaystyle x=\ln(7-\ln x),$

from which form the iterative equation

$\displaystyle x_{n+1}=\ln(7-\ln(x_{n})).$

A simple sketch graph shows that the root is somewhere around $\displaystyle x = 2$, so taking this to be our starting value $\displaystyle x_{0},$ successive substitution gets us the sequence

1.84164, 1.85463, 1.85353, 1.85362, 1.8536237 etc.