Question 631098
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There is no algebraic solution to this equation.  You have to use numerical methods.


I used a graphing program and then ran some numbers through the function in Excel.  I was able to come up with two intervals: [0.000005,0.000006] and [6,6.1] where the function changes sign from one end of the interval to the other.  The Intermediate value theorem says there has to be a root on each of those intervals.


Then I used the Newton-Raphson method:


I first defined the function as


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \log(2x)\ -\ x\ +\ 5]


Then Newton-Raphson says 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ x_0\ -\ \frac{f(x_0)}{f'(x_0)}]


and then a closer approximation can be obtained from:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_2\ =\ x_1\ -\ \frac{f(x_1)}{f'(x_1)}]


And you can repeat the process as often as you like to get the desired precision.


The derivative of *[tex \LARGE \log(2x)\ -\ x\ +\ 5] is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(x)\ =\ \frac{\log(e)}{x}\ -\ 1]


You can run a calculator as well as I can.  Enjoy.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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