I assume this equation is . You can't solve it exactly, you have to use a numerical method.

Results 1 to 9 of 9

- April 4th 2011, 01:16 AM #1

- Joined
- Apr 2011
- Posts
- 3

- April 4th 2011, 01:24 AM #2

- April 4th 2011, 01:39 AM #3

- April 4th 2011, 04:03 AM #4

- Joined
- Apr 2005
- Posts
- 17,586
- Thanks
- 2235

I will admit that I would have interpreted the question as "if x= 1, what is ", but solving the equation is much more interesting.

Of course, so is the same as and that is the same as . Let y -ln(1.5)x so that and the equation becomes or .

Now, we have y= W(-ln(1.5)) where W is "Lambert's W function", the inverse function to [itex]f(x)= xe^x[/itex].

Finally, then .

Of course, as earboth says, x is not a real number because the domain of the real valued W function, while it includes some negative numbers, does not include any that negative.

- April 4th 2011, 05:19 AM #5

- Joined
- Apr 2011
- Posts
- 3

- April 4th 2011, 06:16 AM #6

- April 4th 2011, 06:54 AM #7

- April 4th 2011, 10:13 AM #8
Of course, you could use any number of methods of approximation. I tried the risky approach of blind iteration, using the following steps (having drawn the graphs first for a rough idea of the roots, obviously):

Starting at

....

We get the solution at approximately the 9th iteration. However I was trying to get the other solution! You could try a slightly better approach, such as interval bisection, Newton-Raphson or linear interpolation or something, but then you probably know all this already.

- April 5th 2011, 05:47 AM #9

- Joined
- Apr 2011
- Posts
- 3

I wrote a program yesterday, which has solved my problem.

You can download it here: http://strategist.fw.hu/program.rar