Hi,

I don't remember how can I solve it for example: x=1,5^x

Help me please,

The Riddler

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- Apr 4th 2011, 01:16 AMRiddlerx=(1.5)^x
Hi,

I don't remember how can I solve it for example: x=1,5^x

Help me please,

The Riddler - Apr 4th 2011, 01:24 AMProve It
I assume this equation is . You can't solve it exactly, you have to use a numerical method.

- Apr 4th 2011, 01:39 AMearboth
- Apr 4th 2011, 04:03 AMHallsofIvy
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. - Apr 4th 2011, 05:19 AMRiddler
My previous example was wrong.

But if: x=(1.15)^x

It has got a real solution: x=22.1723..

How can I solve it? - Apr 4th 2011, 06:16 AMe^(i*pi)
It also has a solution of

Not sure how you'd show it though, I used wolfram - Apr 4th 2011, 06:54 AMProve It
Drawing the graphs of and and noting where they intersect gives you an idea of the solutions...

- Apr 4th 2011, 10:13 AMQuacky
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. - Apr 5th 2011, 05:47 AMRiddler
I wrote a program yesterday, which has solved my problem.

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