thanks I will check out your suggestions. looking for values between 0 >x <1.Except for specific values of y there is no closed form solution. (There is the Lambert W function if you want to give it a try, but that's just a way to bury the numbers.)
Let's recast this as [imath]a = \dfrac{ln(1 - x)}{ln(x)}[/imath]. If you have a specific a value in mind I'd graph it. Find out where the graphs of y = a ln(x) and y = ln(1 - x) intersect.
Again, if you have a specific value of a in mind there are other methods to approximate x. If you know a little Calculus Newton's method would be another.
-Dan
Please explain the context of this equation. Where did come from - class-room, physics research - what?Hi
I wonder if someone can help me.
Is it possible solve this equation for x.
y = ln(1-x)/ln(x)
If so how?
Thanks
Good, because you can't solve this for x.thanks for the reply.
No, I don't know how to go any further.
If your x-values are between 0 and 1, then the correct notation is:looking for values between 0 >x <1
Hi Dan,thanks I will check out your suggestions. looking for values between 0 >x <1.
I took a closer look at this. It seems that I'm wrong in bringing it up. The issue is that your equation reduces to [imath]x^y = 1 - x[/imath]. This is not an exponential equation in x. For it to be that we'd have to have something like [imath]y^x = 1 - x[/imath], which can be solved with the W function (x = 0 is one solution.) Your equation [imath]x^y = 1 - x[/imath] is more like a polynomial equation and so the W function will not work here.Hi Dan,
I've never come across the Lambert W function you mentioned. Looks interesting, so I had a look at the wiki pages and some examples. I had a go at trying to get the formula's x^y = 1-x and y = ln(1-x)/ln(x) into the correct format to use the Lambert W function y.e^y = x
It's probably my algebra skills, but I could not see a way. It requires a lot tricks that maybe I don't know.
How would you go about this?
Cheers
Steve
To confirm, you're trying to determine the ratio vo/vin when ratio tdischrg/tchrg is known. Is that correct?Y [is] ratio of tdis/tcharge. [X] = Vo/Vin
That's ok, no worries. I've learned about new functionI took a closer look at this. It seems that I'm wrong in bringing it up. The issue is that your equation reduces to [imath]x^y = 1 - x[/imath]. This is not an exponential equation in x. For it to be that we'd have to have something like [imath]y^x = 1 - x[/imath], which can be solved with the W function (x = 0 is one solution.) Your equation [imath]x^y = 1 - x[/imath] is more like a polynomial equation and so the W function will not work here.
Sorry for the red herring!
-Dan
Yes that's it. The equation comes from these two equations. If CR is unknown and Vo/Vin is the same in both equations.To confirm, you're trying to determine the ratio vo/vin when ratio tdischrg/tchrg is known. Is that correct?
[imath]\;[/imath]
You're missing two minus signs...Yes that's it. The equation comes from these two equations. If CR is unknown and Vo/Vin is the same in both equations.
Vo = Vin (1 - exp(-Tch/CR))
vo = vin (exp(-Tdis/CR))
I changed the equations around like this
Ln(1 -Vo/Vin) = Tch / CR
Ln(Vo/Vin) = Tdis / CR
[imath]\frac{Tch} {Tdis} = \frac{ln(1-Vout/Vin)}{ln(Vout/Vin)}[/imath]
Lowercase "y" in the original post is actually Tch/Tdis (the reciprocal of the above)Y was a ratio of tdis/tcharge. X was = Vo/Vin
Just "Sorry" won't cut it - Corner... Corner...It seems that I'm wrong in bringing it up
Dang it! I just got out of there!Just "Sorry" won't cut it - Corner... Corner...
oops.. Yeah, you are correct, forgot the minus signs.You're missing two minus signs...
ln(1 - Vo/Vin) = -Tch/CR
ln(Vo/Vin) = -Tdis/CR
However, once you divide they cancel each other out so that your final result is correct...
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Note that the following post is wrong...
Lowercase "y" in the original post is actually Tch/Tdis (the reciprocal of the above)
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Sorry, the above points don't help with your bigger problem. I agree with posts 2 and 8. You can't write an equation that starts with "x=<something in terms of y>" unless you manage to find an obscure mathematical function that few people will have heard about. Or you could define your own mathematical function (and perhaps investigate its properties).
Also, you could find approximate solutions with different degrees of accuracy - you can get VERY accurate if you like. Methods include producing a:- lookup table; graph; spreadsheet (that can perform a goal-seek for you); series expansion or write a quick (computer) algorithm. Would any of these ideas be useful?
Note that if you're implementing a lookup table, then you only need to store half of the full table (for y values between 0 and 1). Let the lookup function be called f(y). If y≤1 then x = f(y) else x = 1 - f(1/y)I have tried the spreadsheet goal-seek which worked. For the application I am thinking about, a program inside a micro, my thought was to use a lookup table. I'm not sure how I would use a series expansion or write an algorithm.