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Thomas Thomassen
Thomas Thomassen
Join Date 01/2003
+10

Math headache

May 1, 2005 3:17:03 PM from WinCustomize ForumsWinCustomize Forums
Right, I'm trying to figure out how to calculate the base number of a power when the result and exponent is known. It's been a while since I did this sort of work and now my brain has come to a complete stop. I have looked at the internet for formulas to help me, but I'm finding them just as hard to understand. (Mind you that I'm Norwegian and some of the terms are unfamiliar as I have never learned the english words for the mathematical terminology.)

So I hoped anyone here could help me in the right direction. The formula I have is:

200 = logb10
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Thomas Thomassen
Essencay
Join Date 01/2003
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Reply #1 May 1, 2005 4:11:14 PM
from WinCustomize ForumsWinCustomize Forums
http://www.math.utah.edu/~alfeld/math/Log.html If it worked right the answer I got was 1.011579

I've been out of school way too long, so I have no clue if that is correct way to display the base number or not. I have forgotten so much.
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Thomas Thomassen
Chris TH
Join Date 02/2003
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Reply #2 May 1, 2005 6:56:30 PM
from WinCustomize ForumsWinCustomize Forums
Looks right to me (FWIW!)

Posted via WinCustomize Browser/Stardock Central
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Thomas Thomassen
Skarny
Join Date 10/2002
+45
Reply #3 May 1, 2005 7:40:20 PM
from WinCustomize ForumsWinCustomize Forums
Getting the function to something that's solvable using most scientific calculators:

200 = logb10
translates to:
b^200 = 10 take the 200th root of both sides ->
b = 10^(1/200) simplify ->
b = 10^0.005
= 1.011579

Just another rearrangement to show some more working on how to get b as the 'focus'. The last line used a calculator (the x^y button) found on most scientific calculators.
A general formula might be: x = logby
b = y^(1/x)
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Thomas Thomassen
tri1ogy
Join Date 01/2005
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Reply #4 May 1, 2005 7:56:36 PM
from WinCustomize ForumsWinCustomize Forums
Ah, Skarnivorous beat me, but has the correct solving method.

His formulas appearance looks much better, so I'll only add this check...
logbx = (log x)/(log = (ln x)/(ln
= ln(10)/ln(1.011579454)
= 200
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Thomas Thomassen
thomassen
Join Date 01/2003
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Reply #5 May 1, 2005 8:00:45 PM
from WinCustomize ForumsWinCustomize Forums
Thank you. I'm making a DX widget where I got some objects that animated as they expand or collapse. However, I wanted something more fluid than just an even animation. I wanted it to start of fast and then slow down right before it reaches it's final size. What I came up with fast was something that started slow and ended fast. Opposite of what I wanted.

When I got back home this summer I'll see if I can find my old math books, if I haven't burned them, and freshen up my memory.
I swear my old math teachers are giggling to them self now, thinking: -That'll teach him for saying "This is rubbish, I'll never have any use for this!"!

Thanks guys.
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Thomas Thomassen
thomassen
Join Date 01/2003
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Reply #6 May 10, 2005 5:42:28 PM
from WinCustomize ForumsWinCustomize Forums
Right, I've got another issue. Similar, just inverted. I'd need a function that will in 10 steps create a curve that goes from a given number (example: 200) to 0. Last time the function generated a curve like the red one. Now a need a function the lives me a curve like the green one.


I'm quite annoyed that I can't figure out this my self yet. I remember doing stuff like this just a couple of years ago..
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Thomas Thomassen
KeithMacDonald
Join Date 11/2004
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Reply #7 May 10, 2005 6:48:47 PM
from WinCustomize ForumsWinCustomize Forums
ok, what kind of math is this?? im confused... im only able to do type II algebra, at the best (distributive property)
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Thomas Thomassen
thomassen
Join Date 01/2003
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Reply #8 May 11, 2005 4:44:12 AM
from WinCustomize ForumsWinCustomize Forums
Logarithm. Never thought I'd have any use for it. Hence why I forgot about it...
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Thomas Thomassen
Skarny
Join Date 10/2002
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Reply #9 May 11, 2005 7:08:42 PM
from WinCustomize ForumsWinCustomize Forums
I did have a lengthy post about decreasing functions and other bits, but it would have still left you with trying to 'fiddle' a function like 1/x or 1/sqrt(x) to make something that goes from 200 to 0 with an 'even' curve, in the shape you wanted. So I just did it for you. This was still done with a bit of trial and error and a graphing program, but it does what you want.

for x = 0 to 200 step 20
y = -40*(x-200)/(x+40)
next

y will decrease from 200 to 0, as x ranges from 0 to 200, the step 20 makes the 10 steps you wanted.
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Thomas Thomassen
Skarny
Join Date 10/2002
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Reply #10 May 11, 2005 7:17:26 PM
from WinCustomize ForumsWinCustomize Forums
as an extra note, a general form of the function:

-a*(x-/(x+a)

a is roughly the closness of the 'curving' point to the origin, so INCREASING this number makes the function a more GRADUAL curve (I recommend sticking around 20-40, but increase this as you increase

b is your limit,
so b = 100 will make the function decrease from 100 to 0 in x = 0 to 100
and b = 200 will make the function decrease from 200 to 0 in x = 0 to 200
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Thomas Thomassen
thomassen
Join Date 01/2003
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Reply #11 May 12, 2005 1:10:36 PM
from WinCustomize ForumsWinCustomize Forums
Ok, the -a*(x-/(x+a) function works perfect. However I have two different functions now. Is there any way that the -a*(x-/(x+a) function can be adapted to create a curve like the red one? I feel I got better control over the curve with this last function.
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Thomas Thomassen
Phoon
Join Date 10/2001
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Reply #12 May 12, 2005 2:30:30 PM
from WinCustomize ForumsWinCustomize Forums
I guess my Jethro Bodine school of Double-Naught Cypherin' diploma is quite useless in this scenario...
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Thomas Thomassen
Skarny
Join Date 10/2002
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Reply #13 May 12, 2005 6:49:56 PM
from WinCustomize ForumsWinCustomize Forums
heh heh, that has a quick solution that might not be equivilent to a mathematical inverse or anything. Think about the curve you have now. Mirror it across the x-axis, it now has the right shape, but it's from -200 to 0 on the y-axis, we want it from 0 to 200, so simply add 200 to every point on the curve! so -200 + 200 -> 0 and 0 + 200 -> 200, and we get the correct y-range.

So, to mirror a function across x-axis, a rough method is to multiply the entire function by -1, and to add 200 to every point, just add 200 to the function.

therefore:
y = 200+40*(x-200)/(x+40)

generally:
y = b+a*(x-/(x+a)

will give you what you want.
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Thomas Thomassen
thomassen
Join Date 01/2003
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Reply #14 May 13, 2005 7:05:48 AM
from WinCustomize ForumsWinCustomize Forums
Ah! Brilliant! Thank you very much for your help. It'll be credited in the widget.
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