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Old 12-24-2004, 10:01 AM   #16
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Re: y = ln(x/m -sa)/ r^2

Quote:
Originally posted by AcrobatMan
y = ln(x/m -sa)/ r^2
yr^2 = ln(x/m-sa)
e^ (yr^2) = x/m - sa
m(e^ (yr^2) + sa ) = x
me^(yr^2) = x-mas
me^(rry) = x-mas



uuummmmmmmm
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Old 12-24-2004, 10:22 AM   #17
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that's incorrect. you went from subtracting sa on one side to multiplying it with m.

and your ry in the left side should be converted to e.



me^(re) = x-m+(sa)

eliminate the m's for balance:

e^(re) = x-m+(sa)

then to solve for x:

x = e^(re)+m-(sa)

solve for m:

m = e^(re)-x-(sa)

solve for s:

sa/a = (e^(re)-x)/a

solve for a:

sa/s = (e^(re)-x)/s

e and r need to be given to solve this i think. i'm not that good at math.
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Old 12-24-2004, 12:49 PM   #18
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Re: y = ln(x/m -sa)/ r^2

Quote:
Originally posted by AcrobatMan
y = ln(x/m -sa)/ r^2
yr^2 = ln(x/m-sa)
e^ (yr^2) = x/m - sa
m(e^ (yr^2) + sa ) = x
me^(yr^2) = x-mas
me^(rry) = x-mas



AcrobatMan Rocks!!!
(Although nightmares of undergraduate math are returning... )
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Old 12-24-2004, 01:30 PM   #19
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I originally thought this was Soul Nation asking us to do her home work AGAIN; Merry Xmas anyway!
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