I know how to do it if A = B with log rules so I get it in the form ln(C) = (x+y)*ln(A) which allows me to solve for x+y by utilizing my data for C,A,B and determining the slope which is x+y. but if A ~= B then I get the form x*lnA + y*lnB on the right hand side which messes it up. I need to solve for x+y without using a system of equations. I could assume the values for x and y since I know what they are already and solve for C or something but i wanted to see if I could fine it otherwise.
Captain Matticus, LandPiratesInc
ln(C) = ln(a^x) + ln(b^y)
ln(c) = x * ln(a) + y * ln(b)
y * ln(b) = ln(c) - x * ln(a)
y * ln(b) = -x * ln(a) + ln(c)
y = (-ln(a) / ln(b)) * x + (ln(c) / ln(b))
y = (ln(1/a) / ln(b)) * x + (ln(c) / ln(b))
The slope is ln(1/a)/ln(b) I don't see why you think you need a system of equations.
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