July 23rd, 2011, 02:15 PM  #1 
Newbie Joined: Jul 2011 Posts: 1 Thanks: 0  Logaritm expression simplification
Hi i´m trying to make a better expression to this logaritm problem, but i couldn´t understand what i´m doing wrong...Can someone help me solvingthis ? The expression is to compute the Hue from a RGB color data: Hue = tan1( (log(R)  log (G)) / ( log (R)  log (G) +2*log(B) ) On the logaritm part i made a error that i can´t solve..i achieve this: ( (log(R)  log (G)) / ( log (R)  log (G) +2*log(B) ) = log2(R/G)/log2(B*B*R/G) ? whit this is wrong ? What is the proper solution for the logaritm part ? The log part must be on a basis of 2, because i´m trying to make a better approach to a assembly function. I achieved the result separating the quotients and dividends like this: quotient = log(R)log(G) = log(R/G) = log2(R/G)*log10(2) dividend = (log(R)log(G)+2*log(B)) = log(R/G)+log(B)+log(B) = log(B*B*R/G) = log2(B*B*R/G)*log10(2) Resulting in: log2(R/G) / log2(B*B*R/G) But this is wrong...why ? Note: After solving the logaritm part..what is the best solution for the tan1(Result expression) ? Where tan1 = tangent of a power of 1 = 1/tan = arctan ? or it is: 1/tan(( (log(R)  log (G)) / ( log (R)  log (G) +2*log(B) )) ??? Source of formula is: http://citeseerx.ist.psu.edu/viewdoc/do ... 1&type=pdf Best Regards, guga 
July 23rd, 2011, 02:41 PM  #2  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond  Re: Logaritm expression simplification Quote:
arctan, that is, tan^(1), is the inverse tangent function and it is not equivalent to 1/tan(x). See http://en.wikipedia.org/wiki/Inverse_tr ... _functions for information on the arctan function and http://www.programmersheaven.com/mb/x86 ... ngarctan/ for information about coding it. (Read the entire topic). See http://www.purplemath.com/modules/logrules.htm for information on handling logarithms.  

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