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Author:
ziwuxin
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SD power node
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Re: SD power node
Reply by
electricsauce
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May 2015
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Adobe Substance
You're right. I though Pow2 was x^2 not 2^n. Thanks for the help.
Re: SD power node
Reply by
electricsauce
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May 2015
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Adobe Substance
I'm not sure that it's working exactly as I want it to. For example 1.5 ^ 3 = 3.375, but using the substance power formula (((log 1.5) / (log 2)) * 3) ^ 2 = 3.079. It would be nice if there was a way to debug each nodes output.
Re: SD power node
Reply by
oblomov
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May 2015
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Nope, that's 2^((log(1.5)/log(2))*3)=3.375, so it works ;) In general, whatever values of x,y,z (x and z > 0), you have x^y=z^((log x/log z)*y)
SD power node
Topic by
electricsauce
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May 2015
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Stupid question. How do I get a standard power node (Base ^ exponent) in a function? Everything I can find is based on a power of 2.
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