Natural Logarithm of Convolution

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If I have a convolution of two variables, say x * y, and I take the natural logarithm of this operation, ln(x*y), do the same properties of logarithms apply?

So, does ln(x*y) = ln(x)+ln(y) ?
 

mathman

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In your notation, what are x, y, x*y and ln? Usually ln(x) is the natural log of a number (x).
 

LCKurtz

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If I have a convolution of two variables, say x * y, and I take the natural logarithm of this operation, ln(x*y), do the same properties of logarithms apply?

So, does ln(x*y) = ln(x)+ln(y) ?
I'm afraid your question doesn't make any sense as stated. The convolution operation is an operation on two functions, not two variables, and it gives another function. And if you ask the same question for functions, the answer is no.
 
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I realize that now since a convolution is an integral... just wishful thinking on my part I guess.
 

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