what is the gradient of log x with base 'b'. ie it's not ln. it's a 'general' base.?

Answer 1

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If : y = log(b) x,

then : y = ( ln x ) / ( ln b )

i.e. : y = ( 1/ ln b )· ln x.

Hence : dy/dx = ( 1/ ln b )· ( 1/ x ) = 1 / ( x· ln b ) ... Ans.

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