We know that $\log (a) - \log (b) = \log {(a/b)}$, suppose if $a$ and $b$ are some values with units say metre, the operation $\log (a)-\log (b)$ results in $x$ output with units, but the$ \log {(a/b)}$ results in cancellation of units. Can someone explain it to me?
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Don't you think that you have just argued that you cannot input something with a unit in a logarithm? – B. Pasternak Nov 12 '17 at 16:44
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A more clear indication that one cannot have a logarithm of something with a unit is the Maclaurin series $\ln(1-x)=-x-\frac{1}{2}x^2-\frac{1}{3}x^3\ldots$. – B. Pasternak Nov 12 '17 at 16:46
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In that case, what you have is actually $\log\frac ab=\log\frac a{\text{m}}-\log\frac b{\text{m}}$. – Nov 12 '17 at 16:50
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@B.Pasternak don't see how your second example makes sense... $1-x$ is dimensionless – spaceisdarkgreen Nov 12 '17 at 16:50
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@B.Pasternak And regarding the first, I think we've really shown that $\log(x)$ where $x$ has units sometimes makes sense provided in the expression there's a $\log(y)$ added to it where $y$'s units cancel out $x$'s – spaceisdarkgreen Nov 12 '17 at 16:55