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What is the result of this expression , It should mention that the log is natural logarithm. $$ \log\left(\exp(-x) - \exp(-y)\right) $$

Could we use the formula which mentioned in wikipedia about logarithmic identities?

$$ \log_{b}(a -c) = \log_b a + \log_b(1- \frac{c}{a}) $$

and does any body know the refrence of the above mentioned formula in wikipedia?

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The formula works well, as $$\log(e^{-x} - e^{-y}) = \log\left(e^{-x}(1-e^{x-y})\right) = -x + \log(1-e^{x-y}).$$ –  Pragabhava Nov 8 '12 at 1:34
    
@Pragabhava: Thanks for your comment. –  ben Nov 8 '12 at 1:43
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up vote 0 down vote accepted

I assume you are familiar with $$\log r+\log s=\log(rs)$$ Now if you replace $r$ with $a$, and $s$ with $1-(c/a)$, you get $$\log a+\log(1-(c/a))=\log(a(1-(c/a)))=\log(a-c)$$ So that's the source of the identity you quote.

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We can't really make it any simpler, sadly.

Also, notationally, we might write $e^{-x}$ or $\exp(-x)$, but not $\exp^{-x}$.

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Means that there is no formula for logarithm of subtraction? –  ben Nov 8 '12 at 0:22
    
Exactly. Neither for addition. 'What operation is before +?' –  Berci Nov 8 '12 at 0:38
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