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?

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

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.


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

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

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

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