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The natural logarithm is the logarithm to the base $e$, where $e$ is an irrational and transcendental constant. $e=\lim_{n\to \infty}\left(1+\frac {1}{n}\right)^n.$

$\ln a=\log_{e} a.$

I know that $\ln {(AB)}=\ln {(A}) + \ln {(B)}$ and $\ln {(A^B)}=B \ln {(A)}$.

  1. Is there any difference between $\ln {(AB)}$ and $\ln {(A\cdot B)}$ ?

  2. Is there any other way to solve $\ln {(A+B)}$ just like $\ln {(AB)}$ ?

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    @Arturo Magidin i solve it with Lambert W Function2012-06-08

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There is no difference between $\ln (AB)$ and $\ln(A\cdot B)$. Just like there is no difference between "$2x$" and "$2\cdot x$". Multiplication is often denoted by juxtaposition.

No, there is no way to "solve" or simplify $\ln(A+B)$ in a manner similar to that of the product.