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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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    What does the dot mean in $\ln({A.B})$?2012-06-07
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    There's only a difference between $\ln(AB)$ and $\ln(A\cdot B)$ if there's a difference between $AB$ and $A\cdot B$. Usually, both of those things denote multiplication, i.e. $A$ times $B$. What do you mean by $A\cdot B$ if not that?2012-06-07
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    What in your post relates to the "undefined" in your title?2012-06-07
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    @Arturo Magidin $\ln {(x+a)}=xb$ do you know how to find x without opening logarithm? i think this is still Undefined2012-06-07
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    @Mr Are: Provided that $x\neq 0$, there always exists a $b$ such that $\ln(x+a) = xb$. As to solving such an equation *given* $a$ and $b$, it probably requires the use of Lambert's W function (see Wikipedia). But this is nowhere in your post, so there is **nothing** in your post that is "undefiend" (and there is nothing in your comment that qualifies as "undefined", either).2012-06-07
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    @Arturo Magidin when i am not able to solve it with easy way in my opinion it is! thats my opinion2012-06-07
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    @MrAres: Your opinion it may be, but "undefined" is the wrong thing to call it. It may be your opinion that when you cannot solve a problem then the problem should be called a "movie", but it doesn't make it a movie. Just because you cannot solve an equation does not mean that the equation is "undefined."2012-06-07
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    @MrAres: Finally, if you are actually interested in solving a problem like $\ln(x+a) = xb$ for given $a$ and $b$, then you should post that question in the body of your post, not buried in the comments. Then someone might tell you how to solve it.2012-06-07
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    @Arturo Magidin [Lambert W Function](http://en.wikipedia.com/Lambert_W_function)2012-06-07
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    $a\cdot b=a_1b_1+a_2b_2+...+a_nb_n$2012-06-07
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    @Arturo Magidin "post that question in the body of your post" $$\int \frac {dx}{(ax+b)}=a^{-1}\ln {(ax+b)}$$2012-06-07
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    @MrAres: First, that is not the body of your post, it's a comment. Second, that is not a question, it's a not-quite-correct indefinite integral. Third, don't confuse the dot product of vectors with multiplication of scalars. Fourth, *I* know about the Lambert W function; I'm the one who directed you to Wikipedia. Fifth: I have no idea what it is you believe your point is, but it certainly has not made it all the way to here, assuming you've actually mentioned it.2012-06-08
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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.