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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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1  
What does the dot mean in $\ln({A.B})$? –  Sean Jun 7 '12 at 20:19
2  
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? –  Zev Chonoles Jun 7 '12 at 20:21
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What in your post relates to the "undefined" in your title? –  Arturo Magidin Jun 7 '12 at 20:23
    
@Arturo Magidin $\ln {(x+a)}=xb$ do you know how to find x without opening logarithm? i think this is still Undefined –  user29646 Jun 7 '12 at 20:30
    
@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). –  Arturo Magidin Jun 7 '12 at 20:33

1 Answer 1

up vote 3 down vote accepted

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.

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