
That's only an equivalent way of defining it.
I hadn't entirely read all of the post. So, to make it a bit more understandable: The log of 0 is the area under the curve 1/x from x = 0 to x = 1. Notice, however, that this region contains the point (0, 1/0). Hence, that point, x = 0 is not in the domain.

How does that help you solve for a value of ln?
the integral of 1/x is lnx. Doesn't really help you solve for a given value. Also what about other log bases?
Also you care to link anything that defines log in that manner?

Okay. Say you have a value M of which you want to find the natural logarithm of (or at least, approximate). Given these two options, which do you take? 1) Randomly guess values of x such that e^x = M and hope you get close eventually, 2) take the curve 1/x and integrate from 1 to M using geometric approximations for the integral. More to the point, which is a better algorithm for doing this.
You have to multiply the integrand by a constant, namely, 1/ln(b) where b is the base.
Well, it shouldn't take a link to convince you that the two definitions are equivalent. I'd hope you'd be able to see that since the two definitions produce functions which a) have the same domain, b) take on the value at at least one place, and c) have the same properties (hence taking identical values over the entire domain), that they are in fact the same. If you insist on a link, however, here was one which was a small Google search away: http://mathworld.wolfram.com/NaturalLogarithm.html
*edit*
And with the definition I gave, the integral is ln(x) not ln(x) otherwise you are integrating through a discontinuity, so the domain is necessarily restricted to x > 0.

Well personally if I need to calculate it I bust out the trusty calculator or when programming use the built in libraries. I'm not gonna waste my time doing enough iterations of the trapezoidal or simpon's rules to get an accurate value ;)
What I was looking for is for a reference that said that logs (not just ln) were defined in such a manner. Yes it works for e and from that we can get the other ones. But its not the actual defination of logs.

Granted, I would break out the calculator too. Though, when I have written algorithms for ln, I have used the integral method.
At any rate, it is a way of defining logs. Yours is too. And, they both (as they should) produce the same thing. :)
Well, I think we are arguing in circles around each other at this point (just don't ask me to define circle ;) ).
Cheers

Nah defining a circle is too easy. Now figuring out why xviddivxoggmp3 keeps posting math questions here is a much harder question ;)

Ok, Thanks for everybodies input. You are correct, that I do not know anything about integral values. As for why I post math questions, I'm just trying to make sure I have a good foundation for calculas and calculas based physics (which I have been advised is needed to be a good programmer). I have received numerous details that have been misleading me on this issue. Again sorry for the elementry nature of the questions I have been posting. I have tried to get answers from my teacher, math lab, and our tutor, but I'm still unclear on some things. I also have been unable to find any math forum that is active. I have found one or two, but no posts at all. Can you think of a valid example of how to use the distributive rule with a logarithm.

You can't use it like you did. Firstly, any function must act on a value (be evaluated somewhere), so you can't factor anything out without putting something else in. Secondly, the property which would allow you to treat it that way is linearity, a property which log does not posess.
Some more about that:
A function T is said to be linear if the following conditions hold:
i) T(0) = 0
ii) T(x+y) = T(x) + T(y)
iii) T(c*x) = c*T(x) for any constant c
For example, let T be multiplication by a constant. So, say T(x) = a*x.
Condition one holds: T(0) = a*0 = 0
Condition two holds: T(x+y) = a*(x+y) = a*x + a*y = T(x) + T(y)
Condition three holds: T(c*x) = a*(c*x) = (a*c)*x = c*(a*x) = c*T(x)
Now, log violates all of these conditions. Now, if you look at the set of positive real numbers multiplicatively (call this set R*), then a slight variation of the conditions do hold. Firstly, note that if x is in R*, then so is its inverse (multiplicatively), and 1 (the multiplicative identity) is also in R*.
i) log(1) = 0
ii) log(a*b) = log(a) + log(b)
iii) log(a^b) = b*log(a)
You can see a correlation between the multiplicative properties inside the logarithm, and the additive properties outside the logarithm. Note, however, that there are no special properties when addition takes place inside the logarithm. So, for instance, log(a+b) has no special qualities about it. For this reason, you cannot use the distributive property.
You can play around with the properties listed above, though:
log(a^c * b^c) = log(a^c) + log(b^c) = c*log(a) + c*log(b) = c*(log(a) + log(b))
And of course:
log(a) = log(1*a) = log(1) + log(a) = 0 + log(a) = log(a)

Zach,
thanks for the clarification.
So from what you are saying I can only factor the power after it has been logged.
i.e.
3^x = 2^x
x log 3 = x log 2
x log 3  x log 2 = 0
x (log 3  log 2) = 0
and would it also be correct to say you can do the following.
4^2 = 2^4
2 log 4 = 4 log 2
2 log 4  4 log 2 = 0
2 ( log 4  2 log 2) = 0
When I run these in my calculator, all forms produce 0, making me think this is valid.
Again thank you everybody for the help. My understanding of logarithms have been clarified.:)
I wish I would have had this understanding prior to my last exam. :rolleyes:

You are correct in both cases.