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Calculus/"e" and log
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Question
well I've been having a lot of trouble with this, cause I can't find a way to
demonstrate this is equal to 1, without using l'Hospital, and what I mean with
this, is Lim x->0 [Log(1+x)]/x
I am use to writing logarithms in two different ways. Log usually refers to using base ten, whereas ln refers to using bas e. I think that you are using log to really refer to what I call ln.
The log of a number is the power of ten it takes to get there.
Log(10)=1, Log(100)=2, Log(1000)=3, Log(10^n)=n.
Method 1
The natural log of a number, ln, refers to the power of e that it takes to get there. By definition, ln(e^2)=2, ln(e^3)=3, and ln(e^4)=4.
Thus, by definition, ln(e)=1.
Method 2
Another way to say this would be to solve ln(x)=1. To invert the ln function, you should make both sides an exponet of e. This will give e^ln(x) = e^1.
Note that e^ln(x) reduces to x (by definition of the functions).
Also not that e^1 reduces to e, since anything to the 1 is itself.
What this equation turns into is x = e. Looking back at the first equation in Method 2, we have ln(e)=1.
Summary
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