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arildno

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I assume that you know that an antiderivative is ln(x), but you don't know how to get there?

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whoa crap, ln (x) didn't cross my mind. Thanks alot........arildno said:I assume that you know that an antiderivative is ln(x), but you don't know how to get there?

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ln(x) will never "cross your mind." You have to either memorize it or know how to derive it.

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[tex]\frac{dx}{dy}=e^y=x[/tex]

[tex](1/x)dx=dy[/tex]

[tex]\int (1/x) dx=\int dy = y+constant = lnx+constant[/tex]

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StatusX

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[tex] \int_1^x t^{-1 + \epsilon} dt [/tex]

For [tex]\epsilon[/tex] greater than 0. Obviously this would be:

[tex] \frac{x^{\epsilon} - 1}{\epsilon} [/tex]

And you can see by applying l'hopitals that this goes to ln(x) as [tex]\epsilon[/tex] goes to 0.

Now if you want to get more technical, and not assume anything, you can see that the derivative of a^x is:

[tex] \lim_{h \rightarrow 0} \frac{a^{x+h} - a^x}{h} = a^x \lim_{h \rightarrow 0} \frac{a^h - 1}{h} [/tex]

Now if the limit exists, which you can see it clearly does by looking at the graph, this means that the derivative is some function of a times a^x. You can easily check that for a less than one, this function of a is negative, its equal to 0 at one, and its greater than 0 for a greater than one, and continues getting bigger and bigger. For some a, it will be exactly equal to one, and we'll call this e. So the derivative of e^x is just e^x. The inverse of exponentiation is logarithm, and ln(x) is defined as the log base e. Then the derivative of a^x is the derivative of e^(ln(a)*x) which you can see from the chain rule is ln(a)*e^(ln(a)*x), or just ln(a)*a^x. So the function of a was just ln(a). This can be used in the limit above:

[tex] \lim_{\epsilon \rightarrow 0} \frac{x^{\epsilon} - 1}{\epsilon} = \lim_{\epsilon \rightarrow 0} \frac{ln(x) \cdot x^{\epsilon}}{1} = ln(x) [/tex]

Actually, you don't even have to use l'hopitals at this point, because if you look carefully you'll see this limit and the funtion of a are indentical, just with a replaced by x and h replaced by [tex]\epsilon[/tex].

For [tex]\epsilon[/tex] greater than 0. Obviously this would be:

[tex] \frac{x^{\epsilon} - 1}{\epsilon} [/tex]

And you can see by applying l'hopitals that this goes to ln(x) as [tex]\epsilon[/tex] goes to 0.

Now if you want to get more technical, and not assume anything, you can see that the derivative of a^x is:

[tex] \lim_{h \rightarrow 0} \frac{a^{x+h} - a^x}{h} = a^x \lim_{h \rightarrow 0} \frac{a^h - 1}{h} [/tex]

Now if the limit exists, which you can see it clearly does by looking at the graph, this means that the derivative is some function of a times a^x. You can easily check that for a less than one, this function of a is negative, its equal to 0 at one, and its greater than 0 for a greater than one, and continues getting bigger and bigger. For some a, it will be exactly equal to one, and we'll call this e. So the derivative of e^x is just e^x. The inverse of exponentiation is logarithm, and ln(x) is defined as the log base e. Then the derivative of a^x is the derivative of e^(ln(a)*x) which you can see from the chain rule is ln(a)*e^(ln(a)*x), or just ln(a)*a^x. So the function of a was just ln(a). This can be used in the limit above:

[tex] \lim_{\epsilon \rightarrow 0} \frac{x^{\epsilon} - 1}{\epsilon} = \lim_{\epsilon \rightarrow 0} \frac{ln(x) \cdot x^{\epsilon}}{1} = ln(x) [/tex]

Actually, you don't even have to use l'hopitals at this point, because if you look carefully you'll see this limit and the funtion of a are indentical, just with a replaced by x and h replaced by [tex]\epsilon[/tex].

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StatusX

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[tex] \ln(x) = \lim_{\epsilon \rightarrow 0} \frac{x^{\epsilon} - 1}{\epsilon} [/tex]

Now noting that e^x is the inverse of ln(x):

[tex] y = e^x [/tex]

[tex] x = \ln(y) = \lim_{\epsilon \rightarrow 0} \frac{y^{\epsilon} - 1}{\epsilon} [/tex]

rearranging (and noting that these expressions are only valid in the limit as [tex]\epsilon[/tex] goes to 0):

[tex] \epsilon \cdot x = y^\epsilon - 1 [/tex]

[tex] y^\epsilon = 1 + x \cdot \epsilon [/tex]

[tex] y = (1 + \epsilon \cdot x)^{1/\epsilon} [/tex]

which leaves:

[tex] e^x =\lim_{\epsilon \rightarrow 0} (1 + \epsilon \cdot x)^{1/\epsilon}[/tex]

or, plugging in 1 for x, and replacing [tex]\epsilon[/tex] with 1/n:

[tex] e = \lim_{n \rightarrow \infty} (1 + 1/n)^n[/tex]

Which is the standard way of defining e. I'm sorry if this is much more than you wanted. I never really thought about this stuff, just accepting it at face value, and your question prompted me to try to derive it myself. I was suprised how nicely it all works out.

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Don't forget that in calculus they use log[x] as the natural log, instead of ln[x].misogynisticfeminist said:whoa crap, ln (x) didn't cross my mind. Thanks alot........

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dav2008

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Can you explain that...?Chrono said:Don't forget that in calculus they use log[x] as the natural log, instead of ln[x].

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Hey, thanks for showing how the integral for 1/y was obtained, that was helpful....

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Pretty much every calculus book uses log[x] instead of ln[x] to represent the natural log with base e. Remeber that in high school they told you that if you had log[x] without a base it was understood to be 10? Well, that's not ture in calculus. Log[x] has a base e. Is that better?dav2008 said:Can you explain that...?

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dav2008

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I have never seen log(x) used as the natural logChrono said:Pretty much every calculus book uses log[x] instead of ln[x] to represent the natural log with base e. Remeber that in high school they told you that if you had log[x] without a base it was understood to be 10? Well, that's not ture in calculus. Log[x] has a base e. Is that better?

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I see it used all the time. Even Mathematica uses it like that.dav2008 said:I have never seen log(x) used as the natural log

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I don't go near engineers, because they smell of booze and modernism, but I'm told that they use log to mean log to the base 10.

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Galileo

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The author writes:

My (university) calculus book uses ln, for log base e. Simply to seperate it from the other bases.Note: All logarithms in this book are natural logarithms - The base 10 logarithm is as much of a historical curiosity as a slide rule.)

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My statistical mechanics book (Kittel) also uses log to mean ln. It has something to do with clarity. ln can be confused easily as a product of l and n or something...DeadWolfe said:

I don't go near engineers, because they smell of booze and modernism, but I'm told that they use log to mean log to the base 10.

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BobG

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Base 10 log is used to measure gain in power amplification, used to measure sound levels, used to measure earthquakes, used to measure power of radio signals and gain of antennas among other things.Note: All logarithms in this book are natural logarithms - The base 10 logarithm is as much of a historical curiosity as a slide rule.)

Even more importantly,

What do they mean slide rules are only a historical curiosity!! I still use mine!

:grumpy:

And what's all this talk of smelling like booze Geez, I've gotta find a stronger aftershave

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matt grime

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log_{10} is useful since it is, give or take a small number, the number of digits in the original number. but it's still only factor of 2 and a bit away from log proper....

as we were told:

log will always mean log base e from now on. After all, why on earth would you want to use anything else?

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As for the integration part you just have to remember that any equation with 1/x or x^-1 will have a ln in it's answer.

I alway remember by doing this problem.

what is the antiderivative of 1/chair ?

I'll let you figure it out.

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Galileo

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Which is actually curious. They should've used the natural logarithms.BobG said:Base 10 log is used to measure gain in power amplification, used to measure sound levels, used to measure earthquakes, used to measure power of radio signals and gain of antennas among other things.

Who was this dopey guy who invented this dB scale? I think it was Alexander Graham Bell, who invented the telephone.

Well, that means he was an engineer so that figures :rofl:

...I was just kidding BobG, no need to get your underwear into a wrinkle.

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I've gotten used to using lg as log_{10} and ln as log_{e}.

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I just think it's safe to always assume log[x] as being the natural log, unless otherwise specified.

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vsage

I think engineers are right in this case. Ln is the notation for log base e in other countries (and not log), so I am told by my spanish (from spain) linear alg. prof. They don't call it the natural log though, they call it the Naperian log. That's curious because Napier lived well before Euler ever conceived of e.

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Alkatran

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It shouldn't be base e because ln() already has that value. What's the point of having two notations for the same thing when one of them used to do something different? The peopel trying to change it are just setting everyone up for troubles.

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