Ln x as a logarithmic function

  • #1
My book finds a function of x say ln(x). It is the area under 1/x. Having the properties (d/dx) ln x = 1/x and ln 1 = 0. It says it determines ln(x) completely. It satisfies the laws of logarithms, but why can I regard it as a logarithm just because it satisfies those laws?
 
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  • #2
SammyS
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My book finds a function of x say ln(x). It is the area under 1/x. Having the properties (d/dx) ln x = 1/x and ln 1 = 0. It says it determines ln(x) completely. It satisfies the laws of logarithms, but why can I regard it as a logarithm just because it satisfies those laws?
It's like the Duck Test.
"If it looks like a duck, swims like a duck, and quacks like a duck, then it probably is a duck."​
.
 
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  • #3
SteamKing
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My book finds a function of x say ln(x). It is the area under 1/x. Having the properties (d/dx) ln x = 1/x and ln 1 = 0. It says it determines ln(x) completely. It satisfies the laws of logarithms, but why can I regard it as a logarithm just because it satisfies those laws?
Waterfowl aside, you can use the Fundamental Theorem of the Calculus with the integral definition of the natural log to verify that ln (x) obeys the properties of a logarithmic function. See:

https://en.wikipedia.org/wiki/Logarithm

Specifically the section called "Integral representation of the natural logarithm".
 
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  • #4
Ssnow
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As you said ##\frac{d}{dx}\ln{x}=\frac{1}{x}## so the antiderivative of ##\frac{1}{x}## is ##\ln{x}+c##. If you consider a definite integral, the interpretation as area under the graph follows.
 
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  • #5
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My book finds a function of x say ln(x). It is the area under 1/x. Having the properties (d/dx) ln x = 1/x and ln 1 = 0. It says it determines ln(x) completely. It satisfies the laws of logarithms, but why can I regard it as a logarithm just because it satisfies those laws?
Based on that definition, you can prove that the function satisfies the laws of logarithms.
 
  • #6
Based on that definition, you can prove that the function satisfies the laws of logarithms.
Yes. I guess I didn't need to give all this information. The question is: Why if I satisfy the laws of logarithms does it have to be a logarithm.
I mean, logarithms have to satisfy the laws of logarithms, but can't there exist something that satisfies these same properties but that is not a logarithm?
 
  • #7
SteamKing
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Yes. I guess I didn't need to give all this information. The question is: Why if I satisfy the laws of logarithms does it have to be a logarithm.
I mean, logarithms have to satisfy the laws of logarithms, but can't there exist something that satisfies these same properties but that is not a logarithm?
Maybe there is. But what is it? Mathematics doesn't seem to have stumbled over this beast yet.
 
  • #8
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Yes. I guess I didn't need to give all this information. The question is: Why if I satisfy the laws of logarithms does it have to be a logarithm.
I mean, logarithms have to satisfy the laws of logarithms, but can't there exist something that satisfies these same properties but that is not a logarithm?
No, because you can also prove the opposite direction.

If a function f: R+ -> R satisfies f(a*b)=f(a)+f(b) for all positive a,b, and (a>b => f(a)>f(b)), then f(x)=c*ln(x) with c>0.
Note that you do not even have to require f(1)=0, because it follows from f(1*b)=f(1)+f(b).
 
  • #9
No, because you can also prove the opposite direction.

If a function f: R+ -> R satisfies f(a*b)=f(a)+f(b) for all positive a,b, and (a>b => f(a)>f(b)), then f(x)=c*ln(x) with c>0.
Note that you do not even have to require f(1)=0, because it follows from f(1*b)=f(1)+f(b).
Where does the c from c*ln(x) come from? And how do those two prior conditions imply f(x) = c*ln(x). Sorry for my ignorance, I'm not understanding.

Also when you say ln(x) do you already know it is a logarithmic function. Because if it you do, your not proving it. And if you don't, I don't understand what is telling me that it is in fact a logarithmic function?
 
  • #10
Yes. I guess I didn't need to give all this information. The question is: Why if I satisfy the laws of logarithms does it have to be a logarithm.
I mean, logarithms have to satisfy the laws of logarithms, but can't there exist something that satisfies these same properties but that is not a logarithm?
Surely that's contradiction. If it satisfies the laws of logarithms then it is by definition a logarithm!
 
  • #11
Surely that's contradiction. If it satisfies the laws of logarithms then it is by definition a logarithm!
Sorry I don't understand the logic, can I have a few extra steps?
 
  • #12
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Where does the c from c*ln(x) come from?
f(x)=2*ln(x), f(x)=523.35*ln(x) and so on all satisfy the logarithm laws. To narrow down f to f(x)=ln(x), you have to add one more requirement, like f(e)=1 or f'(0)=1.
And how do those two prior conditions imply f(x) = c*ln(x). Sorry for my ignorance, I'm not understanding.
Via some mathematical proof that I don't know in detail, but I know it exists. Basically, you show that any function that deviates from c*ln(x) at any point has to violate the logarithm laws somewhere.
Also when you say ln(x) do you already know it is a logarithmic function. Because if it you do, your not proving it. And if you don't, I don't understand what is telling me that it is in fact a logarithmic function?
You can use any definition of ln that you like there, the integral over 1/x, or the inverse function of the exponential function, or whatever. The function is unique, you can prove that the logarithm laws are satisfied by exactly one function (up to the constant discussed above), and you can also show that this function satisfies all the other properties you expect from it.
 
  • #13
Sorry I don't understand the logic, can I have a few extra steps?
If we agree to define a logarithm as a function having certain properties, and we then find a function which has these defined properties then it must be a logarithm, by definition.
You said it yourself:
Yes. I guess I didn't need to give all this information. The question is: Why if I satisfy the laws of logarithms does it have to be a logarithm.
I mean, logarithms have to satisfy the laws of logarithms, but can't there exist something that satisfies these same properties but that is not a logarithm?
You can't have something which satisfies all the laws of logarithms which isn't a logarithm! It's a contradiction.
 
  • #14
If we agree to define a logarithm as a function having certain properties, and we then find a function which has these defined properties then it must be a logarithm, by definition.
You said it yourself:


You can't have something which satisfies all the laws of logarithms which isn't a logarithm! It's a contradiction.
Without a proof of that not necessarily. All logarithms have to satisfy all logarithmic laws, but not the other way around, unless we can prove reciprocity.
All bananas are big and yellow, but there are other fruits that are big and yellow and are not bananas (e.g. mango).
 
  • #15
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All logarithms have to satisfy all logarithmic laws, but not the other way around, unless we can prove reciprocity.
You cannot prove a definition. You can prove that the logarithm laws uniquely define a function (up to the constant mentioned before). This function is called "logarithm".
 
  • #16
You cannot prove a definition. You can prove that the logarithm laws uniquely define a function (up to the constant mentioned before). This function is called "logarithm".
Well yeah I didn't say to prove a definition, and I thought reciprocity implied uniqueness.

On another note, can you direct me on where to find this proof, like the one you mentioned above about deviating from c*ln(x).
 
  • #17
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Every good textbook should have that.

Here is a sketch, if you replace the "increasing" by "continuous" (should also work with increasing):

Take an arbitrary continuous function f: R+ -> R that satisfies f(a*b)=f(a)+f(b) for all real a,b>0
Consider b=1: f(a*1)=f(a)+f(1), therefore 0=f(1).
Define c as f(e)=c.
$$c=f(e)=f(\sqrt e \sqrt e ) = f(\sqrt e )+f(\sqrt e ) = 2 f(\sqrt e)$$
therefore ##f(\sqrt e)=\frac{c}{2}##.
$$\frac c 2 =f(e^{1/4} \cdot e^{1/4} ) = 2 f(e^{1/4})$$
##f(e^{1/4})=\frac{c}{4}##
Doing that inductively fixes the function at a dense subset of the real interval [1,e] (all that can be expressed as ##e^{a/2^n}## for some a,n). With f(a*e)=f(a)+f(e)=f(a)+1 and 0=f(1)=f(a*(1/a))=f(a)+f(1/a) we can extend that to all positive numbers.

Let f and g be two (potentially different) functions that satisfy the criteria above, with the same constant c. Consider f-g. It is 0 everywhere on the dense subset, and as f and g are continuous it is continuous as well. The only function that satisfies those conditions is zero everywhere, therefore f=g. The function we found is unique.
 
  • #18
Without a proof of that not necessarily. All logarithms have to satisfy all logarithmic laws, but not the other way around, unless we can prove reciprocity.
All bananas are big and yellow, but there are other fruits that are big and yellow and are not bananas (e.g. mango).
Ah yes, but this means that big and yellow is not a 'good' definition of a banana. If we agree that it IS a good definition of a banana, then a mango would have to be called a banana. So if we agree on a definition of a logarithm, then any function which lives up to this definition must be a logarithm otherwise the world is a crazy place and we might as well give up!
 

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