Exploring Logarithmic Laws: When x Has Dimensions

[theperthvan]

Say you have the function f(x)=log(x).

If x had dimensions of metres, say, then you can't have x=2m or whatever because dimensions can't be put into the equation to give log(2m).

What about if you have x1=4m and x2=2m and put
log(4m)-log(2m)
Each term doesn't exist by itself, but using log laws, this would be log(4m/2m)=log(2).
So how can they exist together but not by themself?

 

[christianjb]

That's a good question!

I don't have the answer- but note that

log(4m)-log(2m)=log(4)-log(2)+(log(1m)-log(1m))=log(4)-log(2)

So the units cancel when you write it this way too- as they should.

 

[whatta]

okay how about 0 exist but 0/0 does not? or, 3+4i and 5+4i are not real but (5+4i) - (3+4i) = 2 is?

 

[theperthvan]

that's true. Sill seems a bit strange though.

 

[lalbatros]

log(4m/1m)-log(2m/1m) makes sense

and

log(4m/1m)-log(2m/1m) = log(4m/2m)

no mystery

 

[whatta]

no? so what exactly log(2m) is?

 

[arevolutionist]

Correct me if I am incorrect.

log(Cm) = log(C) + log(m) = log(C) + 0 = log(C)

 

[sutupidmath]

Correct me if I am incorrect.

log(Cm) = log(C) + log(m) = log(C) + 0 = log(C)

how do you know wheather log(m) is 0 or not??

 

[sutupidmath]

log(4m/1m)-log(2m/1m) makes sense

and

log(4m/1m)-log(2m/1m) = log(4m/2m)

no mystery

However this is not exactly what theperthvan asked!

 

[whatta]

Correct me if I am incorrect.

log(Cm) = log(C) + log(m) = log(C) + 0 = log(C)

so Cm = C and m = 1. and why we even need a word meters.

 

[arevolutionist]

how do you know wheather log(m) is 0 or not??

I am using the logic that a meter is 1. By asking what log(m) is, I must ask what a meter is. In the mathematics a meter is basically just 1. By displaying 1 by an m we make it clear that this 1 is referring to a physical object.

 

[sutupidmath]

yeah, as m is merely a unit i think it should not need be written at all.

 

[theperthvan]

I am using the logic that a meter is 1. By asking what log(m) is, I must ask what a meter is. In the mathematics a meter is basically just 1. By displaying 1 by an m we make it clear that this 1 is referring to a physical object.

Nope, the argument must be dimensionless. You can't just make a dimension a number.

log(4m/1m)-log(2m/1m) makes sense

and

log(4m/1m)-log(2m/1m) = log(4m/2m)

no mystery

yeah, but then you get log(4m)-log(1m)-log(2m)+log(1m) which is back to the original question.
Pretty much, the argument must be dimensionless, because the Taylor expansion will have x+x^2+x^3+... (well not for this function obviously, but the idea is the same). So that is equivalent to adding say 1m+1m^2+1m^3. i.e you cannot add length to area to volume to whatever comes next.

 

[whatta]

maybe you would get dimension units in Taylor coefficients.

 

[HallsofIvy]

Understand that "meters" or "yards" or whatever are NOT "mathematical" entities. They are rather, parts of applications of mathematics to physics, geography, or whatever. The question you should be asking is not "what does log(3 meters) mean?" but rather "Are the any physics (or geography or...) formulas that require the logarithm of a distance?"- and if so, what does the physics say it means?

I know several formula that use logarithms but they all take the logarithm of a ratio so that the units cancel.


Let me ask you this: suppose on a quiz, you have a problem that defines f(x) to be x² and then asks you to find f(3). You think "no problem, whip out your calculator and find that f(3)= 3²= 9. On the very next question, g(x) is defined to be sin(x) and now you are asked to find g(3). You take your calculator, enter sin(3) and it gives you 0.0532359562 (approximately).

But when you get the quiz paper back, you find that problem has been marked wrong and your instructor tells you that the correct answer is 0.1411200081 and that you should have had your calculator in "radian mode" instead of "degree mode"! Should you complain? Go to the department chair? The president? The police? The problem didn't say anything about "degrees" or "radians"! Shouldn't either answer be acceptable?

Of course, the first question, "find f(3) when f(x)= x²" didn't say anything about units either. Mathematical functions deal with numbers, not measurements and so not units. That's one reason why the sine and cosine functions used in Calculus and above are not defined in terms of right triangles.

 

[arevolutionist]

Nope, the argument must be dimensionless. You can't just make a dimension a number.

Though, implementing my logic again, m with dimensions is just 1 in that dimension. m^2 is just 1 squared area. m^3 is just 1 cube.

 

[arevolutionist]

Also, log_a(x) is the exponent of base a. Thus, when we say log_a(4m) and a = 2 (for ease of use), implementing that logic that m is 1 to obtain the exponent, it functions correctly when producing x.

log_2(4m) = log(4) = 2

2^2m = 4m

Thus, it functions fine.

 

[Gib Z]

The only reason it functions fine is because you are treating m as a number! For all we know it is an unknown value, so the logarithm operator can be applied to it. However you are using it as meters, a dimension, but operating on it like a number!

 

[whatta]

2^2m = 4m

what a f?

by definition of units, we have equations like 1 * meter = 39.3700787 * inch, or 1 hectare = 10 000 meter * meter. please show me how do you arrive at 2^(2 * meter) = 4 * meter from there?

 

[Gib Z]

Ahh no you misinterpreted that quote, not at your own fault however. The notation was ambiguous. He meant (2^2)m=4m, not 2^(2m)=4m.

 

[Data]

The only time you'll ever need to take the logarithm of a dimensionful quantity is when an annoying lab coordinator tells you to do so. In such cases I suggest ignoring units altogether, and possibly filling in some rather unfriendly remarks on said coordinator's next student evaluation.

In any properly derived relation the units will cancel.