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Proving logarithm rules

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We're doing logarithms right now (grade 10 honours math, the US math curriculum goes incredibly slowly... but that's another topic), and one assignment was to prove logbx - logby = logbx/y This is what I did: logbx - logby = logbx/y let m = logbx <=> bm = x (by definition of logarithms) let n = logby <=> bn = y (by definition of logarithms) logbbm - logbbn ? logbbm/bn (by substitution) logbbm - logbbn ? logbbm - n (by law of exponents) m - n = m - n (by definition of logarithms, i.e. "b to what power equals bz, the answer is z") Q.E.D. Now, from my point of view, this is nice, simple, and logical. From my teacher's point of view, it's a very very poor proof because I lack experience with logarithms. Now, clearly, I haven't been using them for years or anything, but that doesn't sound like a very logical reason as to why it's a poor proof... Would someone please chime in with their point of view? I don't care so much if I'm wrong, but I care as to why I'm wrong so I understand things better. (I know I had another thread like this previously, but I'm always getting into debates with my math teacher (who is also my APCS teacher) and I can't find anyone else to ask as to who's right)

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Quote:
Original post by Ravuya
I'm gonna move this to Math & Physics.

Gotcha, just thought that this:
Quote:

Discussion of math and physics, primarily as they apply to games.

discouraged non-game-applying math discussion [grin]

Quote:
Original post by Gil Grissom
What exactly do you mean when you write
logbbm - logbbn ? logbbm/bn (by substitution)?


I mean that I substitute x and y with bm and bn into the original equation and use a question mark instead of an equal sign since we don't know if it's true yet.

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Your reasoning is right, but your proof is wrong. You can't just start with the statement under dispute and massage it into a statement that is trivially true. Look, here's a proof that 0=1:

0 = 1
0*0 = 1*0
0 = 0

...which is trivially true. But not useful. To prove a statement, you must START with trivially true statements, and transform them into the statement under dispute.

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Quote:
Original post by Sneftel
To prove a statement, you must START with trivially true statements, and transform them into the statement under dispute.


OK, gotcha. I can't understand why my teacher couldn't just have stated that.

Thanks [smile].

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Quick follow up - does that mean I can essentially reverse what I had? Like so:

prove: logbx - logby = logbx/y

proof:
let m = logbx <=> bm = x (by definition of logarithms)
let n = logby <=> bn = y (by definition of logarithms)

m - n = m - n

logbbm - logbbn = logbbm - n (by definition of logarithms, this is equivalent to the above equation)

logbbm - logbbn = logbbm/bn (exponent rules)

logbx - logby = logbx/y (substitution)

Q.E.D.

Would that be valid, since I started with a trivially true statement and worked up to the original?

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Quote:
Original post by nullsquared
I mean that I substitute x and y with bm and bn into the original equation and use a question mark instead of an equal sign since we don't know if it's true yet.

So my problem with this proof is that it goes in the wrong direction. You start with the premise, manipulate it in some ways and get a true statement (m-n = m-n). As I understand, you then assume that the premise itself is proved, which of course it is not since you can start with a false premise and still get a true conclusion. Your idea itself is OK, but you need to work it out a bit to get an actual proof. I wouldn't call it "lack of experience with logarithms", though -- maybe more "lack of experience with formal proofs".

I'm not sure what your teacher meant by lack of experience with logarithms. Did he give an example of what would be a good proof? Maybe, say, you already know that log x + log y = log xy, and he just wanted you to use that for a much shorter proof.

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Quote:
Original post by Gil Grissom
Quote:
Original post by nullsquared
I mean that I substitute x and y with bm and bn into the original equation and use a question mark instead of an equal sign since we don't know if it's true yet.

So my problem with this proof is that it goes in the wrong direction. You start with the premise, manipulate it in some ways and get a true statement (m-n = m-n). As I understand, you then assume that the premise itself is proved, which of course it is not since you can start with a false premise and still get a true conclusion.

Yup, as Sneftel mentioned. Now my question is, is my new version of the proof (which is essentially a reversal of what I originally had) better? See the post above yours.

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Quote:
Original post by nullsquared
Would that be valid, since I started with a trivially true statement and worked up to the original?

Yes, it looks valid, as long as by "substitution" you mean "the equation one before last is true for any m, n, and, therefore, for m = log x and n = log y in particular".

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Quote:
Original post by Gil Grissom
Quote:
Original post by nullsquared
Would that be valid, since I started with a trivially true statement and worked up to the original?

Yes, it looks valid, as long as by "substitution" you mean "the equation one before last is true for any m, n, and, therefore, for m = log x and n = log y in particular".


Thanks, appreciate it. If anyone else wants to chime in, feel free.

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Quote:

m - n = m - n

logbbm - logbbn = logbbm - n (by definition of logarithms, this is equivalent to the above equation)
You have not shown that this is true.

Namely, you have yet to demonstrate that log b^m - log b^n = log b^(m-n)
You're still assuming that which you have to prove.

Here's a better way:
log x - log y ? log x/y
Let m = log x <=> b^m = x
Let n = log y <=> b^n = y

m - n ? log x/y
b^(m-n) ? log x/y
log b^(m-n) ? log x/y
log b^m/b^n ? log x/y
log x/y ? log x/y
log x/y = log x/y

Thus: log x - log y = log x/y

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Quote:
Original post by SticksandStones
Quote:

m - n = m - n

logbbm - logbbn = logbbm - n (by definition of logarithms, this is equivalent to the above equation)
You have not shown that this is true.

Namely, you have yet to demonstrate that log b^m - log b^n = log b^(m-n)
You're still assuming that which you have to prove.

Well, by definition of logs, the question is b to the what power is bm? and the answer is clearly m because bm = bm. Therefore, logbbm is just m, and the same goes for n and (m - n). Perhaps if I explicitly write out bm = bm it would work?

<edit> Actually, my part might be better rewritten as "since logarithms and exponentiation are inverse operations, logbbm is equivalent to m" </edit>

Quote:

Here's a better way:
log x - log y ? log x/y
Let m = log x <=> b^m = x
Let n = log y <=> b^n = y

m - n ? log x/y
b^(m-n) ? log x/y
log b^(m-n) ? log x/y
log b^m/b^n ? log x/y
log x/y ? log x/y
log x/y = log x/y

Thus: log x - log y = log x/y

I understand your proof, but don't your first and third steps do exactly what you told me I can't do? You change m - n into logbbm-n, which is essentially what I did before.

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If you want to use the same steps as your first proof, start by assuming that the statement is not true:

logbx - logby != logbx/y

Standard proof by contradiction.

However, assuming you already have some other rules for logarithms, there's a more direct proof. Here's a hint: u - v = u + (-v)

I think this leads to a better proof since it's more direct, leads to a better understanding of why the relation is true, and, to a teacher, shows a better understanding of the underlying concepts.

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My book does it neatly by first proving this property :

ln(ab) = ln(a) + ln(b)

Then letting a = 1/b :

ln(1/b * b) = ln(1/b) + ln(b)

since ln(1/b * b ) = ln(1) = 0;

then

0 = ln(1/b) + ln(b)

ln(1/b) = -ln(b)


and we know that ln(1/b) = -ln(b).

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Here's how I would do it. Let's forget about b for a second an use natural logarithms (it mostly just simplifies notation):

k := log(x)-log(y)
exp(k) = exp(log(x)-log(y))
exp(k) = exp(log(x))/exp(log(y))
exp(k) = x/y
k = log(x/y)

Therefore,
log(x)-log(y) = log(x/y)

Now you can divide everything by log(b) and get the same result with base-b logarithms.

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Quote:
Original post by alvaro
Here's how I would do it. Let's forget about b for a second an use natural logarithms (it mostly just simplifies notation):

k := log(x)-log(y)
exp(k) = exp(log(x)-log(y))
exp(k) = exp(log(x))/exp(log(y))
exp(k) = x/y
k = log(x/y)

Therefore,
log(x)-log(y) = log(x/y)

Now you can divide everything by log(b) and get the same result with base-b logarithms.


Oh, that's incredibly simple and elegant. Thanks for pointing it out!

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it seems liek you are getting interested in mathematics. I have a large number of electronic resources i have collected over the years that go along with the classes at MIT OCW and stanford online etc etc.. I feel like getting to see some real mathematics would be great for you at this early stage. If you are interested email me at i.be.brett@gmail.com

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