# 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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I'm gonna move this to Math & Physics.

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What exactly do you mean when you write
logbbm - logbbn ? logbbm/bn (by substitution)?

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Quote:
 Original post by RavuyaI'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 GrissomWhat exactly do you mean when you writelogbbm - 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 SneftelTo 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 nullsquaredI 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 nullsquaredI 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 nullsquaredWould 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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