Question

Proof that \((ab)^n = a^n b^n\) using logarithms

Answer 1

\[\large (ab)^n = a^n b^n\]
\[\large n \log (ab) = \log (a^n b^n)\]
\[\large n \log a + n\log b = \log (a^n b^n)\]
\[\large \log a^n + \log b^n = \log (a^n b^n)\]
\[\large \log (a^n b^n) = \log (a^n b^n) \checkmark\]

Q.E.D.

Answer 2

hmmm

Answer 3

Lgba... you're doing a proof assuming the thing you're setting out to prove.

Answer 4

Your first statement is
(ab)^n=a^nb^n
and everything else follows from there....

Answer 5

lol that "hnnn" really makes anyone nervous =)))

i was actually asking if i did was right...so what should be done?

Answer 6

Well, in proofs, we our first statement (called our assumption) has to be something completely obvious and accepted. From there, we make further and further statements and ARRIVE at the conclusion which wasn't originally obvious.

If you begin with an assumption that isn't obvious, then you have a problem.

Answer 7

hmm so i think this one is obvious

Answer 8

Well, your starting statement is
(ab)^n = a^nb^n
which is obvious to you and me, sure. However, the idea is that we're trying to prove that statement, so to call it obvious is... cheating.

Answer 9

lol i was referring to your statement that the assumption should be obvious

Answer 10

You can always say that you're doing it by contradiction. ;P

Answer 11

Haha I didn't want to get into any fancy proofs with him yet. You should know, I'm describing the most basic kind of proof. There are other proofs in which your assumption isn't some obvious statement.

Answer 12

I mean, just state that\[(ab)^n\neq a^nb^n\]and show otherwise (you already did).

Answer 13

No he didn't.

Answer 14

I've never liked this kind of proofs, really. I mean, IMHO, saying that\[(ab)^n=a^nb^n\implies n\log(ab)=\log(a^nb^n)\]requires a justification.

Answer 15

lol i hate proving too =))) succeeding steps need to be generally accepted

Answer 16

otherwise it leads to proving the proof

Answer 17

Across, are you referring to the middle step he left out?
that is
\[(ab)^n = a^nb^n \implies \log((ab)^n) = \log(a^nb^n) \implies nlog(ab) = \log(a^nb^n)\]

Answer 18

so i guess i have to prove log thingies too :(

Answer 19

I meant the general\[a=b\implies\log a=\log b\]

Answer 20

Yeah, well. To me that depends on the context. It's certainly worth justifying at a lot of levels, but I don't want to justify it every time I use it =/

Answer 21

In a proof about exponent rules though, yeah, I think it's a pretty hefty assumption to make.

Answer 22

so it's not wise to use logs to prove exponents?

Answer 23

Since logarithms are the inverses of exponents, the rules for logarithms are derived from the rules for exponents. It would be better to use the rules for exponents to prove the rules for logarithms.

Answer 24

haha lol i see what you did there :P

Answer 25

What are some problems that you guys have run into with logarithms?

Answer 26

The only issue that I can recall is complex integration over branch cuts. For example:\[\int_\Gamma\frac{1}{z}dz,\]where \(\Gamma\) is the contour going from \(-1-i\) to \(-1+i\).