Question

Need help in proving ...

Answer 1

please be trig proving

Answer 2

\[(ab)^{n} = a ^{n}.b ^{n} \]

Answer 3

isnt this the rule of exponents?

Answer 4

I know but I guess there should be a way of proving it? That's what the question says anyway ....

Answer 5

\[\large (ab)^n \implies n\log (ab) \implies n\log a + n\log b \implies \log a^n + \log b^n\\\large \implies \log(a^nb^n)\]

Answer 6

now how to remove log

Answer 7

Easy.
(ab)^n = (ab)(ab)(ab)(ab) (n times)
by associativity and commutativity of multiplication:
(ab)(ab)(ab)(ab) (n times) =
a*a*a*a*a*b*b*b*b*b (n times each) =
a^n*b^n

Answer 8

Basically the idea is, drop the parenthesis, rearrange the terms.

Answer 9

That seems quite easy .... but I'm eager to know how @lgbasallote solves that one.

Answer 10

hah i can do this =_=

Answer 11

oh lol of course!

Answer 12

Overly complicated proof is overly complicated.

Answer 13

i logged both sides

Answer 14

\[(ab)^n = a^n b^n\]
\[n \log (ab) = \log (a^n b^n)\]
then i proved n log (ab) = log (a^n b^n)

Answer 15

so (ab)^n = a^n b^n

QED

Answer 16

I'm new to log but I suppose that second step involves a property right?

Answer 17

yup it's the product law
\[\log (xy) \implies \log x + \log y\]

Answer 18

if that's the step you were talking about

Answer 19

OK, I got it. Thanks!!

Answer 20

The problem is that you've used logarithm rules to prove exponent rules... which is like using subtraction rules to prove addition rules.

Answer 21

lol well it worked :> idk

Answer 22

It's like using division rules to prove multiplication rules.

Answer 23

i cant imagine anything else on how to prove it lol

Answer 24

So
\[(a/b)^{n} = a ^{n}/b ^{n} \]
Must also be done in a similar way?

Answer 25

i think so

Answer 26

Thanks guys!! But to whom should I give the medal?

Answer 27

\[\large( \frac{a}{b})^n = \frac{a}{b}*\frac{a}{b}*\frac{a}{b}*\frac{a}{b}*... = \frac{a*a*a*a*a...}{b*b*b*b*b...} = \frac{a^n}{b^n}\]