Algebraic structure : Group

naoufelabs

Please I have a problem with natural log for group set as follow:


a*b=e^ln(a)*ln(b)

1- Show that the group law * is associative and commutative
2- Show that the group law * accept an element e (Identity element)

Thank you !

lavinia

what are you stuck on?

naoufelabs

Sorry, I mean to achieve to result that a*b = b*a, and a*e=a.
But I'm stuck on.
Thanks

spamiam

You answered lavinia's question...by not answering it. As a start, use the definition you wrote in your first post to write out what a*b, b*a, and a*e are.

Deveno

is e^lna*lnb the same as, or different than e^lnb*lna?

show us your work so far

naoufelabs

I tested with Maple e^ln(a)*ln(b), then the result is a^ln(b).
so I tested with 2 different numbers, example:
3^ln(2) = 2^ln(3) ==> Gives the same result.
So I found that the e^ln(a)*ln(b) is commutative.

micromass

No, you have tested it with two arbitrary numbers. You have to test it for all numbers!!!!
Just checking it with two numbers does not suffice at all!!

You must show that a*b=b*a. Write out the definition of * and show us what it means.

Deveno

get a piece of paper, and use your brain. Maple won't solve this one for you.

Matt Benesi

Read exponentiation identities and properties over at wikipedia. Specifically the third identity.

naoufelabs

Thanks for all