## Algebraic structure : Group

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

a*b=eln(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 !

 PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug

Recognitions:
 Quote by naoufelabs Please I have a problem with natural log for group set as follow: a*b=eln(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 !
what are you stuck on?

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

## Algebraic structure : Group

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.

 Recognitions: Science Advisor is elna*lnb the same as, or different than elnb*lna? show us your work so far
 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.

Blog Entries: 8
Recognitions:
Gold Member
Staff Emeritus
 Quote by 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.
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.

Recognitions: