# Algebraic structure : Group

1. Oct 10, 2011

### 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 !

Last edited: Oct 10, 2011
2. Oct 10, 2011

### lavinia

what are you stuck on?

3. Oct 10, 2011

### naoufelabs

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

4. Oct 10, 2011

### 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.

5. Oct 12, 2011

### Deveno

is elna*lnb the same as, or different than elnb*lna?

show us your work so far

6. Oct 12, 2011

### 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.

7. Oct 12, 2011

### 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.

8. Oct 13, 2011

### Deveno

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

9. Oct 13, 2011

### Matt Benesi

Last edited by a moderator: Apr 26, 2017
10. Oct 14, 2011

### naoufelabs

Thanks for all