# 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 !

Last edited:

lavinia
Gold Member
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

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 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.

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.

Deveno