Is the Group Law * Associative and Commutative in Algebraic Structures?

[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]

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 !

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]

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]

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.

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

 

[Matt Benesi]

Read http://en.wikipedia.org/wiki/Exponentiation#Identities_and_properties" Specifically the third identity.

 

[naoufelabs]

Thanks for all