Homework Help: Algebra-Did I even begin correctly?

Posted by Matt on Monday, December 13, 2010 at 7:03pm.

How would I write log(6)27-2log(6)3 as a single logarithm

Would I start by getting rid of the (-2)- I know its the same log6 but what would I do with the 27 and the 3
I'm really confused on this one and I have to solve several similar ones with no clue on how to it-I've reread my notes but I'm still mixed up
Thanks for any help you can provide

Algebra-Did I even begin correctly? - DrBob222, Monday, December 13, 2010 at 7:19pm
I believe that will be
log₍₆₎(27/3^2). Perhaps Reiny, or Mathmate, or Bob Pursley will double check it. It has been so long since I did one of these.
Algebra-Did I even begin correctly? - Matt, Monday, December 13, 2010 at 7:23pm
would 27/3^2 = 3 then in the equation so I would write it as log(6)3

Thanks for trying to help-I appreciate it-maybe I can repost for mathmate or Dr. Bob or Reiny
Algebra-Mathmate, Reiny or Dr. Bob-please double check-Did I even begin correctly? - Matt, Monday, December 13, 2010 at 7:25pm
Please check the above problem-Thank you
Algebra-Did I even begin correctly? - MathMate, Monday, December 13, 2010 at 7:33pm
Dr. Bob was right.

log(6)27-2log(6)3
=log₆(27) - log₆3²
=log₆(27/3²)
=log₆3

Appendix:
Laws of exponents:
x^a*x^b = x^a+b
x^a/x^b = x^a-b
(x^a)^b = x^ab
x^-a = 1/x^a
x^1/a = ath root of x
Algebra-Did I even begin correctly? - Matt, Monday, December 13, 2010 at 7:45pm
Thank you

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