Posted by Matt on Monday, December 13, 2010 at 7:03pm.
How would I write log(6)272log(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 itI've reread my notes but I'm still mixed up
Thanks for any help you can provide

AlgebraDid 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.

AlgebraDid 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 helpI appreciate itmaybe I can repost for mathmate or Dr. Bob or Reiny

AlgebraMathmate, Reiny or Dr. Bobplease double checkDid I even begin correctly?  Matt, Monday, December 13, 2010 at 7:25pm
Please check the above problemThank you

AlgebraDid I even begin correctly?  MathMate, Monday, December 13, 2010 at 7:33pm
Dr. Bob was right.
log(6)272log(6)3
=log_{6}(27)  log_{6}3²
=log_{6}(27/3²)
=log_{6}3
Appendix:
Laws of exponents:
x^{a}*x^{b} = x^{a+b}
x^{a}/x^{b} = x^{ab}
(x^{a})^{b} = x^{ab}
x^{a} = 1/x^{a}
x^{1/a} = ath root of x

AlgebraDid I even begin correctly?  Matt, Monday, December 13, 2010 at 7:45pm
Thank you
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