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December 20, 2014

December 20, 2014

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

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 7:19pmI 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 7:23pmwould 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 7:25pmPlease check the above problem-Thank you

- Algebra-Did I even begin correctly? -
**MathMate**, Monday, December 13, 2010 at 7:33pmDr. Bob was right.

log(6)27-2log(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^{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 7:45pmThank you

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