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Algebra-Did I even begin correctly?

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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? - ,

    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? - ,

    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? - ,

    Please check the above problem-Thank you

  • Algebra-Did I even begin correctly? - ,

    Dr. Bob was right.

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

    Appendix:
    Laws of exponents:
    xa*xb = xa+b
    xa/xb = xa-b
    (xa)b = xab
    x-a = 1/xa
    x1/a = ath root of x

  • Algebra-Did I even begin correctly? - ,

    Thank you

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