# 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

• 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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