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

Question

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

Answer 1

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.

Answer 2

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

Answer 3

Please check the above problem-Thank you

Answer 4

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

Answer 5

Thank you