Algebra-Did I even begin correctly?
posted by Matt on .
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
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.
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
Please check the above problem-Thank you
Dr. Bob was right.
=log6(27) - log63²
Laws of exponents:
xa*xb = xa+b
xa/xb = xa-b
(xa)b = xab
x-a = 1/xa
x1/a = ath root of x