SOLUTION: I NEED HELP....THIS IS THE LAST ONE TROUBLING ME....STUDYING FOR FINAL: write as single logarithm: -2 log base3 (1/x) + 1/3 log base3 square rt(x) APPRECIATE ALL YOU ALL DO.

Algebra ->  Algebra  -> Exponential-and-logarithmic-functions -> SOLUTION: I NEED HELP....THIS IS THE LAST ONE TROUBLING ME....STUDYING FOR FINAL: write as single logarithm: -2 log base3 (1/x) + 1/3 log base3 square rt(x) APPRECIATE ALL YOU ALL DO.      Log On

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Question 139973: I NEED HELP....THIS IS THE LAST ONE TROUBLING ME....STUDYING FOR FINAL:
write as single logarithm:
-2 log base3 (1/x) + 1/3 log base3 square rt(x)
APPRECIATE ALL YOU ALL DO..THIS IS A LIFE SAVER WHAT YA'LL ARE DOING TO HELP...THANKS SO MUCH!!!

Answer by ankor@dixie-net.com(12706) About Me  (Show Source):
You can put this solution on YOUR website!
-2 log base3 (1/x) + 1/3 log base3 square rt(x)
:
rewrite using exponents, the reciprocal gets rid of the negative:
:
logb3(x^2) + (1/3)*logb3(x^(1/2)); note that the sqrt%28x%29 = x^(1/2)
:
logb3(x^2) + logb3(x^(1/6)); multiply the exponents 1/3*1/2
:
Plus means multiply, add exponents: (2 + 1/6 = 12/6 + 1/6)
:
logb3(x^2*x^1/6)) = logb3(x^(13/6))