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 ->
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
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(22740) (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 = 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))
