SOLUTION: This is the question: log(base 5) 125^1/2 + log(base 11) 121^1/3= 13/6 What I tried: log(base 5) 125^1/2 + log(base 11) 121^1/3 = 13/6 log(base 5) 5*5^1/2+ log (base11) 121^

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: This is the question: log(base 5) 125^1/2 + log(base 11) 121^1/3= 13/6 What I tried: log(base 5) 125^1/2 + log(base 11) 121^1/3 = 13/6 log(base 5) 5*5^1/2+ log (base11) 121^      Log On


   

Question 581754: This is the question:
log(base 5) 125^1/2 + log(base 11) 121^1/3= 13/6
What I tried:
log(base 5) 125^1/2 + log(base 11) 121^1/3 = 13/6
log(base 5) 5*5^1/2+ log (base11) 121^1/3= 13/6
Yup,unfortunately, I'm stuck at here.
Thanks in advance ^0^!!
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
log(base 5) 125^1/2 + log(base 11) 121^1/3= 13/6
log(base 5) (5^3)^1/2 + log(base 11) (11^2)^1/3= 13/6
log(base 5) (5^3/2) + log(base 11) (11^2/3)= 13/6
3/2 + 2/3= 13/6
9/6 + 4/6= 13/6
13/6 = 13/6