Let log√5125√5 = x
Then write this into exponential notation: √5x = 125√5
Now, I want to write 125 with a base of √5:
--- first, 125 = 53
--- also, (√5)2 = 5
--- so: 125 = 53 = [ (√5)2 ]3 = (√5)6
--- and: 125√5 = (√5)6 ·√5 = (√5)7
Thus: √5x = (√5)7 ---> x = 7