Solve for x

log₅ (5^1/x + 125) = log₅ 6 + 1 + 1/2x

Log_{(5^1/x + 125) = log(6) + log(5) + log (5^1/2x)}

                     = log(6.5.5^1/2x)

                     = log(30.5^1/2x)

cancelling log from both sides we get

 5^2/2x + 125 = 30.5^1/2x   

  now put 5^1/2x = y ,

   we get,

y²  + 125 = 30y

y² - 30y + 125 = 0

(y - 5)(y - 25) = 0

from here we get y = 5 or 25

 so x can be 1/2 or 1/4 .

Dude, you cannot just cancel log.. Appy 5 power on both sides