I am trying to find the value of x in the following problem, I have to solve it without logarithm.
Problem : $ \dfrac {27 ^ {(2x+1)} } { 3 ^ {(x+1){5}}} = \dfrac{1}{3} $
EDIT: My work so far:
$ \dfrac {3^{3(2x+1)} } { 3 ^ {(x+1)5}} = 3^{-1} $
I know the formula $ b^{u} = b^{v} <=> u = v $ but I am not able to use it with this problem.
Thanks for help !