Find x in the following equation 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 !

Let's combine all the hints (from above comments) to write:
$\dfrac{27^{(2x + 1)}}{3^{5(x + 1)}} = \dfrac{(3^3)^{(2x + 1)}}{3^{(5x + 5)}} = \dfrac{3^{3(2x + 1)}}{3^{(5x + 5)}} = \dfrac{3^{(6x + 3)}}{3^{(5x + 5)}} = 3^{(6x + 3) -(5x + 5)} = 3^{-1}$
