Solving $ \frac {27 ^ {(2x+1)} } { 3 ^ {(x+1){5}}} = \frac{1}{3} $

Question

I am trying to find the value of $x$ in the following problem, I have to solve it without logarithm.

Problem : $$ \frac {27 ^ {(2x+1)} } { 3 ^ {(x+1){5}}} = \frac{1}{3} $$

EDIT: My work so far:

$$ \dfrac {3^{3(2x+1)} } { 3 ^ {(x+1)5}} = 3^{-1} $$

I know the formula $b^{u} = b^{v} \Longleftrightarrow u = v$, but I am not able to use it with this problem.

Thanks for help !

Answer 1

Let's combine all the hints (from above comments) to write:

$\dfrac{27^{(2x + 1)}}{3^{5(x + 1)}} = \dfrac{(3^3)^{(2x + 1)}}{3^{(5x + 5)}} = \dfrac{3^{3(2x + 1)}}{3^{(5x + 5)}} = \dfrac{3^{(6x + 3)}}{3^{(5x + 5)}} = 3^{(6x + 3) -(5x + 5)} = 3^{-1}$