Solve for $n$ in $18^{n+1} = 2^{n+1} \cdot 27$ 
Solve for $n$: $$18^{n+1} = 2^{n+1} \cdot 27$$

I tried:
$$18^{n+1} = 2^{n+1} \cdot 27  \Leftrightarrow 18^n \cdot 18 = 2^n \cdot 54 \Leftrightarrow \frac{18^n}{54} = \frac{2^n}{18} \Leftrightarrow \frac{18 \cdot 18^n - 54 \cdot 2^n}{972} = 0 \Leftrightarrow \\ 18  \cdot 18^n - 54 \cdot 2^n = 0 \Leftrightarrow ???$$
What do I do next? Am I doing it right?

$18^{n+1}=2^{n+1}⋅27$
$\implies (3^{2}⋅2)^{n+1}=2^{n+1}⋅3^{3}$
$\implies 3^{2(n+1)}⋅2^{n+1}=2^{n+1}⋅3^{3}$
$\implies 3^{2(n+1)}=3^{3}$
$\implies 2(n+1) = 3$
