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?

$18^{n+1}=2^{n+1}\times 27$
$3^{2n+2}\times 2^{n+1}=2^{n+1}\times3^{3}$ 
$3^{2n+2}=3^3$
$2n+2=3$
$n=1/2$
