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?

$$18^{n+1} = 2^{n+1}\cdot27$$
$$\Leftrightarrow \frac{18^{n+1}}{2^{n+1}} = 27$$
$$\Leftrightarrow \left(\frac{18}{2}\right)^{n+1} = 27$$
$$\Leftrightarrow 9^{n+1} = 27$$
$$\Leftrightarrow 9^{n} = 3$$
$$\Leftrightarrow n =\log_9(3) =\frac {\log 3}{\log 9} =\frac {\log 3}{2\cdot \log 3} = \frac 1 2$$
