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?
 A: Divide both sides by $2^{n+1}$.
$$9^{n+1}=27$$
$$n+1=\log_9(27)=\frac 32$$
$$n=\frac 1 2$$
A: The problem with what you did
What you have done so far are valid algebraic manipulations, but you haven't made any progress. You ended up with the equation
$$
18 \cdot 18^n - 54 \cdot 2^n = 0
$$
which you will notice is the exact same equation you had in the second step, $18 \cdot 18^n = 54 \cdot 2^n$, so all those intermediate steps (where you subtracted and found a common denominator) didn't get anywhere.
Extended hint
To get somewhere (make progress), you want to isolate $n$ which means bring all the things with $n$ to one side. So you have
$$
18 \cdot 18^n = 54 \cdot 2^n
$$
and you want to bring all the $n$s to one side, so you can divide by $2^n$:
$$
\frac{18^n}{2^n} = \frac{54}{18}
$$
I will leave it to you to simplify $\frac{54}{18}$. What about the other side? Well,
$$
\frac{18^n}{2^n} = \left( \frac{18}{2} \right)^n = ?
$$
A: $18^{n+1}=2^{n+1}⋅27$
$\implies (3^{2}⋅2)^{n+1}=2^{n+1}⋅3^{3}$
$\implies 3^{2(n+1)}⋅2^{n+1}=2^{n+1}⋅3^{3}$
$\implies 3^{2(n+1)}=3^{3}$
$\implies 2(n+1) = 3$
A: $$18^{n+1} = 2^{n+1}\cdot27$$
$$\Leftrightarrow \frac{18^{n+1}}{2^{n+1}} = 27$$
$$\Leftrightarrow \left(\frac{18}{2}\right)^{n+1} = 27$$
$$\Leftrightarrow 9^{n+1} = 27$$
$$\Leftrightarrow 9^{n} = 3$$
$$\Leftrightarrow n =\log_9(3) =\frac {\log 3}{\log 9} =\frac {\log 3}{2\cdot \log 3} = \frac 1 2$$
A: Since $18=9\cdot 2$, we can rewrite the equation as$$(9\cdot2)^{n+1}=2^{n+1}\cdot 27\tag1$$
Simplifying $(1)$, we can eliminate $2^{n+1}$ to obtain$$3^{2n+2}=3^3\tag2$$
Which can be solved by $2n+2=3$. Solving, we get $n=\frac 12$.
A: $18^{n+1}=2^{n+1}\times 27$
$3^{2n+2}\times 2^{n+1}=2^{n+1}\times3^{3}$ 
$3^{2n+2}=3^3$
$2n+2=3$
$n=1/2$
A: Well, your not doing it wrong.
For better or worse, you reached:
$18*18^n -2^n*54=0$
Continue:
$2^n (18*9^n-54)=0$
$9*9^n-27=0$
$9^n-3=0$
$3^{2n}=3^1$
$2n=1$
$n=1/2$
So keep going and you'll get an answer.
But once done and in hindsight, it might become apparent although you were doing it a not wrong way, you weren't necessarily doing it the "right" way.
