Find $n$ if $\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$ Find $n$ if $$\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$$

In this video they show a shortcut and say $n=-1/2$ without any explanation. 
Key observation here is that the geometric mean of $9$ and $4$ is $6$.
It seems numerator and denominator are partial sums of geometric series, but I don't know how to proceed. Any help?
 A: $ \frac{9^{n+1} + 4^{n+1}}{9^n + 4^n} = 6 $
$ 9^{n+1} + 4^{n+1} = 6\cdot 9^n + 6 \cdot 4^n $
$3 \cdot 9^n = 2 \cdot 4^n $
$ 3^{2n+1} = 2^{2n+1}$
$(\frac{3}{2})^{2n+1} = 1$
So, $2n+1 = 0 $
$ n = -\frac{1}{2}$
A: Start by dividing top and bottom by $4^n$:
$$\frac{\frac{9^{n+1}}{4^n} + \frac{4^{n+1}}{4^n}}{\frac{9^n}{4^n} + 1} = 6.$$
Substitute $u = \frac{9^n}{4^n}$ to get
$$\frac{9u + 4}{u + 1} = 6 \iff 9u + 4 = 6(u + 1) \iff u = \frac{2}{3}.$$
Therefore,
$$\left(\frac{9}{4}\right)^n = \frac{2}{3} \implies n = -\frac{1}{2}.$$
A: Hint $\,\ \overbrace{\{ 9^{\large n+1},\, 4^{\large n+1}\} = \{6\cdot9^{\large n}\!,\, 6\cdot4^{\large n}\}}^{\large \text{by equal sum & product}}\,\ $  so $\,\ 9^{\large n+1}\! = 6\cdot 4^{\large n}\Rightarrow\, \left(\dfrac{3}2\right)^{\!\large 2n}\!\! = \dfrac{2}3$
A: Another idea: $$\frac{9^{n+1}+4^{n+1}}{9^n+4^n}=\frac{5\cdot9^n+4\left(9^n+4^n\right)}{9^n+4^n}=5\frac{9^n}{9^n+4^n}+4\implies$$
$$\frac{9^{n+1}+4^{n+1}}{9^n+4^n}=6\iff5\frac{9^n}{9^n+4^n}=2\implies\frac1{\frac{9^n+4^n}{9^n}}=\frac25\implies\frac1{1+\left(\frac49\right)^n}=\frac25\implies1+\left(\frac49\right)^n=\frac52\implies$$
$$\left(\frac49\right)^n=\frac32=\left(\frac23\right)^{-1}\iff\left(\frac23\right)^{2n}=\left(\frac23\right)^{-1}\implies 2n=-1\implies n=-\frac12$$
A: This is a possible way: 
$$\frac{9^{n+1}+4^{n+1}}{9^n+4^n}=6$$
Now: $$3^{2(n+1)}+2^{2(n+1)}=3\cdot 2\cdot 2^{2n}+2\cdot 3\cdot 3^{2n}$$ 
I obtain: $$3^{2(n+1)}+2^{2(n+1)}=3\cdot2^{2n+1}+2\cdot3^{2n+1}$$ 
In other words: $$3^{2n+1}(3-2)=2^{2n+1}(3-2)$$
The solution is $n=-\frac{1}{2}$.
A: As Theo well says, your equation is equivalent to
$$
{\left(\frac{9}{4}\right)}^n = \frac23
$$
On the other hand, for each $y \in (0 , \infty)$, the function
$$
\begin{array}{ccl}
\mathbb{R} & \to & (0 , \infty)
\\
x & \mapsto & y^x
\end{array}
$$
is strictly creasing for $y > 1$ (like exponential function) and strictly uncreasing for $y < 1$. It is the reason why for each $y , z \in (0 , \infty)$, $y \neq 1$, the equation
$$
y^x = z, \quad x \in \mathbb{R},
$$
admits only one solution, which is $x = \log_y(z)$. In your case, as $y = \frac94$ and $z = \frac23$, your only solution for your equation is $n = - \frac12$.
A: $$\dfrac{9(9^n) + 4(4^n)}{9^n + 4^n} = 6 $$ divide numerator and denominator by $9^n$:
$$ \dfrac{9 + 4\left(\frac{4}{9}\right)^n}{1 + \left(\frac{4}{9}\right)^n} = 6$$
Let $\left(\frac{4}{9}\right)^n = a$:
$$ \dfrac{9+ 4a}{1 + a} = 6 $$
$$a = \frac{3}{2} $$
$$ \left(\frac{4}{9}\right)^n =\left(\frac{2}{3}\right)^{2n} = \left(\frac{2}{3}\right)^{-1}$$
$$ n = -\frac{1}{2}$$
