Given that $\log_a 8 + 3\log_a N = 3$ and $N > \frac{1}{4}$, find the range of values of $a$. Solve the equation $\log_3 x - \log_x 3x - 2 = 0$.

1. Combine the logarithmic terms into a single logarithm.

$\log_a 8 + \log_a N^3 = 3$
$\log_a (8N^3) = 3$

2. Convert the logarithmic equation to exponential form and solve for $a$ in terms of $N$.

$8N^3 = a^3$
$(2N)^3 = a^3$
$2N = a$
$N = \frac{a}{2}$

3. Use the given inequality for $N$ to find the inequality for $a$.

Since $N > \frac{1}{4}$,
$\frac{a}{2} > \frac{1}{4}$
$a > \frac{1}{2}$

4. Apply the conditions for the base of a logarithm to state the final range of $a$.

Since $a$ is the base of a logarithm, $a > 0$ and $a \neq 1$.
The condition $a > \frac{1}{2}$ satisfies $a > 0$, but we must exclude $a = 1$.
Therefore, $a > \frac{1}{2}, a \neq 1$.

1. Express $\log_x 3x$ in terms of base 3 logarithms

$\log_x 3x = \frac{\log_3 3x}{\log_3 x} = \frac{\log_3 3 + \log_3 x}{\log_3 x} = \frac{1 + \log_3 x}{\log_3 x}$

2. Substitute back into the equation and form a quadratic equation

$\log_3 x - \frac{1 + \log_3 x}{\log_3 x} - 2 = 0$

Let $u = \log_3 x$:
$u - \frac{1 + u}{u} - 2 = 0$
$u^2 - (1 + u) - 2u = 0$
$u^2 - 3u - 1 = 0$

3. Solve the quadratic equation for $u$

$u = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)}$
$u = \frac{3 \pm \sqrt{13}}{2}$
$u \approx 3.3028$ or $u \approx -0.3028$

4. Solve for $x$

$\log_3 x = 3.3028 \implies x = 3^{3.3028} \approx 37.7$
$\log_3 x = -0.3028 \implies x = 3^{-0.3028} \approx 0.717$