The variables $x$ and $y$ are defined such that $\log_{2} x - 3\log_{4} y = \log_{9} 3$.

1. State the reason based on the definition of logarithms

For $\log_a(z)$ to be defined, the argument $z$ must be greater than $0$. Therefore, $x > 0$ and $y > 0$.

1. Evaluate the constant term on the right hand side

$\log_{9} 3 = \log_{3^2} 3 = \frac{1}{2}$

2. Change the base of the second term to base 2

$\log_{4} y = \frac{\log_{2} y}{\log_{2} 4} = \frac{\log_{2} y}{2}$

3. Substitute back into the equation and combine the logarithmic terms

$\log_{2} x - \frac{3}{2}\log_{2} y = \frac{1}{2}$
$\log_{2} x - \log_{2} (y^{\frac{3}{2}}) = \frac{1}{2}$
$\log_{2} \left( \frac{x}{y^{\frac{3}{2}}} \right) = \frac{1}{2}$

4. Convert from logarithmic form to exponential form to express $x$ in terms of $y$

$\frac{x}{y^{\frac{3}{2}}} = 2^{\frac{1}{2}}$
$x = 2^{\frac{1}{2}} y^{\frac{3}{2}} = \sqrt{2} y^{\frac{3}{2}}$

1. Rewrite the equation using a substitution to form a quadratic equation.

Let $u = 5^x$.
$5(5^x) + \frac{2}{5^x} = 11$
$5u + \frac{2}{u} = 11$
$5u^2 - 11u + 2 = 0$

2. Solve the quadratic equation for $u$.

$(5u - 1)(u - 2) = 0$
$u = \frac{1}{5}$ or $u = 2$

3. Solve for $x$ when $u = \frac{1}{5}$.

$5^x = \frac{1}{5}$
$5^x = 5^{-1}$
$x = -1$

4. Solve for $x$ when $u = 2$ and express in logarithmic form.

$5^x = 2$
$x = \log_5 2$