The solution of the equation $3^{2x-1}=4(3^x)$ can be expressed as $\dfrac{\ln p}{\ln q}$.

1. Rewrite the equation using laws of indices to isolate terms in $3^x$

$3^{2x} \cdot 3^{-1} = 4(3^x)$
$\frac{1}{3}(3^x)^2 = 4(3^x)$

2. Use substitution to form a quadratic equation

Let $u = 3^x$.
$\frac{1}{3}u^2 = 4u$
$u^2 - 12u = 0$

3. Solve the quadratic equation for $u$ and reject the invalid solution

$u(u - 12) = 0$
$u = 0 \text{ or } u = 12$
Since $u = 3^x > 0$, $u = 0$ has no solution.
$3^x = 12$

4. Solve for $x$ using logarithms and identify $p$ and $q$

$\ln(3^x) = \ln 12$
$x \ln 3 = \ln 12$
$x = \frac{\ln 12}{\ln 3}$
Comparing with $\frac{\ln p}{\ln q}$, we have $p = 12$ and $q = 3$.

1. Change the base of the logarithm from 16 to 4.

$\log_{16}(3y-1) = \frac{\log_4(3y-1)}{\log_4 16} = \frac{\log_4(3y-1)}{2}$

2. Substitute back into the equation and clear the denominator.

$\frac{1}{2}\log_4(3y-1) + \frac{1}{2} = \log_4(3y)$
$\log_4(3y-1) + 1 = 2\log_4(3y)$

3. Use laws of logarithms to combine terms into a single equation without logarithms.

$\log_4(3y-1) + \log_4 4 = \log_4((3y)^2)$
$\log_4(4(3y-1)) = \log_4(9y^2)$
$4(3y-1) = 9y^2$
$12y - 4 = 9y^2$

4. Solve the resulting quadratic equation for $y$.

$9y^2 - 12y + 4 = 0$
$(3y - 2)^2 = 0$
$3y - 2 = 0$
$y = \frac{2}{3}$