Show that the solution of the equation $3^{2x+4}\times5^{x}=3^{3x}\times25^{x}$ is $x=\dfrac{\lg81}{\lg15}$. Express the equation $\log_{2} x + \log_{4} (x+6) = 2$ as a cubic equation in $x$.

1. Express all terms with prime bases and apply index laws to separate the constant and variable powers.

$3^{2x} \cdot 3^4 \cdot 5^x = 3^{3x} \cdot (5^2)^x$
$3^{2x} \cdot 81 \cdot 5^x = 3^{3x} \cdot 5^{2x}$

2. Group the terms with $x$ on one side of the equation.

$81 = \frac{3^{3x} \cdot 5^{2x}}{3^{2x} \cdot 5^x}$
$81 = 3^{3x-2x} \cdot 5^{2x-x}$

3. Simplify the expression to a single base to the power of $x$.

$81 = 3^x \cdot 5^x$
$81 = (3 \cdot 5)^x$
$81 = 15^x$

4. Take logarithms on both sides to solve for $x$.

$\lg 81 = \lg(15^x)$
$\lg 81 = x \lg 15$
$x = \frac{\lg 81}{\lg 15}$

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

$\log_{2} x + \frac{\log_{2} (x+6)}{\log_{2} 4} = 2$

2. Evaluate the denominator and simplify the equation

$\log_{2} 4 = 2$
$\log_{2} x + \frac{1}{2}\log_{2} (x+6) = 2$

3. Use logarithm laws to combine the terms

$2\log_{2} x + \log_{2} (x+6) = 4$
$\log_{2} x^2 + \log_{2} (x+6) = 4$
$\log_{2} [x^2(x+6)] = 4$

4. Convert from logarithmic to exponential form and expand to get a cubic equation

$x^2(x+6) = 2^4$
$x^3 + 6x^2 = 16$
$x^3 + 6x^2 - 16 = 0$