Start with the equation:

\(2 \ln(x + 4) - \ln x = \ln(x + a)\)

Apply the logarithm property \(a \ln b = \ln(b^a)\):

\(\ln((x + 4)^2) - \ln x = \ln(x + a)\)

Use the property \(\ln a - \ln b = \ln\left(\frac{a}{b}\right)\):

\(\ln\left(\frac{(x + 4)^2}{x}\right) = \ln(x + a)\)

Since the logarithms are equal, set the arguments equal:

\(\frac{(x + 4)^2}{x} = x + a\)

Clear the fraction by multiplying through by \(x\):

\((x + 4)^2 = x(x + a)\)

Expand both sides:

\(x^2 + 8x + 16 = x^2 + ax\)

Subtract \(x^2\) from both sides:

\(8x + 16 = ax\)

Rearrange to solve for \(x\):

\(ax - 8x = 16\)

\(x(a - 8) = 16\)

Divide by \(a - 8\):

\(x = \frac{16}{a - 8}\)

Start with the given equation:

\(\ln(y + 1) - \ln y = 1 + 3 \ln x\)

Use the logarithm quotient rule: \(\ln \left( \frac{y+1}{y} \right) = 1 + 3 \ln x\)

Exponentiate both sides to remove the logarithm:

\(\frac{y+1}{y} = e^{1 + 3 \ln x}\)

Use the property \(e^{a+b} = e^a \cdot e^b\):

\(\frac{y+1}{y} = e \cdot (e^{\ln x})^3\)

Since \(e^{\ln x} = x\), this becomes:

\(\frac{y+1}{y} = e \cdot x^3\)

Rearrange to solve for \(y\):

\(y + 1 = e x^3 y\)

\(1 = e x^3 y - y\)

\(1 = y(e x^3 - 1)\)

\(y = \frac{1}{e x^3 - 1}\)

Thus, \(y = (e^{x^3} - 1)^{-1}\).

Start with the equation \(x = 4(3^{-y})\).

Take the natural logarithm of both sides: \(\ln x = \ln(4 \cdot 3^{-y})\).

Apply the logarithm of a product: \(\ln x = \ln 4 + \ln(3^{-y})\).

Use the logarithm of a power: \(\ln(3^{-y}) = -y \ln 3\).

Substitute back: \(\ln x = \ln 4 - y \ln 3\).

Rearrange to solve for \(y\): \(y = \frac{\ln 4 - \ln x}{\ln 3}\).