(a) Start with the given equation: \(x = A(3^{-y})\).

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

Simplify using properties of logarithms: \(\ln x = \ln A - y \ln 3\).

Rearrange to express \(y\) in terms of \(\ln x\): \(y = -\frac{1}{\ln 3} \ln x + \frac{\ln A}{\ln 3}\).

This equation is in the form \(y = mx + c\), where the gradient \(m = -\frac{1}{\ln 3}\).

(b) Given that the line intersects the \(y\)-axis where \(y = 1.3\), substitute \(\ln x = 0\) and \(y = 1.3\) into the equation:

\(1.3 = -\frac{1}{\ln 3} \cdot 0 + \frac{\ln A}{\ln 3}\).

Simplify to find \(\ln A = 1.3 \ln 3\).

Exponentiate both sides to solve for \(A\): \(A = e^{1.3 \ln 3}\).

Calculate \(A\) to 2 decimal places: \(A = 4.17\).