(a) Start with the equation \(2^y = 3^{1-2x}\). Take the natural logarithm of both sides:

\(y \ln 2 = (1 - 2x) \ln 3\).

Rearrange to express \(y\) in terms of \(x\):

\(y = \frac{\ln 3}{\ln 2} - \frac{2x \ln 3}{\ln 2}\).

This is in the form \(y = mx + c\), where the gradient \(m = -\frac{2\ln 3}{\ln 2}\).

(b) To find the intersection with \(y = 3x\), set \(3x = \frac{\ln 3}{\ln 2} - \frac{2x \ln 3}{\ln 2}\).

Solve for \(x\):

\(3x + \frac{2x \ln 3}{\ln 2} = \frac{\ln 3}{\ln 2}\).

\(x(3 + \frac{2 \ln 3}{\ln 2}) = \frac{\ln 3}{\ln 2}\).

\(x = \frac{\ln 3}{3\ln 2 + 2\ln 3}\).

Recognize that \(3\ln 2 + 2\ln 3 = \ln(2^3 \cdot 3^2) = \ln 72\), so:

\(x = \frac{\ln 3}{\ln 72}\).