\(Given the equation xⁿy² = C, take the natural logarithm of both sides:\)

\(n \ln x + 2 \ln y = \ln C\)

The equation of the line is \(y = mx + c\), where \(m\) is the gradient.

Calculate the gradient \(m\) using the points (0.31, 1.21) and (1.06, 0.91):

\(m = \frac{0.91 - 1.21}{1.06 - 0.31} = \frac{-0.3}{0.75} = -0.4\)

Since \(m = \frac{-1}{2}n\), equate and solve for \(n\):

\(-0.4 = \frac{-1}{2}n\)

\(n = 0.8\)

Substitute \(n = 0.8\) back into the equation:

\(0.8 \ln x + 2 \ln y = \ln C\)

Using the point (0.31, 1.21):

\(0.8 \times 0.31 + 2 \times 1.21 = \ln C\)

\(0.248 + 2.42 = \ln C\)

\(\ln C = 2.668\)

Calculate \(C\):

\(C = e^{2.668} \approx 14.41\)

(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: \(A = 4.17\).

(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:

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

\(Given the equation y² = Ae^kx, take the natural logarithm of both sides:\)

\(2 ln y = ln A + kx.\)

This is in the form of a straight line equation y = mx + c, where the slope m is k and the intercept c is ln A.

Using the points (1.5, 1.2) and (5.24, 2.7), calculate the slope k:

k = \(\frac{2.7 - 1.2}{5.24 - 1.5} \approx 0.80\).

Substitute k = 0.80 into the equation 2 ln y = ln A + 0.80x and use one of the points to find ln A:

\(At (1.5, 1.2), 2(1.2) = ln A + 0.80(1.5).\)

\(2.4 = ln A + 1.2.\)

\(ln A = 2.4 - 1.2 = 1.2.\)

A = \(e^{1.2} \approx 3.31\).

\((i) Taking the natural logarithm of both sides of the equation yⁿ = Ax³, we have:\)

\(n \ln y = \ln A + 3 \ln x\)

This equation is of the form \(Y = mX + C\), where \(Y = \ln y\), \(X = \ln x\), \(m = 3\), and \(C = \ln A\). Therefore, the graph of \(\ln y\) against \(\ln x\) is a straight line.

(ii) Using the given points \((x_1, y_1) = (1.20, 2.58)\) and \((x_2, y_2) = (2.51, 9.49)\), substitute into the equation:

For \((1.20, 2.58)\):

\(n \ln(2.58) = \ln A + 3 \ln(1.20)\)

For \((2.51, 9.49)\):

\(n \ln(9.49) = \ln A + 3 \ln(2.51)\)

Solving these two equations simultaneously, we find:

\(n = 1.70\)

\(A = 2.90\)