Given \(y = ab^x\), taking natural logarithms gives \(\ln y = \ln a + x \ln b\).

For the point (1, 3.7):

\(3.7 = \ln a + 1 \cdot \ln b\)

For the point (2.2, 6.46):

\(6.46 = \ln a + 2.2 \cdot \ln b\)

We have two equations:

1. \(3.7 = \ln a + \ln b\)

2. \(6.46 = \ln a + 2.2 \ln b\)

Subtract equation 1 from equation 2:

\(6.46 - 3.7 = (\ln a + 2.2 \ln b) - (\ln a + \ln b)\)

\(2.76 = 1.2 \ln b\)

\(\ln b = \frac{2.76}{1.2} = 2.3\)

\(b = e^{2.3} \approx 9.97\)

Substitute \(\ln b = 2.3\) into equation 1:

\(3.7 = \ln a + 2.3\)

\(\ln a = 3.7 - 2.3 = 1.4\)

\(a = e^{1.4} \approx 4.06\)

(i) Substitute \(x = \ln 3\) and \(y = \ln 2\) into the equation:

\(6e^{2\ln 3} + ke^{\ln 2} + e^{2\ln 2} = c\).

Simplify using \(e^{2\ln 3} = 9\), \(e^{\ln 2} = 2\), and \(e^{2\ln 2} = 4\):

\(6 \times 9 + k \times 2 + 4 = c\).

\(54 + 2k + 4 = c\).

\(58 + 2k = c\).

(ii) Differentiate the left-hand side with respect to \(x\):

\(\frac{d}{dx}(6e^{2x} + ke^y + e^{2y}) = 0\).

\(12e^{2x} + ke^y \frac{dy}{dx} + 2e^{2y} \frac{dy}{dx} = 0\).

Substitute \(x = \ln 3\), \(y = \ln 2\), and \(\frac{dy}{dx} = -6\):

\(12 \times 9 + 2k(-6) + 2 \times 4(-6) = 0\).

\(108 - 12k - 48 = 0\).

\(60 = 12k\).

\(k = 5\).

Substitute \(k = 5\) into \(58 + 2k = c\):

\(58 + 2 \times 5 = c\).

\(c = 68\).

To solve part (i), start with the given equation:

\(y^3 = Ae^{2x}\)

Take the natural logarithm of both sides:

\(\ln(y^3) = \ln(Ae^{2x})\)

Using logarithm properties, this becomes:

\(3\ln(y) = \ln(A) + 2x\)

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

For part (ii), substitute \(x = 0\) and \(\ln(y) = 0.5\) into the equation:

\(3(0.5) = \ln(A) + 2(0)\)

\(1.5 = \ln(A)\)

Exponentiate both sides to solve for \(A\):

\(A = e^{1.5}\)

Calculate \(A\) to two decimal places:

\(A \approx 4.48\)

\((i) Start with the equation xⁿy = C. Taking natural logarithms on both sides gives:\)

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

Substitute the given values:

\(For x = 1.10, y = 5.20:\)

\(n \ln(1.10) + \ln(5.20) = \ln C\)

\(For x = 3.20, y = 1.05:\)

\(n \ln(3.20) + \ln(1.05) = \ln C\)

Solving these two equations simultaneously, we find:

\(n = 1.50\)

Substitute back to find \(C\):

\(C = 6.00\)

(ii) The equation \(n \ln x + \ln y = \ln C\) is linear in \(\ln y\) and \(\ln x\), which means the graph of \(\ln y\) against \(\ln x\) is a straight line.

The given equation is \(y = Ax^n\). Taking natural logarithms on both sides, we get:

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

This is in the form of a straight line \(y = mx + c\), where \(m = n\) and \(c = \\ln A\).

From the graph, estimate the y-intercept (\(\\ln A\)) and the gradient (\(n\)).

Estimate the y-intercept from the graph, which is approximately 0.7. Therefore, \(\\ln A = 0.7\), giving \(A = e^{0.7} \approx 2.0\).

Calculate the gradient using two points from the graph. For example, using points (0.5, 0.5) and (2.5, 1.0):

Gradient \(n = \frac{1.0 - 0.5}{2.5 - 0.5} = 0.25\).

Thus, \(A \approx 2.0\) and \(n = 0.25\).