(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 \(n = 0.25\).

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

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

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:

\(n = 1.70\)

\(A = 2.90\)

To solve for \(C\) and \(a\), we start by considering the equation \(y = C(a^x)\). Taking the natural logarithm of both sides gives:

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

This equation is in the form of a straight line \(y = mx + c\), where \(\ln y\) is plotted against \(x\), \(\ln C\) is the y-intercept, and \(\ln a\) is the slope.

Plot the given points \((x, \ln y)\):

• (0.9, 1.7)
• (1.6, 1.9)
• (2.4, 2.3)
• (3.2, 2.6)

Draw a straight line through these points. From the line, determine the slope \(\ln a\) and the y-intercept \(\ln C\).

Using the mark scheme, we find:

\(\ln C = \ln 3.7\)

\(\ln a = \ln 1.5\)

Thus, \(C = 3.7\) and \(a = 1.5\).

(i) Start with the equation \(3^y = 4^{2-x}\).

Take the natural logarithm of both sides: \(\ln(3^y) = \ln(4^{2-x})\).

Using the logarithm power rule, this becomes \(y \ln 3 = (2-x) \ln 4\).

Rearrange to get \(y \ln 3 = 2 \ln 4 - x \ln 4\).

This can be written as \(y = -\frac{\ln 4}{\ln 3} x + \frac{2 \ln 4}{\ln 3}\), which is in the form \(y = mx + c\), showing it is a straight line.

The gradient \(m\) is \(-\frac{\ln 4}{\ln 3}\).

(ii) To find the intersection with \(y = 2x\), set \(-\frac{\ln 4}{\ln 3} x + \frac{2 \ln 4}{\ln 3} = 2x\).

Rearrange to get \(\frac{2 \ln 4}{\ln 3} = 2x + \frac{\ln 4}{\ln 3} x\).

Factor out \(x\): \(x \left(2 + \frac{\ln 4}{\ln 3}\right) = \frac{2 \ln 4}{\ln 3}\).

Solve for \(x\): \(x = \frac{\ln 4}{\ln 6}\).

\(Given the equation y = Ae^-kx2, taking the natural logarithm of both sides gives:\)

\(\ln y = \ln A - kx^2\)

This is in the form of a straight line \(y = mx + c\) with \(\ln y\) as the dependent variable and \(x^2\) as the independent variable. The slope \(m = -k\) and the intercept \(c = \ln A\).

Using the points (0.64, 0.76) and (1.69, 0.32), calculate the slope:

\(m = \frac{0.32 - 0.76}{1.69 - 0.64} = \frac{-0.44}{1.05} = -0.419\)

Thus, \(k = 0.42\).

Substitute one of the points into the equation \(\ln y = \ln A - kx^2\) to find \(\ln A\):

Using (0.64, 0.76):

\(0.76 = \ln A - 0.42 \times 0.64^2\)

\(0.76 = \ln A - 0.42 \times 0.4096\)

\(0.76 = \ln A - 0.172032\)

\(\ln A = 0.76 + 0.172032 = 0.932032\)

Therefore, \(A = e^{0.932032} \approx 2.80\).