METHOD 1 Analytical approach
Recognizing that the linear equation must be expressed in log form (M1)
\( \log y = m \log x + \log c \)
Use of slope formula (must involve logs) (M1)
\( m = \frac{\log 34.822 – \log 13.1951}{\log 4 – \log 2} = \frac{1.5417 – 1.1203}{0.6021 – 0.3010} = \frac{0.4214}{0.3011} \approx 1.4 \) (A1)
Attempt to substitute a value (M1)
\( \log c = \log 13.1951 – 1.4 \times \log 2 = 1.1203 – 1.4 \times 0.3010 = 1.1203 – 0.4214 = 0.6989 \)
\( \Rightarrow c = 10^{0.6989} \approx 5 \) (A1)
Equation: \( y = c \times x^m \implies y = 5 \times x^{1.4} \) (M1)
Result: \( y = 5 x^{1.4} \) [6]

METHOD 2 Regression analysis
Recognizing that a log-log graph results in a power function model (M1)
\( y = a \times x^b \)
Attempt to find a power regression model using the given two points (M1)
Set up equations: \( 13.1951 = a \times 2^b \), \( 34.822 = a \times 4^b \) (M1)
Divide: \( \frac{34.822}{13.1951} = \frac{a \times 4^b}{a \times 2^b} = 2^b \implies 2^b = 2.639 \implies b = \log_2 2.639 \approx 1.4 \) (A1)
Substitute: \( a = \frac{13.1951}{2^{1.4}} = \frac{13.1951}{2.639} \approx 5 \) (A1)
Equation: \( y = 5 \times x^{1.4} \) (A1)
Result: \( y = 5 x^{1.4} \) [6]