Substitute \( t = 0 \) into \( A(t) = 500 \times e^{-k \times t} \):
\( A(0) = 500 \times e^{0} = 500 \).
Answer: 500 mg (A1 N1)

[1 mark]

Method 1: Direct Logarithmic Solution

Given: \( A(3) = 280 \), so \( 280 = 500 \times e^{-k \times 3} \).
Simplify: \( e^{-3 \times k} = \frac{280}{500} = 0.56 \) (M1)

Take natural logarithm: \( -3 \times k = \ln(0.56) \),
\( k = -\frac{\ln(0.56)}{3} \approx -\frac{-0.579818}{3} \approx 0.193279 \).
Correct to 3 significant figures: \( k \approx 0.193 \) (A1 A1 N2)

Method 2: Using Logarithm Property

From: \( 280 = 500 \times e^{-3 \times k} \),
\( \frac{280}{500} = e^{-3 \times k} \implies \ln\left(\frac{280}{500}\right) = -3 \times k \) (M1).

Solve: \( \ln\left(\frac{280}{500}\right) = \ln(0.56) \approx -0.579818 \),
\( k = -\frac{\ln(0.56)}{3} \approx 0.193279 \).
Correct to 3 significant figures: \( k \approx 0.193 \) (A1 A1 N2)

[3 marks]

Given: \( A(T) = 140 \), \( k \approx 0.193279 \), so \( 140 = 500 \times e^{-0.193279 \times T} \).
Simplify: \( e^{-0.193279 \times T} = \frac{140}{500} = 0.28 \) (M1)

Take natural logarithm: \( -0.193279 \times T = \ln(0.28) \approx -1.272965 \),
\( T = -\frac{\ln(0.28)}{0.193279} \approx \frac{1.272965}{0.193279} \approx 6.58637 \).
Correct to two decimal places: \( T \approx 6.59 \) hours (A1 A1 N2)

[3 marks]