Step 1 :Define the exponential decay function: \( N(t) = N_0 \cdot e^{(-kt)} \)
Step 2 :Set \( N_0 = 40,000 \) deaths for the year 1995 (\( t = 0 \))
Step 3 :Convert the average percentage decrease per year to a decimal: \( k = \frac{3.7}{100} = 0.037 \)
Step 4 :Write the exponential decay function with the given values: \( N(t) = 40,000 \cdot e^{(-0.037t)} \)
Step 5 :Set \( N(t) \) to 1,033 to find the year when the number of TB deaths is approximately 1,033: \( 1,033 = 40,000 \cdot e^{(-0.037t)} \)
Step 6 :Divide both sides by 40,000: \( \frac{1,033}{40,000} = e^{(-0.037t)} \)
Step 7 :Calculate the left side of the equation: \( 0.025825 = e^{(-0.037t)} \)
Step 8 :Take the natural logarithm of both sides: \( \ln(0.025825) = \ln(e^{(-0.037t)}) \)
Step 9 :Simplify the equation using the property of logarithms: \( \ln(0.025825) = -0.037t \)
Step 10 :Solve for \( t \): \( t = \frac{\ln(0.025825)}{-0.037} \)
Step 11 :Calculate \( t \) using a calculator: \( t \approx \frac{-3.6542}{-0.037} \)
Step 12 :Round \( t \) to two decimal places: \( t \approx 98.76 \)
Step 13 :Add \( t \) to the base year 1995 to find the approximate year: \( 1995 + 98.76 \approx 1995 + 99 \)
Step 14 :\(\boxed{\text{Year} \approx 2094}\)