Answer to Question #229389 in Algebra for Vffd

Question

The concentration of pollutants , in grams per liter, in the river Buriganga is approximated by P(x) = .04e−4x, where x is the number of miles downstream from a tannery industry that the measurement is taken. Find each of the following: • P(0.5), P(1) • The concentration of pollutants 2 mi downstream • the number of miles downstream where the concentration of pollutants is .002g per liter.

Answer 1

$P(x) = 0.04e^{-4x} \\ P(0.5) = 0.04e^{-4 \times 0.5}= 0.04e^{-2} \\ = 0.04 \times 0.1353 \\ = 0.005412 \;g/l = 5.412 \; mg/l\\ P(1) = 0.04e^{-4 \times 1} \\ = 0.04 \times 0.0183 \\ = 0.000732 \;g/l = 0.732 \; mg/l$

The concentration of pollutants 2 mi downstream

$P(2) = 0.04e^{-4 \times 2} \\ =0.04 \times 0.0003354 \\ = 0.0000134 \;g/l = 1.34 \; \mu g/l$

The number of miles downstream where the concentration of pollutants is .002g per liter.

$0.002 = 0.04e^{-4x} \\ \frac{0.002}{0.04}=e^{-4x} \\ 0.05 = e^{-4x} \\ ln(0.05) = -4x \\ -2.995 = -4x \\ x = \frac{-2.995}{-4}=0.748 \;miles$