Question

the pressure of the air in the earths atmosphere decreases exponentially with altitude above the surface of the earth. the pressure at the earths surface (sea level) is about 14.7 pounds per square inch (psi) and the pressure at 3500 feet is approximately 12.8 psi. if we let y be the air pressure in psi and x be the altitude in feet, then the pressure can be estimated as a function of altitude by using y = 14.7(0.99996)^x. (a) predict the pressure in mexico city, mexico (altitude 7350 ft): psi. round to one decimal place. (b) find the altitude when the air pressure is 4.92 psi. round the answer to the nearest integer. ft.

Explanation:

Step1: Substitute altitude into formula for part (a)

Given $y = 14.7(0.99996)^{x}$, and $x = 7350$. Then $y=14.7\times(0.99996)^{7350}$.
First, calculate $(0.99996)^{7350}$. Let $a = 0.99996^{7350}$. Using the formula for exponentiation, $a = e^{7350\ln(0.99996)}$.
$\ln(0.99996)\approx - 4\times10^{-5}$, and $7350\times(-4\times10^{-5})=-0.294$. So $a = e^{- 0.294}\approx0.745$.
Then $y = 14.7\times0.745\approx10.9$.

Step2: Solve for altitude in part (b)

Given $y = 14.7(0.99996)^{x}$ and $y = 4.92$. Then $4.92=14.7(0.99996)^{x}$.
First, divide both sides by 14.7: $\frac{4.92}{14.7}=(0.99996)^{x}$. So $0.3347=(0.99996)^{x}$.
Take the natural - logarithm of both sides: $\ln(0.3347)=x\ln(0.99996)$.
We know that $\ln(0.3347)\approx - 1.098$ and $\ln(0.99996)\approx - 4\times10^{-5}$.
Then $x=\frac{\ln(0.3347)}{\ln(0.99996)}=\frac{-1.098}{-4\times10^{-5}} = 27450$.

Answer:

(a) $10.9$
(b) $27450$