The formula $P=14.7 e^{-0.21 x}$ gives the average atmospheric pressure $P$, in pounds per square inch, at an altitude $x$, in miles above sea level. Find the elevation of a jet if the atmospheric pressure outside the jet is $6.7 \mathrm{lb} / \mathrm{in}^{2}$. The elevation of the jet is approximately miles. (Do not round until the final answer. Then round to the nearest tenth as needed.)

Step 1 ：We are given the pressure outside the jet and we need to find the altitude. This means we need to solve the equation \(P=14.7 e^{-0.21 x}\) for \(x\) when \(P=6.7\).

Step 2 ：First, we can divide both sides of the equation by 14.7 to isolate the exponential term: \(\frac{P}{14.7} = e^{-0.21 x}\). Substituting \(P = 6.7\), we get \(\frac{6.7}{14.7} = e^{-0.21 x}\).

Step 3 ：Then we can take the natural logarithm of both sides to solve for \(x\): \(-0.21 x = \ln\left(\frac{6.7}{14.7}\right)\).

Step 4 ：Solving for \(x\), we get \(x = \frac{\ln\left(\frac{6.7}{14.7}\right)}{-0.21}\).

Step 5 ：Calculating the above expression, we find that \(x \approx 3.741618892322715\).

Step 6 ：Rounding to the nearest tenth, we get \(x \approx 3.7\).

Step 7 ：Final Answer: The elevation of the jet is approximately \(\boxed{3.7}\) miles.