Problem 76 - College Algebra with Corequisite Support

Question

Atmospheric pressure \(P\) in pounds per square inch is represented by the formula \(P=14.7 e^{-0.21 x},\) where \(x\) is the number of miles above sea level. To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of 8.369 pounds per square inch? (Hint: there are 5280 feet in a mile)

Step-by-Step Solution

Answer

The mountain peak is approximately 14191 feet high.

Step 1: Substitute Known Values

Plug the given atmospheric pressure value into the equation. Since we know the pressure:\[ P = 8.369 \] and the formula is \[ P = 14.7 e^{-0.21x} \], substitute 8.369 for \( P \): \[ 8.369 = 14.7 e^{-0.21x} \].

Step 2: Isolate the Exponential Expression

Divide both sides of the equation by 14.7 to isolate the exponential term. This gives: \[ e^{-0.21x} = \frac{8.369}{14.7} \].

Step 3: Solve for the Exponent

Calculate \( \frac{8.369}{14.7} \) to simplify the equation:\[ e^{-0.21x} \approx 0.56904 \]. Take the natural logarithm on both sides of the equation to solve for \( -0.21x \):\[ \ln(e^{-0.21x}) = \ln(0.56904) \].This simplifies to: \[ -0.21x = \ln(0.56904) \].

Step 4: Calculate the Logarithm

Calculate \( \ln(0.56904) \), which gives:\[ -0.21x \approx -0.5649 \].

Step 5: Solve for x

Divide both sides by \(-0.21\) to solve for \(x\):\[ x = \frac{-0.5649}{-0.21} \].This results in \( x \approx 2.690 \) miles.

Step 6: Convert Miles to Feet

Since there are 5280 feet in a mile, convert \( x \) from miles to feet by multiplying by 5280:\[ \text{Height in feet} = 2.690 \times 5280 \]. This results in approximately 14191.2 feet.

Step 7: Round to the Nearest Foot

Round 14191.2 feet to the nearest whole number, giving: Approximately 14191 feet.

Key Concepts

Logarithms

Logarithms are essential mathematical tools used to solve exponential equations. They help us turn multiplication into addition, which simplifies complex calculations. In the context of the exercise, logarithms are used to solve for the exponent when the base of an exponential equation is known.
When dealing with exponential equations like \[ e^{-0.21x} = 0.56904 \],we employ logarithms to find the value of \(x\). By taking the natural logarithm on both sides, we transform the equation into a linear form, \[ -0.21x = \ln(0.56904) \],which is easier to handle.

• The logarithm function is the inverse of an exponential function.
• It is denoted as \(\log_b(a)\), where \(b\) is the base.
• The most common types are the common logarithm (base 10) and the natural logarithm (base \(e\)).

Logarithms provide a way to "unwrap" an exponential expression, allowing us to isolate unknown variables. In this exercise, taking the logarithm simplifies solving for the height of the mountain relative to atmospheric pressure.

Atmospheric Pressure

Atmospheric pressure is the force exerted by the weight of air in the atmosphere on a surface. It's a critical concept in meteorology and physics, as it affects weather patterns and altitude-related phenomena. In this exercise, atmospheric pressure is expressed as a function of altitude using the equation\[ P = 14.7 e^{-0.21x} \].This equation helps estimate atmospheric pressure at different altitudes, useful in real-world scenarios like determining mountain heights.

• Higher altitudes typically exhibit lower atmospheric pressure.
• Pressure is measured in units such as pounds per square inch (psi).
• Exponential functions often model atmospheric pressure concerning elevation.

Understanding atmospheric pressure is vital for calculating and predicting changes in environmental conditions. In this exercise, it helps relate pressure readings to specific altitudes, crucial for accurate calculations in this context.

Unit Conversion

Unit conversion is an important skill in mathematics and science, allowing us to express measurements in different units. This conversion is crucial when measurements don't match the desired or required units for calculations or insights.
In this exercise, the altitude of the mountain is initially calculated in miles, which is a standard unit for measuring distances. However, the final answer is required in feet, necessitating a conversion.

• 1 mile is equivalent to 5280 feet.
• Conversion involves multiplying the number of miles by 5280 to convert to feet.
• Ensuring units are correctly converted is essential for accurate results.

By converting miles to feet, we adjust our solution to the preferred measurement unit. This ensures that results are understandable and applicable in the context of other related data and interpretations.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is a logarithm with the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. It is a fundamental function in calculus and often used when dealing with exponential growth or decay.
In this exercise, the natural logarithm plays a key role in solving for the unknown variable \(x\) in the equation\[ e^{-0.21x} = 0.56904 \].Taking the natural logarithm of both sides, we convert the exponential equation into a linear form: \[ -0.21x = \ln(0.56904) \].

• The natural logarithm gives the time constant in continuous growth or decay processes.
• It simplifies complex calculations involving exponents.
• The natural logarithm is widely used in many fields, due to its natural properties around exponential functions.

By using the natural logarithm, we effectively strip away the exponential base \(e\), making the equation manageable to solve. Understanding this helps in comprehending how exponential and logarithmic relationships simplify complex real-world mathematical problems.