Problem 46 - Algebra A Combined Function

Question

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. Use this formula to solve. Round to the nearest tenth. Find the elevation of a remote Himalayan peak if the atmospheric pressure atop the peak is 6.5 pounds per square inch.

Step-by-Step Solution

Answer

The elevation of the peak is 3.9 miles above sea level.

Step 1: Set the Given Value as P

We are given that the atmospheric pressure \( P \) is 6.5 pounds per square inch. We need to substitute this value into our equation \( P = 14.7 e^{-0.21 x} \).

Step 2: Substitute and Rearrange the Equation

Substitute \( P = 6.5 \) into the equation: \( 6.5 = 14.7 e^{-0.21 x} \). Now, we need to isolate the exponential part: divide both sides by 14.7 to get \( e^{-0.21 x} = \frac{6.5}{14.7} \).

Step 3: Calculate the Division

Calculate the division \( \frac{6.5}{14.7} \). The result is approximately \( 0.4422 \). Therefore, the equation now is \( e^{-0.21 x} = 0.4422 \).

Step 4: Use the Natural Logarithm

To solve for \( x \), take the natural logarithm of both sides: \( \ln(e^{-0.21 x}) = \ln(0.4422) \). This simplifies to \( -0.21 x = \ln(0.4422) \) because \( \ln(e^y) = y \).

Step 5: Solve for x

Calculate \( \ln(0.4422) \). The result is approximately \( -0.8155 \). Thus, \( -0.21 x = -0.8155 \). Divide both sides by -0.21 to solve for \( x \): \( x = \frac{-0.8155}{-0.21} \).

Step 6: Calculate the Final Answer

Calculate \( \frac{-0.8155}{-0.21} \). The value of \( x \) is approximately \( 3.88 \), which rounds to \( 3.9 \) miles. Therefore, the elevation of the peak is 3.9 miles above sea level.

Key Concepts

Natural Logarithms

Natural logarithms play a crucial role when dealing with exponential functions, especially in real-world applications like atmospheric pressure at varying altitudes. The natural logarithm, denoted as \(\ln\), is essentially the reverse of the natural exponential function \(e^x\). It helps us "undo" the exponential, making it very useful for solving equations where the variable is in the exponent.

• The base of the natural logarithm is the irrational number \(e\), approximately 2.718.
• This value is central in mathematical models involving growth and decay.
• For instance, if you have \(e^y = C\), then \(y = \ln(C)\).

Natural logarithms provide a method to simplify equations, transforming multiplicative relationships into additive ones, which are generally easier to handle. In our solution, we used \(\ln(0.4422)\) to solve for the variable \(x\). This step unravels the exponential part \(e^{-0.21x}\) by turning it into a linear equation.

Solving Equations

Solving equations involving exponential functions may seem daunting at first. However, using known mathematical operations like natural logarithms can simplify the process significantly. Here's how it works:

• First, ensure the equation is in a standard form, like a function equal to a constant.
• Isolate the exponential function by dividing through by any coefficients not part of the exponential term already, like dividing \(6.5\) by \(14.7\) in our example.
• The next step is crucial: use the natural logarithm to "bring down" the exponent, transforming the structure from \(e^{ax}\) to \(ax\).
• Once the exponential function is simplified, solve for the variable using regular algebraic methods like division or multiplication.

This technique streamlines the process, making it manageable despite the complexity of exponential relationships in the original equation. It's effective because it leverages the inherent relationships between exponential and logarithmic functions.

Atmospheric Pressure

Atmospheric pressure is the force exerted by the weight of the air above us. It decreases with height above the Earth's surface, as there is less air and thus fewer air molecules at higher altitudes. This concept is vital for environmental sciences, meteorology, and even aviation.

• At sea level, the standard atmospheric pressure is approximately 14.7 pounds per square inch (psi).
• As you ascend, the atmospheric pressure drops because the air becomes less dense.
• This relationship is described by exponential functions, allowing us to model how pressure changes with altitude.

In our example, the given equation \(P = 14.7 e^{-0.21 x}\) outlines how pressure \(P\) changes concerning altitude \(x\), helping us determine unknown elevations based on observed pressures.

Altitude Calculations

Altitude calculations often involve understanding how different quantities, like atmospheric pressure, change with height. They are particularly useful in activities like mountaineering, aviation, and meteorology.

• Given a specific atmospheric pressure, you can rearrange the model equation to solve for the altitude (\(x\) value).
• This requires algebraic manipulation and understanding of exponential decay, as pressure decreases rapidly at lower altitudes before leveling off.
• In the example, solving for \(x\) involved dividing given pressure values and using natural logarithms to find the equivalent altitude.

These calculations offer insights not just for adventure seekers, but for anyone monitoring the atmosphere's changing dynamics. Knowing how to perform these calculations equips you with the ability to understand and predict environmental conditions at various heights.