Problem 86 - Algebra and Trigonometry

Question

Suppose you’re driving your car on a cold winter day \(\left(20^{\circ} \mathrm{F} \text { outside) and the engine overheats (at }\right.\) about \(220^{\circ} \mathrm{F}\) ). When you park, the engine begins to cool down. The temperature \(T\) of the engine \(t\) minutes after you park satisfies the equation $$\ln \left(\frac{T-20}{200}\right)=-0.11 t$$ (a) Solve the equation for \(T\) . (b) Use part (a) to find the temperature of the engine after \(20 \min (t=20) .\)

Step-by-Step Solution

Answer

The temperature of the engine after 20 minutes is approximately 46.6°F.

Step 1: Isolate the Logarithmic Term

Given the equation \( \ln \left(\frac{T-20}{200}\right) = -0.11t \), we start by isolating the logarithm. We have it in the form of \( \ln(u) = a \), where \( u = \frac{T-20}{200} \) and \( a = -0.11t \).

Step 2: Remove the Logarithm by Exponentiation

Exponentiate both sides of the equation to eliminate the natural logarithm: \( e^{\ln\left(\frac{T-20}{200}\right)} = e^{-0.11t} \). This simplifies to \( \frac{T-20}{200} = e^{-0.11t} \).

Step 3: Solve for T

To solve for \( T \), multiply both sides by 200: \( T - 20 = 200 \cdot e^{-0.11t} \). Then, add 20 to both sides to isolate \( T \): \( T = 200 \cdot e^{-0.11t} + 20 \).

Step 4: Calculate T after 20 Minutes

Substitute \( t = 20 \) into the equation from Step 3: \( T = 200 \cdot e^{-0.11 \cdot 20} + 20 \). Calculate the exponent \( -0.11 \cdot 20 = -2.2 \) and then find \( e^{-2.2} \). Finally, compute \( T = 200 \cdot e^{-2.2} + 20 \) using a calculator to evaluate the exponential term.

Key Concepts

Logarithmic Equations

Logarithmic equations are mathematical expressions where the unknown variable is found inside a logarithm. These types of equations often look something like \( \ln(x) = a \) and can be solved by using exponentiation to eliminate the logarithm. When dealing with natural logs (loge), we use the base \( e \). This base is an irrational number approximately equal to 2.71828. To clear the logarithm from the equation \( \ln(u) = a \), we can exponentiate both sides, which results in \( u = e^a \).

• The key step in solving logarithmic equations is isolating the logarithmic term on one side of the equation.
• After isolation, exponentiate to remove the log function and solve for the variable inside.

Understanding how to handle logarithmic terms is crucial in many fields like physics, engineering, and finance because many natural processes and growth models are derived using logarithmic functions.

Natural Logarithms

Natural logarithms are a special type of logarithm with a base \( e \), which is Euler's number. This type of logarithm is incredibly useful for solving growth and decay problems, much like in the exercise where we calculate the engine temperature over time. In mathematical terms, the natural logarithm of a number \( x \) is written as \( \ln(x) \). Its main property is:

• \( \ln(e) = 1 \)
• \( \ln(1) = 0 \)
• For any \( x > 0 \), \( \ln(x) \) provides the power to which \( e \) must be raised to obtain \( x \).

For solving problems, converting a natural logarithm function into its exponential form is often the first step. For instance, if you have \( \ln(u) = a \), by converting it we get \( u = e^a \). This makes it easier to solve for \( u \), helping simplify and solve equations that involve exponential decay or growth, like cooling scenarios.

Solving Equations

To solve equations like the one in the exercise, where the cooling of an engine is modeled, it's important to follow systematic steps:Start by isolating complex expressions where variables are located. This often involves moving terms across an equation to get all instances of the unknown on one side. In our original example, this involves isolating the logarithmic component.

• Transform the equation to a more approachable form; in this case, this involves removing the logarithm by exponentiating both sides: \( \ln(u) = a \) turns into \( u = e^a \).
• Solve for the variable of interest. For our exercise, once we have \( \frac{T-20}{200} = e^{-0.11t} \), we modify it by multiplying through to get \( T \).
• The final part is substituting given values to find specific results—in this case, calculating the temperature after a given time by inserting \( t = 20 \). Applying the exponential value \( e^{-2.2} \) helps finalize the answer.

Each step is essential for reaching the solution efficiently and correctly. Solving similar equations is applicable in real-world processes such as cooling engines, chemical reactions, and even population growth models.