Problem 20 - Contemporary Precalculus

Question

Solve the equation. First express your answer in terms of natural logarithms (for instance, \(x=(2+\ln 5) /(\ln 3)) .\) Then use a calculator to find an approximation for the answer. $$27 e^{-x / 4}=67.5$$

Step-by-Step Solution

Answer

Question: Solve the equation \(27 e^{-x / 4}=67.5\) and approximate the value of \(x\). Answer: The approximate solution for \(x\) in the given equation is \(x \approx -3.47\).

Step 1: Isolate the exponential term

We will start by isolating the exponential term \(e^{-x/4}\) on one side of the equation. To do this, divide both sides by 27: $$e^{-x/4} = \frac{67.5}{27}$$

Step 2: Simplify the equation

Next, we'll simplify the equation by simplifying the fraction on the right side: $$e^{-x/4} = \frac{67.5}{27} = 2.5$$

Step 3: Apply the natural logarithm to both sides

To get rid of the exponential term, we'll apply the natural logarithm (denoted by \(\ln\)) to both sides of the equation: $$\ln(e^{-x/4}) = \ln(2.5)$$

Step 4: Use the logarithm property to simplify the left side

Using the property \(\ln(a^b) = b \ln(a)\), we can simplify the left side of the equation: $$\frac{-x}{4}\ln(e) = \ln(2.5)$$ We know that \(\ln(e) = 1\), so the left side simplifies to: $$\frac{-x}{4} = \ln(2.5)$$

Step 5: Solve for x

Now we'll solve for \(x\) by multiplying both sides of the equation by \(-4\): $$x = -4\ln(2.5)$$ This is the solution in terms of natural logarithms. To get an approximate value for \(x\), use a calculator to evaluate the numerical expression: $$x \approx -4(\ln(2.5)) \approx -3.47$$ So, the approximate solution for \(x\) is \(x \approx -3.47\).

Key Concepts

Exponential Equations

Exponential equations often involve variables appearing in the exponent, which can make them tricky to solve using simple algebraic techniques. In the exercise given, the exponential expression is in the form of \[e^{-x/4}\]. In this case, we needed to efficiently isolate this term to solve for the variable in question. Key steps to solve such equations include:- **Identify**: Recognize that the variable is in the exponent of an exponential function.- **Isolate**: Move the exponential part to one side of the equation.- **Apply logarithms**: Use logarithms to access and solve the power where the variable resides.In this problem, after moving all numerical expressions aside (specifically the 27 dividing sides), the exponential term was uncovered and simplified. This simplification leads us to the next step of applying the natural logarithm.

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions, allowing us to solve for variables in exponents effectively. The natural logarithm, denoted by \(\ln\), is especially useful when dealing with the base 'e'.Key properties include:

• \(\ln(a^b) = b \cdot \ln(a)\)
• \(\ln(e) = 1\)

These properties allow us to solve exponential equations by transforming the equation into a more manageable linear form. For instance, after applying \(\ln\) to both sides of the equation \(\ln(e^{-x/4}) = \ln(2.5)\), it enabled the transformation using the property \[\frac{-x}{4} \cdot \ln(e) = \ln(2.5)\]Given \(\ln(e) = 1\) simplifies the equation, leaving us with \[\frac{-x}{4} = \ln(2.5)\]. Ultimately, this allows us to solve for \(x\) in a straightforward manner.

Precalculus Problem Solving

Precalculus involves a variety of techniques that help bridge basic algebra and calculus. Problem-solving processes become essential when tackling functions, equations, and transformations, as seen in this exercise.This particular problem demonstrates a few valuable techniques:- **Reorganizing**: By isolating terms and applying principles of logarithms and exponentials, we streamlined complex algebra into manageable pieces.- **Exact vs Approximate Solutions**: It highlights instances when exact expressions are obtained via natural logarithms and then evaluates them numerically using calculators for a real-world estimation.The final solution's process consists of multiplying terms: \[x = -4 \cdot \ln(2.5)\]This grants us a logical and consistent method to derive an answer both in expression form, \(x = -4\ln(2.5)\), and as a numerical approximation \(x \approx -3.47\).Understanding these core concepts is crucial, as the techniques extracted here can be applied to a multitude of similar problems within precalculus and calculus.