Problem 79 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the exponential equation algebraically. Round your result to three decimal places. Use a graphing utility to verify your answer. $$\frac{40}{1-5 e^{-0.01 x}}=200$$

Step-by-Step Solution

Answer

The solution to the equation is \(x \approx -124.803\)

Step 1: Subtract 200 from both sides

Start by subtracting both sides by 200, to isolate the fraction on one side:\(\frac{40}{1 - 5e^{-0.01x}} - 200 = 0\), this simplifies to:\(-160 = -5e^{-0.01x}\)

Step 2: Divide both sides by -5

Proceed by dividing both sides by -5 to isolate \(e^{-0.01x}\) on one side of the equation: \(32 = e^{-0.01x}\)

Step 3: Take the natural logarithm of both sides

Take the natural logarithm of both sides to remove the exponential: \(\ln{32} = -0.01x\)

Step 4: Solve for x

Finally, solve for x by dividing both sides by -0.01: \(x = \frac{\ln{32}}{-0.01}\)

Key Concepts

Algebraic Methods in Exponential Equations

When working with exponential equations, we often encounter expressions where the variable is in the exponent. This can seem daunting at first, but by applying algebraic manipulations, these equations can be transformed into more manageable forms. For instance, consider the textbook exercise where we have the equation \(\frac{40}{1-5 e^{-0.01 x}}=200\). The first step involves getting rid of the denominator by isolating the exponential term, a standard algebraic method. By subtracting 200 from both sides, we simplify the equation before isolating the exponential function \(e^{-0.01x}\).

Dividing both sides by -5 is a continuation of the principle to isolate the term containing the variable, in this case, the exponent. These initial steps are critical as they set the stage for utilizing natural logarithms to solve for the variable, which highlights the interconnectedness of algebraic techniques with other mathematical operations such as logarithms.

Natural Logarithms

Among the most powerful tools to solve exponential equations containing the natural base \(e\) is the natural logarithm, denoted as \(\ln\). The natural logarithm is the inverse operation of taking an exponent with base \(e\), which means that \(\ln(e^x) = x\) and \(e^{\ln(x)} = x\). This property becomes particularly handy in solving for variables in the exponent, as seen in the textbook solution where we have \(e^{-0.01x}\).

By taking the natural logarithm of both sides of the equation \(32 = e^{-0.01x}\), the exponential term is effectively 'canceled out', leaving us with \(\ln{32} = -0.01x\). This direct application of logarithms simplifies the equation to a point where it becomes a straightforward algebra problem, allowing us to solve for \(x\) by dividing both sides of the equation by -0.01. Understanding the relationship between exponents and logarithms is pivotal for students to tackle exponential equations confidently.

Graphing Utility Verification

After solving exponential equations algebraically, it is a good practice to verify the solutions using graphing utilities. These digital tools can provide a visual representation of the equation, allowing students to confirm if their calculated solutions are accurate. In our exercise, entering the original equation \(\frac{40}{1-5 e^{-0.01 x}}=200\) into a graphing utility and graphing it alongside the line \(y=200\) would show where they intersect, which corresponds to the solution for \(x\).

The intersection on the graph provides a visual confirmation of the solution derived through algebraic methods. This step is crucial because it not only reinforces the student's understanding of the solution but also introduces them to a method of verification that can be invaluable for more complicated equations or when checking the feasibility of results. Incorporating graphing utilities into the problem-solving process can enhance a learner's appreciation for the intersection of algebra and technology in mathematics.