Problem 19 - Precalculus Mathematics for Calculus

Question

Find the solution of the exponential equation, rounded to four decimal places. $$e^{2 x+1}=200$$

Step-by-Step Solution

Answer

\( x \approx 2.1492 \)

Step 1: Understand the Equation

We need to solve the exponential equation \( e^{2x+1} = 200 \). The goal is to find the value of \( x \) that satisfies this equation.

Step 2: Isolate the Exponential Term

To handle the equation \( e^{2x+1} = 200 \), first isolate the exponential term by dividing by the base of the exponent, \( e \,\). However, since \( e \) is a constant, we directly deal with changes using logarithms.

Step 3: Apply Logarithms

Take the natural logarithm of both sides to eliminate the exponential term: \( \ln(e^{2x+1}) = \ln(200) \). This simplifies to \( 2x + 1 = \ln(200) \) because \( \ln(e^y) = y \).

Step 4: Solve for x

After simplification, the equation becomes \( 2x + 1 = \ln(200) \). Solve for \( x \) by first subtracting 1 from both sides: \( 2x = \ln(200) - 1 \). Then, divide by 2 to isolate \( x \): \( x = \frac{\ln(200) - 1}{2} \).

Step 5: Calculate the Solution

Use a calculator to compute \( \ln(200) \). Then substitute back into the equation \( x = \frac{\ln(200) - 1}{2} \). The value of \( \ln(200) \) is approximately 5.2983. Thus, \( x = \frac{5.2983 - 1}{2} \).

Step 6: Approximate and Round the Result

Calculate \( x = \frac{4.2983}{2} \), which equals approximately 2.1492 when rounded to four decimal places.

Key Concepts

Natural Logarithms

Natural logarithms are a fundamental concept in mathematics used to solve exponential equations, especially when the base of the exponent is the number \(e\). The natural logarithm of a number \(x\), denoted as \(\ln(x)\), is the power to which \(e\) must be raised to obtain \(x\). In simpler terms, if \(y = \ln(x)\), then \(e^y = x\). This is particularly useful in simplifying equations involving \(e\).

When faced with an exponential equation like \(e^{2x+1} = 200\), we can take the natural logarithm of both sides to simplify the expression. Doing so allows us to use the property \(\ln(e^y) = y\), making the calculations much more manageable. This is a common technique in solving exponential equations, which we'll delve into next.

Solving Exponential Equations

Solving exponential equations often involves using logarithms to simplify and solve for the unknown variable. Let's break down the process using our example: \(e^{2x+1} = 200\).

• Isolate the exponential term: Since the equation is already set to 200, we proceed directly by taking the natural logarithm of both sides.
• Apply natural logarithms: This changes the equation to \(\ln(e^{2x+1}) = \ln(200)\). Using the property \(\ln(e^y) = y\), this simplifies to \(2x + 1 = \ln(200)\).
• Solve for \(x\): Rearrange the equation to solve for \(x\). Subtract 1 from both sides: \(2x = \ln(200) - 1\). Then divide by 2: \(x = \frac{\ln(200) - 1}{2}\).
• Compute the solution: Calculate \(\ln(200)\) using a calculator; it’s approximately 5.2983. Substituting back, we find \(x = \frac{5.2983 - 1}{2}\).

This method allows you to solve for the unknown variable precisely, setting the stage for the final steps of approximating and rounding numbers.

Rounding Numbers

Rounding numbers is a useful skill that helps make large numbers more manageable and easier to communicate. When solving equations, final results often require rounding to meet specified precision levels. For our equation \(x = \frac{4.2983}{2}\), calculating directly gives us a result of approximately 2.14915.

However, the problem requires the number to be rounded to four decimal places. To do this, follow these steps:

• Identify the fourth decimal place: In this instance, it is 9.
• Decide on rounding: Since the following digit (1) is less than 5, the 9 remains unchanged.

Thus, rounding 2.14915 to four decimal places results in 2.1492.

Rounding ensures results are presented neatly and according to standard conventions, which is especially important in scientific and engineering contexts.