Problem 61 - 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. $$500 e^{-x}=300$$

Step-by-Step Solution

Answer

The solution to the exponential equation \(500e^{-x} = 300\) rounded to three decimal places is \(x = -\ln(0.6)\). Compute this value using a calculator to get the numerical answer. Verify by graphing the two functions \(y=500e^{-x}\) and \(y=300\) and locating the intersection.

Step 1: Identify the Given Equation

We are given the exponential equation \(500e^{-x} = 300\). We need to solve this equation for \(x\).

Step 2: Rewrite the Equation

Divide the whole equation by 500, this changes the equation to \(e^{-x} = \frac{300}{500}\). Which simplifies to \(e^{-x} = 0.6\).

Step 3: Take Natural Logarithm

Next, apply the natural logarithm (also known as ln in most calculators) on both sides. It would result in \(-x = \ln(0.6)\).

Step 4: Solve for x

Solving for \(x\) involves multiplying the equation by -1, after which the equation becomes, \(x = -\ln(0.6)\). Now, this expression can be calculated using any scientific calculator and round the result to three decimal places. Note that the natural logarithm function of 0.6 will yield a negative number, but we are multiplying by -1 making \(x\) positive.

Step 5: Verification with a Graphing Utility

We can verify the solution graphically. Plot the equation \(y=500e^{-x}\) and \(y=300\). The \(x\) value at the point of intersection is our solution.

Key Concepts

Understanding Natural Logarithms

Natural logarithms are a fundamental part of solving exponential equations. They are a specific type of logarithm that use the constant \( e \) (approximately 2.718) as the base. Since \( e \) frequently appears in exponential growth processes, the natural logarithm is widely used in various fields such as mathematics, physics, and engineering.
When we encounter an exponential equation like \( e^x \), taking the natural logarithm (abbreviated as \( \ln \)) of both sides helps isolate the variable. This is because of the inverse relationship between exponentials and logarithms. It's key to remember that \( \ln(e^x) = x \).
For example, when we solved \( 500e^{-x} = 300 \), we simplified it to \( e^{-x} = 0.6 \). When we applied the natural logarithm, it became \(-x = \ln(0.6)\), which further solved to \( x = -\ln(0.6) \) after multiplying by -1.

• Natural logarithms are denoted as \( \ln \).
• The base of natural logarithms is the constant \( e \).
• They simplify solving for variables within exponential equations.

Using Graphing Calculators for Verification

Graphing calculators provide a visual approach to understanding the behavior of equations like \( 500e^{-x} = 300 \). These tools can be extremely helpful in verifying algebraically solved equations by plotting functions and finding intersections.
To verify our earlier solution, we plot two functions: \( y = 500e^{-x} \) and \( y = 300 \). The graphing calculator will display these graphs and the point where they intersect represents the solution to our equation. The \( x \) coordinate of this intersection should match the \( x \) found through algebraic methods.

• Graphing calculators allow us to verify solutions visually.
• They help understand function behaviors and intersections.
• The point of intersection of plots confirms the algebraic solution.

Using a calculator for verification is a practical step that helps reinforce the confidence in the computed results.

Solving Exponential Equations Algebraically

Solving exponential equations typically involves isolating the exponential term and transforming the equation using logarithms. This process requires understanding both the properties of exponents and logarithms.
In the example of solving \( 500e^{-x} = 300 \):

• First, we divided both sides by 500 to isolate the exponential term, resulting in \( e^{-x} = 0.6 \).
• Then, applying the natural logarithm to both sides helps us remove the exponential, resulting in \(-x = \ln(0.6)\).
• Multiplying by -1 gives us \( x = -\ln(0.6) \).

It is important to simplify the expression step by step to avoid errors, especially when preparing for precise results like rounding to three decimal places. This discipline is fundamental to solving equations correctly and efficiently.