Problem 5 - Precalculus Mathematics for Calculus

Question

Find the solution of the exponential equation, rounded to four decimal places. $$e^{-2 x}=7$$

Step-by-Step Solution

Answer

The solution is approximately \( x = -0.9730 \).

Step 1: Understand the Equation

We are given the equation \( e^{-2x} = 7 \). The exponential function here involves the variable \( x \) in the exponent. Our goal is to solve for \( x \).

Step 2: Take the Natural Logarithm

To solve for \( x \), first take the natural logarithm on both sides of the equation. This will help bring the exponent down: \( \ln(e^{-2x}) = \ln(7) \).

Step 3: Simplify Using Logarithm Laws

Using the property of logarithms \( \ln(e^a) = a \), the equation simplifies to \( -2x = \ln(7) \).

Step 4: Solve for x

To find \( x \), divide both sides by \(-2\): \( x = -\frac{\ln(7)}{2} \).

Step 5: Calculate the Value

Use a calculator to find \( \ln(7) \) and divide by \(-2\). \( \ln(7) \approx 1.945910 \), so \( x \approx -\frac{1.945910}{2} \approx -0.972955 \). Rounding to four decimal places, we have \( x \approx -0.9730 \).

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a special type of logarithm that uses the base of the mathematical constant \( e \), where \( e \approx 2.71828 \). It is a powerful tool for solving exponential equations because it 'unwraps' the exponent from the exponential function.
If you see an equation with an exponential part involving \( e \), like \( e^x \), taking the natural logarithm allows us to bring the exponent down and work with it more directly. For instance, in our equation \( e^{-2x} = 7 \), applying the natural logarithm to both sides makes it easier to isolate \( x \).

• It transforms the problem into a simpler algebraic form: \( \ln(e^{-2x}) = -2x \).

By taking the natural logarithm of both sides, we convert a problem that initially seemed complex into something more straightforward to solve.

Logarithm Laws

Logarithm laws are set rules that allow us to simplify and manipulate logarithms more efficiently.
Understanding some basic laws can help greatly in solving equations like the one we're dealing with. One crucial property used in solving our original problem is \( \ln(e^a) = a \). This law states that the natural logarithm of \( e \) raised to any power \( a \) equals \( a \) itself.

• In our equation \( \ln(e^{-2x}) = \ln(7) \), using the property \( \ln(e^a) = a \) enables us to simplify the left-hand side directly to \( -2x \).
• This simplification turns what might appear as a daunting equation into something linear and easy to solve.

Knowing and applying these logarithm laws helps break down problems, making them much more manageable and straightforward to work through.

Rounding Numbers

Rounding numbers is an essential skill in mathematics, especially when expressing solutions to a specific degree of accuracy. It involves adjusting the digits of a number to its nearest value depending on the specified decimal place.
For our solution, we are asked to round to four decimal places, which is important for precise and consistent results in calculations. Here's how you round a number to four decimal places:

• Identify the fourth digit after the decimal point. For instance, in \( -0.972955 \), it's 5.
• Look at the next digit. If it's 5 or higher, increase the fourth digit by one. If it's less than 5, keep the fourth digit the same.

Applying this rule, \( -0.972955 \) becomes \( -0.9730 \) after rounding. Understanding rounding helps in presenting a clean and precise final answer suitable in mathematical contexts.