Problem 19 - Precalculus

Question

Solve the exponential equations. Make sure to isolate the base to a power first. Round our answers to three decimal places. $$9-2 e^{0.1 x}=1$$

Step-by-Step Solution

Answer

The solution is approximately \( x = 13.863 \).

Step 1: Isolate the exponential expression

Start with the equation: \[ 9 - 2e^{0.1x} = 1 \]To isolate the exponential expression, first subtract 9 from both sides:\[ -2e^{0.1x} = 1 - 9 \]This simplifies to:\[ -2e^{0.1x} = -8 \]

Step 2: Solve for the exponential term

To make the exponential term positive, divide both sides by -2:\[ e^{0.1x} = \frac{-8}{-2} \]This simplifies to:\[ e^{0.1x} = 4 \]

Step 3: Take the natural logarithm

To solve for \( x \), take the natural logarithm of both sides:\[ \ln(e^{0.1x}) = \ln(4) \]Using the property of logarithms \( \ln(e^y) = y \), this becomes:\[ 0.1x = \ln(4) \]

Step 4: Solve for x

To isolate \( x \), divide both sides by 0.1:\[ x = \frac{\ln(4)}{0.1} \]Calculate \( \ln(4) \) (approximately 1.386294) and divide by 0.1:\[ x \approx \frac{1.386294}{0.1} = 13.863 \]

Step 5: Final Step: Round the answer

Round \( x \approx 13.863 \) to three decimal places, which is already in this form. Therefore, the approximate solution is \( x = 13.863 \).

Key Concepts

Natural Logarithm

The natural logarithm, often denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. Natural logarithms are particularly useful in solving equations involving exponential growth and decay, as they allow for easy manipulation of powers of \( e \). For example, in our equation \( e^{0.1x} = 4 \), taking the natural logarithm of both sides \( \ln(e^{0.1x}) = \ln(4) \) is an effective way to simplify the process.

• It transforms an exponential equation into a linear one.
• The property \( \ln(e^y) = y \) simplifies equations where \( e \) is the base.

This handy property lets us write \( 0.1x = \ln(4) \), making it simpler to solve for \( x \). Understanding how to use natural logarithms effectively can simplify complex exponential problems.

Isolating Variables

Isolating variables involves strategically rearranging equations to express the variable of interest as a single term on one side of the equation. This is a crucial step in solving equations, particularly those involving exponential expressions.
When working with an equation like \( 9 - 2e^{0.1x} = 1 \), your goal is to isolate the \( e^{0.1x} \) term. By subtracting 9 from both sides, you get \( -2e^{0.1x} = -8 \). To make the exponential term positive, divide both sides by -2, simplifying to \( e^{0.1x} = 4 \).

• Use basic algebraic manipulations: addition, subtraction, multiplication, and division.
• Repeat the process until the variable is isolated.

Once the variable or expression is isolated, you can proceed with operations such as taking the logarithm, making the path to the solution clearer.

Rounding Numbers

Rounding numbers is the process of reducing the digits in a number while keeping its value close to the original number. This is especially important in mathematical calculations where exact precision is not necessary or when the numbers are complex decimals.
In our solution, after finding \( x \approx 13.863 \), we round it to maintain simplicity and readability. Rounding to three decimal places means keeping three digits after the decimal point, making the result \( x = 13.863 \) neat without significant loss of accuracy.

• Rounding helps in making numbers easier to work with.
• It simplifies communication of results.
• Maintains adequate precision for most practical purposes.

Whenever rounding, consider the context in which the number will be used, as varying scenarios may require different levels of precision.