Problem 29 - Precalculus : Building Concepts and Connections

Question

Solve the exponential equation. Round to three decimal places, when needed. $$9-e^{x^{2}-1}=2$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 2.807\) and \(x = -2.807\), rounded to three decimal places.

Step 1: Isolate the Exponential Function

Rearrange the given equation \(9-e^{x^{2}-1}=2\) to isolate \(e^{x^{2}-1}\) on one side of the equation. Subtract 2 from both sides: \(9-2 = e^{x^{2}-1} \Rightarrow e^{x^{2}-1}=7\)

Step 2: Take Natural Logarithm on Both Sides

To get rid of the exponential function and to make \(x\) manageable, take the natural logarithm on both sides of the equation: \(\ln(e^{x^{2}-1}) = \ln(7)\)

Step 3: Simplify using Properties of Logarithms

Using the property that \(\ln(e^{a}) = a\), simplify the left side of the equation. Doing so, we get \(x^{2}-1 = \ln(7)\)

Step 4: Solve for \(x\)

Rearrange the above equation to solve for \(x\). Add 1 to both sides, and then take the square root: \(x^{2} = \ln(7) + 1\), then \(x = \pm\sqrt{\ln(7)+1}\).

Step 5: Round To Three Decimal Places

After solving for \(x\), round the values to three decimal places as required by the question. Also note that we have used \(\pm\) while taking the square root, hence, there will be two solutions for \(x\): \(x = \sqrt{\ln(7)+1}\) and \(x = -\sqrt{\ln(7)+1}\). This gives \(x \approx \pm 2.807\) (rounded to three decimal places)

Key Concepts

Natural Logarithm

The natural logarithm is denoted as \( \ln \) and has a base of \( e \), which is an irrational and transcendental number approximately equal to 2.71828. It is particularly useful in solving exponential equations where the exponential term has \( e \) as the base.

In our exercise, after isolating the term \( e^{x^{2}-1} \), taking the natural logarithm on both sides becomes a key step. This is because the natural logarithm function, \( \ln \), is the inverse of the exponential function \( e^{x} \).

By applying \( \ln \) to \( e^{x^{2}-1} = 7 \), we get \( \ln(e^{x^{2}-1}) = \ln(7) \). This effectively "cancels out" the \( e \) on the left side, leaving us with \( x^{2} - 1 \), and thus simplifying the problem significantly.

Properties of Logarithms

Logarithms come with various properties that simplify complex calculations. One of the most beneficial properties is the power rule: \( \ln(e^{a}) = a \).

This property is pivotal in our step-by-step solution where \( \ln(e^{x^{2}-1}) = \ln(7) \). It simplifies to \( x^{2} - 1 = \ln(7) \), allowing us to easily handle the expression involving \( x \).

• **Product Rule**: \( \ln(AB) = \ln(A) + \ln(B) \)
• **Quotient Rule**: \( \ln\left(\frac{A}{B}\right) = \ln(A) - \ln(B) \)
• **Power Rule**: \( \ln(A^{b}) = b \cdot \ln(A) \)

These rules are powerful tools in algebra and calculus. They are important for transforming and simplifying equations that involve logarithms.

Rounding Numbers

Rounding numbers is a mathematical method used to simplify figures to make them easier to work with. It is especially important in ensuring that results reflect the precision required by a particular problem.

When rounding to three decimal places, we follow these general steps:

• Identify the digit at the third decimal place.
• Look at the digit immediately to the right; if it is 5 or greater, increase the third decimal place by 1.
• If it is less than 5, keep the third decimal place unchanged.

In our exercise, after determining the solutions for \( x \), we rounded the values to get \( x \approx \pm 2.807 \), providing clarity while adhering to the problem's instructions.