Problem 68 - Essential Calculus Early Transcendentals

Question

Solve each inequality for \(x .\) $$ \text { (a) }1

Step-by-Step Solution

Answer

(a) \( \frac{1}{3} < x < \frac{\ln(2) + 1}{3} \), (b) \( x > \frac{1}{e} \)

Step 1: Solving First Inequality Part (a)

Start by isolating the middle term in the double inequality. Given: \ \( 1 < e^{3x-1} < 2 \) \ Convert the double inequality into two separate inequalities: \ 1. \( 1 < e^{3x-1} \) \ 2. \( e^{3x-1} < 2 \)

Step 2: Solving First Part of Inequality (a)

For \(1 < e^{3x-1}\), apply the logarithm to both sides to solve for \(x\). \ \( \ln(1) < 3x-1 \) or \( 0 < 3x-1 \) \ Solve: \ \( 3x > 1 \) \ \( x > \frac{1}{3} \)

Step 3: Solving Second Part of Inequality (a)

For \( e^{3x-1} < 2 \), apply the logarithm to both sides to solve for \( x \). \ \( 3x-1 < \ln(2) \) \ Solve: \ \( 3x < \ln(2) + 1 \) \ \( x < \frac{\ln(2) + 1}{3} \)

Step 4: Combining Results for Inequality (a)

Combine the results from Step 2 and Step 3 to find the solution for the inequality: \ \( \frac{1}{3} < x < \frac{\ln(2) + 1}{3} \)

Step 5: Solving the Inequality Part (b)

Rewrite the inequality: \ \( 1 - 2 \ln x < 3 \) \ Subtract 1 from both sides: \ \(-2 \ln x < 2 \) \ Divide the entire inequality by -2 (remember to reverse inequality sign): \ \( \ln x > -1 \)

Step 6: Solving for x in Part (b)

To solve \( \ln x > -1 \), exponentiate both sides: \ \( x > e^{-1} \) \ The solution for x is \( x > \frac{1}{e} \).

Key Concepts

Logarithmic Functions

Logarithmic functions are crucial in solving inequalities that contain exponential expressions. A logarithm is the inverse operation of exponentiation, meaning it helps to undo the exponential function. If you have an equation with an exponential function, applying \ \( \ln \ \) (natural logarithm) can help isolate the variable.

• Example: For the inequality \ \( 1 < e^{3x-1} \ \), taking the natural logarithm of both sides simplifies it. This is because \ \( \ln(e^{3x-1}) = 3x-1 \ \).
• Remembering that \ \( \ln(1) = 0 \ \) is crucial, as it helps to determine that \ \( 3x-1 > 0 \ \).

Logarithmic functions provide a powerful tool for transforming exponential inequalities into linear inequalities. By converting these equations, we make it easier to apply algebraic techniques to solve for the unknown variable, as seen in \ \( \ln x > -1 \ \).

Exponential Functions

Exponential functions, represented by expressions like \ \( e^{3x-1} \ \), are foundational in calculus and many real-world applications. They exhibit growth patterns, such as population growth or radioactive decay, making them essential in mathematical modeling.

• In inequalities, understanding how these functions behave is critical. For example, knowing that \ \( e^{3x-1} \ \) can be split into \ \( e^a \ \) and \ \( e^b \ \) aids in separating the inequality into manageable parts.
• Recognizing that exponentiating a positive number results in a positive number is important when solving inequalities, such as \ \( e^{3x-1} < 2 \ \).

The natural exponent \ \( e \ \), approximately equal to 2.718, is the base of natural logarithms and is commonly used in calculus due to its unique mathematical properties.

Mathematical Reasoning

Mathematical reasoning involves logically piecing together information to solve problems step by step. It requires using known properties and rules to systematically work through an exercise.

• Breaking down inequalities into simpler parts helps, as demonstrated when converting \ \( 1 < e^{3x-1} < 2 \ \) into two separate inequalities.
• Knowing when to reverse the inequality sign is crucial. For instance, dividing by a negative number, as seen in the inequality \ \( -2 \ln x < 2 \ \), requires reversing the comparison sign.

Solid mathematical reasoning aids in identifying errors, confirming solutions, and explaining steps effectively. Having a systematic approach to inequalities ensures students can follow steps logically to arrive at correct answers.

Calculus Problem Solving

Calculus problem-solving allows us to delve deeper into inequality solutions through the use of exponentials and logarithms. Calculus provides techniques for handling complex expressions and can be applied in inequality problems even if differentiation or integration isn't explicitly required.

• Understanding the behavior of exponential and logarithmic functions through their derivatives helps predict how changes in the variable affect the inequality.
• Concepts from calculus reinforce logical approaches needed to deal with inequalities, especially when considering ranges and domains for solutions.

Applying calculus-based thinking aids in visualizing the problem and using derivatives or integrals as future steps for more complex situations. This approach prepares students for more advanced applications beyond basic inequality solving.