Problem 5 - Thomas Calculus in SI Units

Question

Rewrite the expressions in terms of exponentials and simplify the results as much as you can. $$2 \cosh (\ln x)$$

Step-by-Step Solution

Answer

The expression simplifies to \( x + \frac{1}{x} \).

Step 1: Recall the Definition of Hyperbolic Cosine

The hyperbolic cosine function is defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \). We will use this definition to express \( \cosh(\ln x) \) in terms of exponentials.

Step 2: Substitute \( \ln x \) into the Hyperbolic Cosine

Replace \( x \) in the hyperbolic cosine definition with \( \ln x \):\[ \cosh(\ln x) = \frac{e^{\ln x} + e^{-\ln x}}{2} \]

Step 3: Simplify using the Properties of Logarithms

Using the property \( e^{\ln x} = x \), simplify each term:\[ e^{\ln x} = x, \quad e^{-\ln x} = \frac{1}{x} \]

Step 4: Simplification of the Expression

Substitute the simplified terms back into the expression:\[ \cosh(\ln x) = \frac{x + \frac{1}{x}}{2} \]

Step 5: Final Simplification

Multiply the expression from Step 2 by \(2\) to complete the original problem:\[ 2 \cosh(\ln x) = 2 \left( \frac{x + \frac{1}{x}}{2} \right) = x + \frac{1}{x} \]

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. These functions take the form \( f(x) = a^x \), where \( a \) is a positive constant. In the context of our problem, we often work with the natural exponential function \( e^x \), where \( e \) is Euler's number, approximately 2.718.

• Exponential functions exhibit rapid growth or decay; a small change in the exponent leads to a significant effect on the value of the function.
• They play a crucial role in modeling situations where growth is proportional to present value, such as in population growth, compound interest, or radioactive decay.
• Key properties include \( e^{x+y} = e^x \, e^y \) and \( e^{-x} = \frac{1}{e^x} \).

In hyperbolic functions, exponentials appear prominently. Take the hyperbolic cosine \( \cosh(x) = \frac{e^x + e^{-x}}{2} \), which showcases the addition of exponential terms.

Logarithmic Functions

Logarithmic functions serve as the inverse of exponential functions. For a given base \( a \), the logarithm \( \log_a(x) \) gives the power to which \( a \) must be raised to produce \( x \). The natural logarithm, denoted as \( \ln(x) \), uses the base \( e \).

• Logarithms transform products into sums: \( \ln(xy) = \ln(x) + \ln(y) \).
• They convert quotients into differences: \( \ln\left(\frac{x}{y}\right) = \ln(x) - \ln(y) \).
• Power rules simplify logarithmic manipulation: \( \ln(x^k) = k \ln(x) \).

Logarithmic functions simplify the solution of exponential equations. For instance, recalling \( e^{\ln x} = x \) allows us to express exponential and logarithmic forms interchangeably. This property was pivotal in simplifying \( \cosh(\ln x) \) in the exercise.

Simplification of Expressions

Simplification is a process of rewriting mathematical expressions in a more concise and easily understandable form. It often involves combining like terms, reducing fractions, and utilizing algebraic identities. The aim is to achieve a simpler, yet equivalent, expression.

• Identify and utilize relevant algebraic identities or properties, such as \( e^{\ln x} = x \).
• Combine fractions and like terms whenever possible to reduce complexity.
• Transform expressions into simpler forms, which makes further calculations more manageable.

In our exercise, simplifying \( 2 \cosh(\ln x) \) involves recognizing \( e^{\ln x} \) is simply \( x \), and \( e^{-\ln x} \) simplifies to \( \frac{1}{x} \).
This leads to the final simplification: \( 2 \left( \frac{x + \frac{1}{x}}{2} \right) = x + \frac{1}{x} \), showcasing the elegance of simplified, exact expressions.