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Chapter 7: Problem 5

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

Short Answer

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

Step by step solution

01

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.
02

Substitute \( \ln x \) into the Hyperbolic Cosine

Replace \( x \) in the hyperbolic cosine definition with \( \ln x \):\[ \cosh(\ln x) = \frac{e^{\ln x} + e^{-\ln x}}{2} \]
03

Simplify using the Properties of Logarithms

Using the property \( e^{\ln x} = x \), simplify each term:\[ e^{\ln x} = x, \quad e^{-\ln x} = \frac{1}{x} \]
04

Simplification of the Expression

Substitute the simplified terms back into the expression:\[ \cosh(\ln x) = \frac{x + \frac{1}{x}}{2} \]
05

Final Simplification

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Only one of these calculations is correct. Which one? Why are the others wrong? Give reasons for your answers. a. \(\lim _{x \rightarrow 0^{+}} x \ln x=0 \cdot(-\infty)=0\) b. \(\lim _{x \rightarrow 0^{+}} x \ln x=0 \cdot(-\infty)=-\infty\) c. \(\lim _{x \rightarrow 0^{+}} x \ln x=\lim _{x \rightarrow 0^{+}} \frac{\ln x}{(1 / x)}=\frac{-\infty}{\infty}=-1\) d. \(\lim _{x \rightarrow 0^{+}} x \ln x=\lim _{x \rightarrow 0^{+}} \frac{\ln x}{(1 / x)}\) \(=\lim _{x \rightarrow 0^{+}} \frac{(1 / x)}{\left(-1 / x^{2}\right)}=\lim _{x \rightarrow 0^{+}}(-x)=0\)

Find the length of each curve. $$y=\ln \left(e^{x}-1\right)-\ln \left(e^{x}+1\right) \text { from } x=\ln 2 \text { to } x=\ln 3$$

Arc length Find the length of the graph of \(y=(1 / 2) \cosh 2 x\) from \(x=0\) to \(x=\ln \sqrt{5}\).

Which one is correct, and which one is wrong? Give reasons for your answers. a. \(\lim _{x \rightarrow 3} \frac{x-3}{x^{2}-3}=\lim _{x \rightarrow 3} \frac{1}{2 x}=\frac{1}{6}\) b. \(\lim _{x \rightarrow 3} \frac{x-3}{x^{2}-3}=\frac{0}{6}=0\)

Show that $$\lim _{k \rightarrow \infty}\left(1+\frac{r}{k}\right)^{k}=e^{r}$$
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