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Chapter 5: Problem 36

Evaluate the given expression without using a calculator. $$e^{\ln (\ln 2)}$$

Short Answer

Expert verified
Answer: The simplified form of the given expression is \(\ln 2\).

Step by step solution

01

Recall Properties of Logarithms and Exponentials

Before we begin simplifying the expression, let's recall some properties of logarithms and exponentials: 1. The base of the natural logarithm, denoted as \(\ln\), is the Euler's number (\(e\)). 2. \(\ln(a^b) = b \cdot \ln(a)\). 3. \(e^{\ln(a)} = a\). Now that we have these properties in mind, let's move on to simplifying the given expression.
02

Simplify the Expression Inside the Exponential Function

Given the expression: $$e^{\ln (\ln 2)}$$ We know that \(e^{\ln(a)}=a\), which means that if we have an expression of the form \(e^{\ln(a)}\), we can just replace it with \(a\). In this case, our \(a\) is \((\ln 2)\), so the expression becomes: $$e^{\ln (\ln 2)} = \ln 2$$
03

Final Simplification

We have simplified the expression to: $$\ln 2$$ This is already in its simplest form, and since we're not using a calculator, we don't need to evaluate the natural logarithm any further. So, the final answer is: $$e^{\ln (\ln 2)} = \ln 2$$

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Key Concepts

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

Properties of Logarithms
Understanding the properties of logarithms is fundamental for simplifying exponential expressions. Logarithms are a way of expressing exponentiation in reverse. Here are a few key properties:
  • Power Rule: \( \ln(a^b) = b \cdot \ln(a) \). This means the logarithm of a power translates the exponent as a multiplier outside the logarithm.
  • Logarithms and Exponentials: If you have something in the form \( e^{\ln(a)} \), you can simplify it to \( a \). This property indicates that an exponential and a logarithm with the same base cancel each other out due to their inverse relationship.
These properties allow us to solve complex expressions by breaking them down into simpler terms that are easier to handle.
They provide a structured method for transforming and evaluating logarithmic and exponential forms.
Natural Logarithm
The natural logarithm, denoted as \( \ln x \), is a logarithm to the base \( e \), where \( e \) is approximately 2.71828. This special logarithm has many useful properties:
  • It is used to solve equations where the base is \( e \), simplifying the complexity involved with continuous growth or decay.
  • It is particularly beneficial in calculus because its derivative is 1, making differentiation straightforward.
Because \( e \) is the natural base, the natural logarithm appears frequently in growth models, economics, biology, and even in financial calculations.
It serves as a bridge to understanding the relationship between simple and complex growth processes.
Euler's Number
Euler's number, denoted as \( e \), is a mathematical constant that is the base of the natural logarithm. Its value is approximately 2.71828. This number is fundamental:
  • It appears in various contexts across mathematics, especially in calculus, where it describes exponential growth and decay.
  • \( e \) is known for the unique property that the function \( f(x) = e^x \) has the same rate of increase as its value at any given point; that is, its derivative \( f'(x) = e^x \) is \( e^x \).
  • It is an irrational number, meaning it cannot be expressed as a simple fraction, similar to \( \pi \).
Euler's number is essential in mathematical models that include continuous growth, finance for compounding interest, and even in the field of complex numbers when addressing equations involving growth rates.

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Most popular questions from this chapter

Deal with functions of the form \(f(x)=P e^{k x}\) where \(k\) is the continuous exponential growth rate (see Example 6 ). The present concentration of carbon dioxide in the atmosphere is 364 parts per million (ppm) and is increasing exponentially at a continuous yearly rate of \(.4 \%\) (that is, \(k=.004) .\) How many years will it take for the concentration to reach 500 ppm?

Between 1790 and \(1860,\) the population y of the United States (in millions) in year x was given by \(y=3.9572\left(1.0299^{\circ}\right),\) where \(x=0\) corresponds to \(1790 .\)F ind the U.S. population in the given year. $$1859$$

Approximating Logarithmic Functions by Polynomials. For each positive integer \(n,\) let \(f_{n}\) be the polynomial function whose rule is $$ f_{n}(x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\cdots \pm \frac{x^{n}}{n} $$ where the sign of the last term is \(+\) if \(n\) is odd and \(-\) if \(n\) is even. In the viewing window with \(-1 \leq x \leq 1\) and \(-4 \leq y \leq 1,\) graph \(g(x)=\ln (1+x)\) and \(f_{4}(x)\) on the same screen. For what values of \(x\) does \(f_{4}\) appear to be a good approximation of \(g ?\)

The beaver population near a certain lake in year \(t\) is approximately $$p(t)=\frac{2000}{1+199 e^{-.5544 t}}$$ (a) When will the beaver population reach \(1000 ?\) (b) Will the population ever reach \(2000 ?\) Why?

Determine whether an exponential, power, or logarithmic model (or none or several of these) is appropriate for the data by determining which (if any) of the following sets of points are approximately linear: $$\\{(x, \ln y)\\}, \quad\\{(\ln x, \ln y)\\}, \quad\\{(\ln x, y)\\}$$ where the given data set consists of the points \(\\{(x, y)\\}\) $$\begin{array}{|l|c|c|c|c|c|c|} \hline x & 1 & 3 & 5 & 7 & 9 & 11 \\ \hline y & 2 & 25 & 81 & 175 & 310 & 497 \\ \hline \end{array}$$
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