/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 Evaluate the expression to four ... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the expression to four decimal places using a calculator. $$\ln \sqrt{2}$$

Short Answer

Expert verified
The value of \( \ln \sqrt{2} \) evaluated to four decimal places is 0.3466.

Step by step solution

01

Understand the properties of logarithms

The first step is to recognize that the square root can be expressed as a power of 1/2, so the given expression \(\ln \sqrt{2}\) can be rewritten as \( \ln {2^{1/2}} \).
02

Apply the Logarithmic Rule

According to the property of logarithms, \( \ln {a^b} = b \ln {a} \), so the expression from Step 1 can be simplified to \( \frac{1}{2} \ln {2} \).
03

Evaluate using a calculator

The final step is to plug the expression into the calculator. It's important to check the calculator is in the natural logarithm mode. Calculating \( \frac{1}{2} \ln {2} \), the result should be stated to four decimal places.

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Key Concepts

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

Natural Logarithm
The natural logarithm is a special logarithm with the base \( e \), where \( e \approx 2.71828 \). It is commonly denoted as \( \ln \). The natural logarithm is particularly important in mathematics because of its natural occurrence in calculus, involving growth and decay processes.

Natural logarithms have unique properties that make them useful in solving equations and transforming complex expressions. Essentially, \( \ln x \) tells you the power you need to raise \( e \) to get \( x \). For example, if \( \ln 2 \approx 0.6931 \), it means \( e^{0.6931} \approx 2 \).

Some applications of natural logarithms include exponential growth models such as population growth, radioactive decay, and more. Understanding natural logs can be crucial for anyone studying these kinds of applications or calculus.
Logarithmic Properties
Logarithmic properties are fundamental tools that help us simplify and manipulate expressions involving logarithms. One key property is the power rule, which states that \( \ln{a^b} = b \ln{a} \). This property is essential in evaluating logarithms with exponentiated arguments.

In the given exercise, \( \ln \sqrt{2} \) is transformed using the property that the square root of a number is the same as that number raised to the power of \( \frac{1}{2} \). So, \( \ln \sqrt{2} \) becomes \( \ln{2^{1/2}} \), which simplifies further to \( \frac{1}{2} \ln{2} \) using the power rule.

Other useful properties include:
  • Product Rule: \( \ln{(ab)} = \ln{a} + \ln{b} \)
  • Quotient Rule: \( \ln{(\frac{a}{b})} = \ln{a} - \ln{b} \)
  • Identity: \( \ln(1) = 0 \)
The usage of these properties is pivotal in simplifying expressions, solving equations, and integrating logarithmic functions.
Calculator Usage
Using a calculator efficiently is key to solving expressions involving natural logarithms. Most scientific calculators have a dedicated "ln" button which allows you to compute the natural logarithm of a number directly. Before starting your calculations, ensure that your calculator is in the correct mode for natural logarithms.

For the exercise at hand, you will need to first find \( \ln 2 \) using the calculator. Here's how you can do it:
  • Press the "ln" button.
  • Type the number 2.
  • Hit the "Enter" or "Equals" button to get the result.
Once you have determined \( \ln 2 \), multiply it by \( \frac{1}{2} \) to get \( \frac{1}{2} \ln 2 \). This step will give you the final expression, which is the answer to the original problem. Be sure to round the answer to four decimal places as instructed, for an accurate and precise result.

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Most popular questions from this chapter

The decibel (dB) is a unit that is used to express the relative loudness of two sounds. One application of decibels is the relative value of the output power of an amplifier with respect to the input power. since power levels can vary greatly in magnitude, the relative value \(D\) of power level \(P_{1}\) with respect to power level \(P_{2}\) is given (in units of \(\mathrm{dB}\) ) in terms of the logarithm of their ratio as follows: $$D=10 \log \frac{P_{1}}{P_{2}}$$ where the values of \(P_{1}\) and \(P_{2}\) are expressed in the same units, such as watts \((\mathrm{W}) .\) If \(P_{2}=75 \mathrm{W},\) find the value of \(P_{1}\) at which \(D=0.7\)

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Use the change-of-base formula to evaluate each logarithm using a calculator. Round answers to four decimal places. $$\log _{3} 1.25$$

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