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Chapter 4: Problem 92

In Exercises 81鈥�100, evaluate or simplify each expression without using a calculator. $$ \ln \frac{1}{e^{7}} $$

Short Answer

Expert verified
-7

Step by step solution

01

Use the properties of logarithms

Apply the property of logarithm that \(\ln \frac{1}{x}\) equals \(-\ln x\). So, you rewrite the given expression as \(-\ln e^{7}\).
02

Apply the power rule of logarithms

The power rule/law of logarithms states that \(\ln a^{n}\) equals \(n \cdot \ln a\). Applying this law to the expression from step 1 results in a simplified expression: \( -7*\ln e\).
03

Value of \(\ln e\)

\(\ln e\) is equal to 1, because the natural logarithm of \(e\) (Euler's number, approximately equal to 2.71828) is 1. This implies that 鈭�7 multiplied by 1 equals 鈭�7.

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Key Concepts

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

Logarithm Properties
Understanding logarithm properties is crucial in simplifying expressions like \( \ln \frac{1}{e^7} \). One essential property is that \( \ln \frac{1}{x} \) equals \( -\ln x \). This means if you take the natural logarithm of a fraction, you can simply take the negative logarithm of the denominator.
  • This property helps you handle fractions within logarithms.
  • It's a fundamental tool for rewriting more complex logarithmic expressions.
In the exercise, we used this property to change \( \ln \frac{1}{e^7} \) into \( -\ln e^7 \). This is the starting point for simplifying the expression further.
Power Rule of Logarithms
The power rule of logarithms states that \( \ln a^n \) is equal to \( n \cdot \ln a \). This rule makes it easy to deal with logarithms of exponential expressions by pulling the exponent out in front of the logarithm.
  • If you're working with \( \ln e^7 \), the power rule lets you rewrite this as \( 7 \cdot \ln e \).
  • This simplification reduces the complexity of the expression.
In our step-by-step solution, we applied this rule after using the properties of logarithms to reach the expression \( -7 \cdot \ln e \). This step is vital for making the final calculation straightforward.
Euler's Number
Euler's number, denoted as \( e \), is approximately 2.71828 and holds a special place in mathematics. It is the base of the natural logarithms, making \( \ln e = 1 \). This means that the natural logarithm of \( e \) simplifies to 1 effortlessly.
  • Knowing that \( \ln e = 1 \) is key for calculations involving natural logs.
  • It simplifies expressions by reducing terms that multiply with \( \ln e \).
In the exercise, since \( \ln e \) equals 1, the expression \( -7 \cdot \ln e \) simplifies to \( -7 \), showing the utility and simplicity provided by Euler's number.
Logarithm Simplification
Logarithm simplification combines multiple rules and properties to reduce a complex expression to its simplest form.
  • Start by identifying all applicable properties and rules, such as those for fractions and exponents.
  • Utilize the power rule and properties of \( e \) to reduce terms systematically.
Applying these methods, the given expression \( \ln \frac{1}{e^7} \) simplifies step-by-step:1. Rewrite using fractions property: \( -\ln e^7 \).2. Apply power rule: \( -7 \cdot \ln e \).3. Use the value of \( \ln e \): result is \( -7 \).Each step builds upon the last, showing how fundamental understanding leads to a neat and simple solution.

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Most popular questions from this chapter

Explain how to use your calculator to find \(\log _{14} 283\)

In Example I on page \(520,\) we used two data points and an exponential function to model the population of the United States from 1970 through 2010 . The data are shown again in the table. Use all five data points to solve Exercises \(66-70\). $$ \begin{array}{cc} {x, \text { Number of Years }} & {y, \text { U.S. Population }} \\ {\text { after } 1969} & {\text { (millions) }} \\ {1(1970)} & {203.3} \\ {11(1980)} & {226.5} \\ {21(1990)} & {248.7} \\ {31(2000)} & {281.4} \\ {41(2010)} & {308.7} \end{array} $$ Use your graphing utility's linear regression option to obtain a model of the form \(y=a x+b\) that fits the data. How well does the correlation coefficient, \(r,\) indicate that the model fits the data?

Write an equation in point-slope form and slope-intercept form of the line passing through \((1,-4)\) and parallel to the line whose equation is \(3 x-y+5=0 .\)

Three of the richest comedians in the United States are Larry David (creator of Seinfeld), Matt Groening (creator of The Simpsons), and Trey Parker (co- creator of South Park). Larry David is worth \(\$ 450\) million more than Trey Parker. Matt Groening is worth \(\$ 150\) million more than Trey Parker. Combined, the net worth of these three comedians is \(\$ 1650\) million (or \(\$ 16.5\) billion). Determine how much, in millions of dollars, each of these comedians is worth. (Source: petamovies.com) (Section 1.3, Example 1)

In Example I on page \(520,\) we used two data points and an exponential function to model the population of the United States from 1970 through 2010 . The data are shown again in the table. Use all five data points to solve Exercises \(66-70\). $$ \begin{array}{cc} {x, \text { Number of Years }} & {y, \text { U.S. Population }} \\ {\text { after } 1969} & {\text { (millions) }} \\ {1(1970)} & {203.3} \\ {11(1980)} & {226.5} \\ {21(1990)} & {248.7} \\ {31(2000)} & {281.4} \\ {41(2010)} & {308.7} \end{array} $$ Use the values of \(r\) in Exercises \(66-69\) to select the two model= of best fit. Use each of these models to predict by which yeathe U.S. population will reach 335 million. How do these answers compare to the year we found in Example \(1,\) namel \(=\) \(2020 ?\) If you obtained different years, how do you account fo this difference?
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