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In Exercises \(11-16,\) let \(u=\ln x\) and \(v=\ln y .\) Write the given expression in terms of u and v. For example, $$\ln x^{3} y=\ln x^{3}+\ln y=3 \ln x+\ln y=3 u+v$$ $$\ln \left(x^{2} y^{5}\right)$$

Short Answer

Expert verified
Question: Express the given expression ln(x^2 y^5) in terms of u = ln(x) and v = ln(y) using the natural logarithm properties. Answer: ln(x^2 y^5) can be written in terms of u and v as 2 * u + 5 * v.

Step by step solution

01

Applying the first logarithm property

Apply the first logarithm property ln(a * b) = ln(a) + ln(b) to the given expression ln(x^2 * y^5). So, we have ln(x^2 y^5) = ln(x^2) + ln(y^5)
02

Applying the second logarithm property

Apply the second logarithm property ln(a^k) = k * ln(a) to both terms in the result from Step 1. So, we get 2 * ln(x) + 5 * ln(y)
03

Substituting u and v

Substitute u = ln(x) and v = ln(y) in the expression obtained in Step 2. So, we have 2 * u + 5 * v The given expression ln(x^2 y^5) can be written in terms of u and v as: 2 * u + 5 * v

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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 fundamental mathematical function, denoted as \( \ln(x) \), and it refers to the logarithm to the base \( e \), where \( e \) is the Euler's constant, approximately equal to 2.71828. In simple terms, if \( \ln(x) = y \), then we can understand it as \( e^y = x \). This special logarithm is widely used because of its natural properties in growth and decay processes, such as interest in finance or populations in biology.

Students sometimes struggle with understanding why we use \( e \), but it's because its rate of growth is considered to be the natural rate of increase. For every unit increase in value, you can think of e as the factor by which a quantity multiplies itself. The natural logarithm, therefore, helps us find the time or the rate at which these changes happen.

The ease of differentiating and integrating natural logarithms makes them especially useful in calculus. In the context of the textbook exercise, understanding the natural logarithm is key to operating with logarithmic expressions and employing algebraic manipulation.
Logarithmic Expressions
Working with logarithmic expressions often requires understanding their unique properties and rules. These are tools that help simplify complex logarithms into more manageable parts, or sometimes even single variables. For instance, the first property used in the exercise, \( \ln(a * b) = \ln(a) + \ln(b) \), shows how a product within a logarithm can be broken down into a sum of logarithms.

Key Properties of Logarithms

  • Product Property: \( \ln(a * b) = \ln(a) + \ln(b) \)
  • Quotient Property: \( \ln(\frac{a}{b}) = \ln(a) - \ln(b) \)
  • Power Property: \( \ln(a^k) = k * \ln(a) \)
The exercise demonstrates the use of the power property to separate the exponent from the term inside the logarithm. It's also important to note the base of the logs being used. For the natural logarithm, the base is \( e \), but these properties hold true for logs of any base. When students grasp these properties, they can manipulate logarithmic expressions to simplify or solve them.

Using these properties strategically can transform a dense logarithmic problem into a straightforward algebraic one.
Algebraic Manipulation
Performing algebraic manipulation involves rearranging, simplifying, or transforming algebraic expressions using a variety of techniques and properties of algebra. These techniques can include basic operations such as addition, subtraction, multiplication, and division; distributing and factoring; and using properties of exponents and logarithms.

Applying Algebra to Logarithms

Our exercise is a prime example of using algebraic manipulation to simplify logarithmic expressions. After applying logarithm properties, we use substitution—replacing \( \ln(x) \) with \( u \) and \( \ln(y) \) with \( v \)—to turn a complex expression into a much simpler linear combination of variables: \( 2u + 5v \).

This technique is particularly useful in higher-level mathematics, where simplifying an expression can make problems more tractable. In algebraic manipulation, the goal is often to reduce the problem to a form that is easier to interpret or solve, as shown in the step-by-step solution. Remember, the more practice you get with these manipulations, the better you'll understand how to apply them across a variety of mathematical problems.

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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 ). Between 1996 and \(2004,\) the number of United States subscribers to cell-phone plans has grown nearly exponentially. In 1996 there were 44,043,000 subscribers and in 2004 there were \(182,140,000^{\dagger}\) (a) What is the continuous growth rate of the number of cell-phone subscribers? (b) In what year were there 60,000,000 cell-phone subscribers? (c) Assuming that this rate continuous, in what year will there be 350,000,000 subscribers? (d) In 2007 the United States population was approximately 300 million. Is your answer to part (c) realistic? If not, what could have gone wrong?

Use the fact that \(x^{3}+y^{3}=\) \((x+y)\left(x^{2}-x y+y^{2}\right)\) to rationalize the denominator. $$\frac{1}{\sqrt[3]{3}+1}$$

The number of children who were home schooled in the United States in selected years is shown in the table. (a) Sketch a scatter plot of the data, with \(x=0\) corresponding to 1980 (b) Find a quadratic model for the data. (c) Find a logistic model for the data. (d) What is the number of home-schooled children predicted by each model for the year \(2015 ?\) (e) What are the limitations of each model? $$\begin{array}{|c|c|} \hline \begin{array}{c} \text { Fall of } \\ \text { School Year } \end{array} & \begin{array}{c} \text { Number of } \\ \text { Children (in thousands) } \end{array} \\ \hline 1985 & 183 \\ \hline 1988 & 225 \\ \hline 1990 & 301 \\ \hline 1992 & 470 \\ \hline 1993 & 588 \\ \hline 1994 & 735 \\ \hline 1995 & 800 \\ \hline 1996 & 920 \\ \hline 1997 & 1100 \\ \hline 1999 & 1400 \\ \hline 2000 & 1700 \\ \hline 2005 & 1900 \\ \hline \end{array}$$

Assume that you watched 1000 hours of television this year, and will watch 750 hours next year, and will continue to watch \(75 \%\) as much every year thereafter. (a) In what year will you be down to ten hours per year? (b) In what year would you be down to one hour per year?

Rationalize the denominator and simplify your answer. $$\frac{2}{\sqrt{6}}$$

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