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Chapter 5: Problem 40

Suppose that \(\ln x=t\) and \(\ln y=u\) Write each expression in terms of t and \(u\) (a) \(\ln \left(e^{\ln x}\right)\) (b) \(e^{\ln (\ln x y)}\) (c) \(\ln \left(\frac{e x}{y}\right)-\ln \left(\frac{y}{e x}\right)\) (d) \(\frac{(\ln x)^{3}-\ln \left(x^{4}\right)}{\left(\ln \frac{x}{e^{2}}\right) \ln \left(x e^{2}\right)}\)

Short Answer

Expert verified
(a) t; (b) t + u; (c) 2(t - u); (d) \(\frac{t^3 - 4t}{t^2 - 4}\)

Step by step solution

01

Simplify Expression (a)

Given the expression \( \ln \left(e^{\ln x}\right) \), we use the property that \( \ln(e^a) = a \). Applying this, we get \( \ln(e^{\ln x}) = \ln x \). Since \( \ln x = t \), the expression simplifies to \( t \).
02

Simplify Expression (b)

Consider the expression \( e^{\ln (\ln x y)} \). Using the identity \( e^{\ln a} = a \), the expression becomes \( \ln xy \). Utilize the logarithm property \( \ln(ab) = \ln a + \ln b \) to get \( \ln x + \ln y \). Thus, the expression simplifies to \( t + u \).
03

Simplify Expression (c)

The expression \( \ln \left(\frac{e x}{y}\right)-\ln \left(\frac{y}{e x}\right) \) can be simplified using the logarithmic property \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \). First, for \( \ln \left(\frac{e x}{y}\right) \), we have \( \ln(e x) - \ln y = (1 + \ln x) - \ln y \). Similarly, \( \ln \left(\frac{y}{e x}\right) = \ln y - \ln (e x) = \ln y - (1 + \ln x) \). Subtract these, we get \[(1 + \ln x - \ln y) - (\ln y - 1 - \ln x) = 2 + 2\ln x - 2\ln y\]. This simplifies to \( 2(t - u) \).
04

Simplify Expression (d)

Start with the expression \( \frac{(\ln x)^{3}-\ln(x^4)}{\left(\ln\frac{x}{e^2}\right)\ln(x e^{2})} \). The numerator simplifies as follows: \((\ln x)^3 - \ln(x^4) = t^3 - 4t \). For the denominator, use properties to get \( \ln\frac{x}{e^2} = \ln x - \ln e^2 = t - 2 \) and \( \ln(x e^2) = \ln x + \ln e^2 = t + 2 \). Thus the denominator becomes \((t - 2)(t + 2) = t^2 - 4 \). The whole expression is then \( \frac{t^3 - 4t}{t^2 - 4} \).

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Key Concepts

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

Properties of Logarithms
In algebra, logarithms can seem daunting at first glance, but they are simply the inverse operation of exponentiation. By mastering the **Properties of Logarithms**, we unlock the power to simplify and manipulate logarithmic expressions. There are several key properties that frequently come into play:
  • Product Property: \( \ln(ab) = \ln a + \ln b \). This property implies that the logarithm of a product is the sum of the logarithms of its factors. It appears handy when separating the components of a multiplied term.

  • Quotient Property: \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \). With this rule, the logarithm of a quotient is the difference between the logarithm of the numerator and the denominator.

  • Power Property: \( \ln(a^n) = n\ln a \). This property shows us that the logarithm of a power is simply the exponent multiplied by the logarithm of the base.

In the given exercise, these properties prove instrumental in simplifying complex logarithmic expressions by reducing them into basic forms. They transform cumbersome calculations into more manageable steps, as demonstrated in the problem-solving process.
Exponential Functions
Exponential functions feature prominently in mathematics due to their unique properties and applications. Essentially, an exponential function involves a constant base raised to a variable exponent, expressed generally as \( f(x) = a^x \). A special note is the natural exponential function \( e^x \), where \( e \) is approximately 2.71828. One crucial concept regarding exponential functions is their relationship with logarithms, essentially serving as inverse operations. This relationship can be seen in the identity \( e^{\ln a} = a \), which was used in the exercise to simplify terms containing the exponential function exponentially raised to a natural logarithm.Exponential functions are famously involved in growth and decay processes, such as population growth, radioactive decay, and compound interest calculations. They capture how quantities expand or decrease rapidly over time, making them ideal for modeling a variety of real-world scenarios. Understanding how they interact with logarithmic expressions allows us to navigate and simplify real-life issues modeled by these functions.
Logarithmic Identities
When dealing with logarithms, certain identities help simplify expressions and solve equations efficiently. **Logarithmic Identities** are symbolic equations involving logarithms and serve as fundamental tools in logarithmic calculus. A key identity worth highlighting is the core relationship between logarithms and exponentials: \( \ln(e^x) = x \). The natural logarithm and the exponential function cancel each other out, emphasizing their role as inverse functions.Another identity frequently used, especially in the exercise, is the logarithm of one: \( \ln 1 = 0 \). Since any number to the power of zero is one, the natural log of one is always zero. Also, it鈥檚 important to remember the identity involving bases: switching between logarithms of different bases using the change of base formula, although not directly used in the exercise, is an important tool: \( \log_b a = \frac{\ln a}{\ln b} \).By understanding these identities, one can efficiently break down and simplify logarithmic problems. They are particularly useful when combined with properties of logarithms and provide an intimate understanding of the structures underpinning the relationship between logarithms and exponentials.

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Most popular questions from this chapter

The age of some rocks can be estimated by measuring the ratio of the amounts of certain chemical elements within the rock. The method known as the rubidium-strontium method will be discussed here. This method has been used in dating the moon rocks brought back on the Apollo missions. Rubidium-87 is a radioactive substance with a half-life of \(4.7 \times 10^{10}\) years. Rubidium- 87 decays into the substance strontium- \(87,\) which is stable (nonradioactive). We are going to derive the following formula for the age of a rock: $$T=\frac{\ln \left[\left(\mathcal{N}_{s} / \mathcal{N}_{r}\right)+1\right]}{-k}$$ where \(T\) is the age of the rock, \(k\) is the decay constant for rubidium-87, \(\mathcal{N}_{s}\) is the number of atoms of strontium-87 now present in the rock, and \(\mathcal{N},\) is the number of atoms of rubidium-87 now present in the rock. (a) Assume that initially, when the rock was formed, there were \(\mathcal{N}_{0}\) atoms of rubidium-87 and none of strontium-87. Then, as time goes by, some of the rubidium atoms decay into strontium atoms, but the to tal number of atoms must still be \(\mathcal{N}_{0} .\) Thus, after \(T\) years, we have \(\mathcal{N}_{0}=\mathcal{N}_{r}+\mathcal{N}_{s}\) or, equivalently, $$ \mathcal{N}_{s}=\mathcal{N}_{0}-\mathcal{N}_{r}$$However, according to the law of exponential decay for the rubidium-87, we must have \(\mathcal{N}_{r}=\mathcal{N}_{0} e^{k T} .\) Solve this equation for \(\mathcal{N}_{0}\) and then use the result to eliminate \(\mathcal{N}_{0}\) from equation \((1) .\) Show that the result can be written $$\mathcal{N}_{s}=\mathcal{N}_{r} e^{-k T}-\mathcal{N}_{r}$$ (b) Solve equation (2) for \(T\) to obtain the formula given at the beginning of this exercise.

In 2000 the Philippines and Germany had similar size populations, but very different growth rates. The population of the Philippines was 80.3 million, with a relative growth rate of \(2.0 \%\) /year. The population of Germany was 82.1 million, with a relative "growth" rate of \(-0.1 \% /\) year. Using exponential models, make a projection for the population of Germany in the year when the Philippine population has doubled.

The following extract is from an article by Kim Murphy that appeared in the Los Angeles Times on September 14 1994 CAIRO-Over a chorus of reservations from Latin America and Islamic countries still troubled about abortion and family issues, nearly 180 nations adopted a wide-ranging plan Tuesday on global population, the first in history to obtain partial endorsement from the Vatican. The plan, approved on the final day of the U.N. population conference here, for the first time tries to limit the growth of the world's population by preventing it from exceeding 7.2 billion people over the next two decades. (a) In 1995 the world population was 5.7 billion, with a relative growth rate of \(1.6 \% /\) year. Assuming continued exponential growth at this rate, make a projection for the world population in the year \(2020 .\) Round off the answer to one decimal place. How does your answer compare to the target value of 7.2 billion mentioned in the article? (b) As in part (a), assume that in 1995 the world population was 5.7 billion. Determine a value for the growth constant \(k\) so that exponential growth throughout the years \(1995-2020\) leads to a world population of 7.2 billion in the year 2020 .

Let \(\mathcal{N}=\mathcal{N}_{0 e^{k t}} .\) In this exercise we show that if \(\Delta t\) is very small, then \(\Delta \mathcal{N} / \Delta t \approx k \mathcal{N} .\) In other words, over very small intervals of time, the average rate of change of \(\mathcal{N}\) is proportional to \(\mathcal{N}\) itself. (a) Show that the average rate of change of the function \(\mathcal{N}=\mathcal{N}_{0} e^{t t}\) on the interval \([t, t+\Delta t]\) is given by $$\frac{\Delta \mathcal{N}}{\Delta t}=\frac{\mathcal{N}_{0} e^{k t}\left(e^{k \Delta t}-1\right)}{\Delta t}=\frac{\mathcal{N}\left(e^{k \Delta t}-1\right)}{\Delta t}$$ (b) In Exercise 26 of Section 5.2 we saw that \(e^{x} \approx x+1\) when \(x\) is close to zero. Thus, if \(\Delta t\) is sufficiently small, we have \(e^{k \Delta t} \approx k \Delta t+1 .\) Use this approximation and the result in part (a) to show that \(\Delta \mathcal{N} / \Delta t \approx k N\) when \(\Delta t\) is sufficiently close to zero.

Graph each function and specify the domain, range, intercept(s), and asymptote. $$y=\ln (x+e)$$
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