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Chapter 5: Problem 39

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

Short Answer

Expert verified
(a) \(1 + t\); (b) \(u - t\); (c) \(\frac{3}{2}t + \frac{1}{2}u - 1\); (d) \(2 + t + \frac{1}{2}u\)."

Step by step solution

01

Understanding the given expressions

We are given the expressions involving natural logarithms and need to express them in terms of the variables \(t = \ln x\) and \(u = \ln y\). This involves applying properties of logarithms such as the product rule, quotient rule, power rule, and the logarithm of exponentials.
02

Solving part (a) \(\ln (e x)\)

Using the property \(\ln(a\cdot b) = \ln a + \ln b\), express \(\ln (e x)\) as follows:\[ \ln (e x) = \ln e + \ln x \]Since \(\ln e = 1\), we have:\[ \ln (e x) = 1 + \ln x = 1 + t \]
03

Solving part (b) \(\ln x y - \ln (x^{2})\)

Apply the property \(\ln(a\cdot b) = \ln a + \ln b\) and \(\ln(a^b) = b \ln a\):\[ \ln x y - \ln (x^2) = (\ln x + \ln y) - 2\ln x \]Substituting \(\ln x = t\) and \(\ln y = u\), we get:\[ (t + u) - 2t = u - t \]
04

Solving part (c) \(\ln \sqrt{x y} + \ln (x / e)\)

For \(\ln \sqrt{x y}\), use \(\ln(a^b) = b \ln a\):\[ \ln \sqrt{x y} = \frac{1}{2} \ln (x y) = \frac{1}{2} (\ln x + \ln y) = \frac{1}{2} (t + u) \]For \(\ln (x / e)\), use \(\ln(a / b) = \ln a - \ln b\):\[ \ln (x / e) = \ln x - \ln e = t - 1 \]Combine these:\[ \frac{1}{2} (t + u) + (t - 1) = \frac{1}{2} t + \frac{1}{2} u + t - 1 = \frac{3}{2} t + \frac{1}{2} u - 1 \]
05

Solving part (d) \(\ln (e^{2} x \sqrt{y})\)

Apply \(\ln(a\cdot b) = \ln a + \ln b\), and \(\ln(a^b) = b \ln a\):\[ \ln (e^{2} x \sqrt{y}) = \ln(e^2) + \ln x + \ln \sqrt{y} \]This becomes:\[ 2\ln e + \ln x + \frac{1}{2}\ln y = 2 + t + \frac{1}{2} u \]

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Key Concepts

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

Logarithm Properties
Logarithms have several properties that make them highly useful in mathematical problem solving. These properties are based on the relationships between logarithms of products, quotients, and powers.

For products, we use the rule:
  • \( \ln(a \cdot b) = \ln a + \ln b \)
This means if you have the logarithm of two numbers multiplied together, you can break it into the sum of their individual logarithms. This is particularly helpful for simplifying complex expressions.For quotients, the rule is:
  • \( \ln(\frac{a}{b}) = \ln a - \ln b \)
This means the logarithm of a division can be expressed as the difference of the logarithms of the numerator and the denominator.Lastly, for powers, the rule is:
  • \( \ln(a^b) = b \cdot \ln a \)
This allows us to bring the exponent down as a multiplier. These properties are crucial in simplifying and solving logarithmic expressions.
Natural Logarithms
Natural logarithms are logarithms with base \(e\), where \(e\) is approximately 2.718. The natural logarithm of a number \(x\) is denoted as \(\ln x\).

Natural logarithms arise frequently in calculus due to their relationship with exponential functions. For instance, they are often used to simplify expressions involving exponentials because of the property \(\ln(e^x) = x\). This is because raising \(e\) to a power and taking the natural logarithm of that result are inverse operations.In solving problems, it's crucial to recognize scenarios where \(\ln e\) simplifies calculations, since \(\ln e = 1\). This can make transformations involving \(e\) seem almost trivial and is particularly useful in expressions where \(e\) appears, such as in the solutions of our exercise's parts (a) and (d).
Expression Simplification
Expression simplification using logarithms involves making an expression easier to work with by applying the various properties of logarithms.

Consider the expression \(\ln(e x)\). By applying the product rule of logarithms, it simplifies to \(\ln e + \ln x\). Using the known fact that \(\ln e = 1\) further simplifies this to \(1 + \ln x\). This simplification makes the expression more digestible and can aid in further calculations or substitutions in problem-solving.Similarly, in part (c) of our exercise, we combined multiple rules鈥攑ower and quotient鈥攖o rewrite \(\ln \sqrt{x y} + \ln(x / e)\), making it more workable by reducing it to an expression involving \(t\) and \(u\). Knowing how to apply these simplifications is a critical skill in controlling complex mathematical expressions.
Precalculus Problem Solving
Precalculus involves a variety of mathematical skills that prepare you for calculus. Logarithm problems are a common feature in precalculus.

To solve problems involving logarithms, it's necessary to have a firm understanding of their properties and behaviors. Our exercise demonstrates this through the use of converting expressions such as \(\ln x y - \ln(x^2)\) into simpler terms by leveraging the properties of logarithms. This can help in forecasting more complicated calculus problems.One powerful approach shown in our solutions is breaking down complex expressions into simpler parts, such as identifying products and quotients within logs and rewriting them according to their rules. By practicing these transformations, students become better at recognizing where simplification can occur, which is vital for tackling both precalculus and calculus problems effectively.

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Most popular questions from this chapter

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 .

(a) Use a graphing utility to estimate the root(s) of the equation to the nearest one-tenth (as in Example 6). (b) Solve the given equation algebraically by first rewriting it in logarithmic form. Give two forms for each answer: an exact expression and a calculator approximation rounded to three decimal places. Check to see that each result is consistent with the graphical estimate obtained in part (a). $$e^{1-4 t}=12.405$$

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$6\left(5-1.6^{x}\right) \geq 13$$

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$2^{x}>0$$

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.
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