Problem 14 - Contemporary Precalculus

Question

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

Step-by-Step Solution

Answer

3. \(\ln(a \cdot b) = \ln(a) + \ln(b)\) #tag_title#Step 2: Rewrite the given expression in terms of x and y#tag_content# Given expression: \(e^{\frac{\ln x - \ln y}{\sqrt{xy}}}\) Rewrite \(\sqrt{xy}\) using properties of logarithms: \(\sqrt{xy} = e^{\frac{1}{2}(\ln x + \ln y)}\) (using property 3 and property 1) Now, substitute this back into the expression: \(e^{\frac{\ln x - \ln y}{e^{\frac{1}{2}(\ln x + \ln y)}}}\) #tag_title#Step 3: Rewrite the expression in terms of u and v#tag_content# Let \(\ln x = u\), and \(\ln y = v\). Substitute u and v into the expression: \(e^{\frac{u - v}{e^{\frac{1}{2}(u + v)}}}\) #tag_title#Step 4: Apply the properties of logarithms#tag_content# Now, we can apply property 2 of logarithms: \(\ln a - \ln b = \ln(\frac{a}{b})\) So, \(\frac{u - v}{e^{\frac{1}{2}(u + v)}} = \frac{\ln(\frac{x}{y})}{e^{\frac{1}{2}(u + v)}}\) #tag_title#Step 5: Rewrite the expression in exponential form#tag_content# Finally, rewrite the expression in exponential form: \(e^{\frac{\ln(\frac{x}{y})}{e^{\frac{1}{2}(u+v)}}}\) Thus, the given expression can be rewritten in terms of u and v as \(e^{\frac{\ln(\frac{x}{y})}{e^{\frac{1}{2}(u+v)}}}\). Short Answer: The expression can be rewritten as \(e^{\frac{\ln(\frac{x}{y})}{e^{\frac{1}{2}(u+v)}}}\), where \(u = \ln x\) and \(v = \ln y\).

Step 1: Identify the logarithmic properties

We will be using the following logarithmic properties: 1. \(\ln(a^b) = b\ln(a)\) 2. \(\ln(a/b) = \ln(a) - \ln(b)\)

Step 2: Rewrite the given expression in terms of x and y#tag_content

Given expression: \(e^{\frac{\ln x - \ln y}{\sqrt{xy}}}\) Rewrite \(\sqrt{xy}\) using properties of logarithms: \(\sqrt{xy} = e^{\frac{1}{2}(\ln x + \ln y)}\) (using property 3 and property 1) Now, substitute this back into the expression: \(e^{\frac{\ln x - \ln y}{e^{\frac{1}{2}(\ln x + \ln y)}}}\) #tag_title

Step 3: Rewrite the expression in terms of u and v#tag_content

Let \(\ln x = u\), and \(\ln y = v\). Substitute u and v into the expression: \(e^{\frac{u - v}{e^{\frac{1}{2}(u + v)}}}\) #tag_title

Step 4: Apply the properties of logarithms#tag_content

Now, we can apply property 2 of logarithms: \(\ln a - \ln b = \ln(\frac{a}{b})\) So, \(\frac{u - v}{e^{\frac{1}{2}(u + v)}} = \frac{\ln(\frac{x}{y})}{e^{\frac{1}{2}(u + v)}}\) #tag_title

Step 5: Rewrite the expression in exponential form#tag_content

Finally, rewrite the expression in exponential form: \(e^{\frac{\ln(\frac{x}{y})}{e^{\frac{1}{2}(u+v)}}}\) Thus, the given expression can be rewritten in terms of u and v as \(e^{\frac{\ln(\frac{x}{y})}{e^{\frac{1}{2}(u+v)}}}\). Short Answer: The expression can be rewritten as \(e^{\frac{\ln(\frac{x}{y})}{e^{\frac{1}{2}(u+v)}}}\), where \(u = \ln x\) and \(v = \ln y\).

Key Concepts

Natural Logarithms

Natural logarithms use the base of Euler's number, denoted as \( e \), which is approximately 2.718. They are incredibly important in mathematics due to their unique properties that streamline many calculations. In this exercise, we are using natural logarithms, represented by the function \( \ln x \). Natural logarithms have a close relationship with exponential functions. The definition of a natural logarithm is the inverse of the exponential function; this means that if \( y = \ln x \), then \( x = e^y \). Here, \( e \) serves as a constant base for natural logs and is similar to how we understand base 10 in common logarithms. For example, if you encounter \( \ln(e) \), it simplifies to 1 because \( e^1 = e \). Similarly, \( \ln(1) = 0 \) because \( e^0 = 1 \). This distinct relationship between \( e \) and the log function is crucial in achieving easier manipulation of logarithmic and exponential expressions, providing valuable simplifications in calculus, growth models, and other science applications.

Logarithmic Properties

Logarithmic properties are essential for transforming and simplifying complex expressions, as they help break down complicated logs into simpler terms. In the given exercise, we utilize two primary properties: the power rule and the quotient rule. - **Power Rule:** The power rule states \( \ln(a^b) = b \ln(a) \). This property is useful for turning the exponentiation inside a logarithm into multiplication outside the log function, simplifying calculations significantly.- **Quotient Rule:** The quotient rule shows us that \( \ln(a/b) = \ln(a) - \ln(b) \). This tells us that dividing two numbers inside a log can be expressed as the difference between their individual logs.Using these rules can help convert complex logarithms into more straightforward expressions. For instance, consider \( \ln(\frac{a^c}{b^d}) \). By applying the logarithmic properties, it becomes \( c \ln(a) - d \ln(b) \). These transformations are crucial for solving many logarithmic equations and tuning them into algebraic expressions, facilitating easier computations.

Algebraic Manipulation

Algebraic manipulation involves rearranging expressions to make them easier to work with, solve, or understand. In the context of logarithmic expressions, it becomes particularly useful. After applying logarithmic properties, algebraic manipulation helps further simplify or express answers in terms of desired variables like \( u \) and \( v \).For instance, in the exercise \( \ln\left(\frac{\sqrt{x y}}{y^2}\right) \), you follow a series of steps:1. **Understand the Structure:** Recognize that you have nested expressions—a root, division, and multiplication.2. **Apply the Quotient Rule:** Break the expression into \( \ln(\sqrt{xy}) - \ln(y^2) \).3. **Use the Power and Root Properties:** Rewrite \( \ln(\sqrt{xy}) \) as \( \frac{1}{2}\ln(xy) \) and \( \ln(y^2) \) as \( 2\ln(y) \).4. **Apply the Product Rule:** Further breakdown \( \ln(xy) \) into \( \ln(x) + \ln(y) \).5. **Substitute Known Variables:** Finally, convert these into terms of \( u \) and \( v \), resulting in \( \frac{1}{2}(u + v) - 2v \).This step-by-step approach makes it easier to handle even complex problems, thus solidifying one’s understanding of logarithmic transformations and their algebraic equivalences.