/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Use the properties of logarithms... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the properties of logarithms to expand the expression as a sum, difference, and or constant multiple of logarithms. (Assume all variables are positive.) $$\ln z(z-1)^{2}, z>1$$

Short Answer

Expert verified
After application of properties of logarithms, the expression \( \ln z(z-1)^{2} \) expands to \( \ln z + 2 * \ln (z-1) \).

Step by step solution

01

Recognition of multiplication within the log

Initial expression is \( \ln z(z-1)^{2} \), which can be seen as \( \ln (z*(z-1)^{2}) \). This would suggest use of multiplication property of logs.
02

Applying the multiplication property

Now, applying the multiplication property, \( \ln (z*(z-1)^{2}) \) can be re-written as \( \ln z + \ln (z-1)^{2} \).
03

Applying the power rule of logarithms

Applying the power rule of the logarithms on \( \ln (z-1)^{2} \), the above expression becomes \( \ln z + 2 * \ln (z-1) \). This expression is a sum of logarithms, which was our initial goal.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Multiplication Property of Logarithms
When dealing with logarithmic expressions involving products, the multiplication property of logarithms comes in handy. This property states that the logarithm of a product can be expressed as the sum of the logarithms of the individual factors. For example, if you have an expression like \( \ln (a \cdot b) \), it can be expanded to \( \ln a + \ln b \). This transforms complex calculations into much simpler ones. In our problem, we started with the expression \( \ln z(z-1)^2 \), which can be seen as \( \ln (z \cdot (z-1)^2) \). By applying the multiplication property of logarithms, this was rewritten as \( \ln z + \ln (z-1)^2 \). This step is crucial as it breaks down the expression into more manageable parts, allowing us to handle each logarithmic term individually. Breaking down expressions in this way often simplifies the process of solving or transforming them. Overall, this property is a powerful tool for expanding logarithmic expressions and is often the first step in simplifying or solving an expression.
Power Rule of Logarithms
Once the expression is broken down using the multiplication property, the power rule of logarithms comes into play. The power rule states that the logarithm of an expression raised to a power can be rewritten by bringing the power down as a coefficient in front of the logarithm. Mathematically, this translates to \( \ln(a^n) = n \cdot \ln a \). For the given expression \( \ln (z-1)^2 \), applying the power rule allows us to rewrite it as \( 2 \cdot \ln (z-1) \). Here, the exponent 2 becomes a multiplier for the logarithm of \( z-1 \). This transformation is vital because it creates a simpler linear expression, which is often easier to work with or solve. The power rule of logarithms is intuitive once you understand that logarithms and power functions are inherently linked through their operations. It's like borrowing the power factor and placing it in front of the log, simplifying the whole expression.
Expanding Logarithmic Expressions
Expanding logarithmic expressions often combines several properties and rules to make the expression simpler or more workable. Starting with a given logarithmic expression, we seek to express it as a sum, difference, or multiple of individual logarithms. This process takes advantage of key properties like the multiplication property and power rule. In the context of our exercise, after recognizing and applying these properties, we expanded \( \ln z(z-1)^2 \) to finally become \( \ln z + 2 \cdot \ln (z-1) \). Each step involved careful transformation using characterized properties of logarithms. Expanding is not merely an algebraic exercise; it enhances understanding by breaking complex expressions into simpler parts. This can reveal underlying relationships or simplify calculations involved with logarithmic equations. Learning to master this form of expansion equips you with tools that can solve real-world problems efficiently. By mastering these techniques, you also gain a deeper appreciation of how logarithms reveal themselves in natural growth patterns and various scientific calculations. It's a skill that proves to be highly valuable in both academic and practical scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility. $$2 x^{2} e^{2 x}+2 x e^{2 x}=0$$

A cup of water at an initial temperature of \(78^{\circ} \mathrm{C}\) is placed in a room at a constant temperature of \(21^{\circ} \mathrm{C}\). The temperature of the water is measured every 5 minutes during a half-hour period. The results are recorded as ordered pairs of the form \((t, T),\) where \(t\) is the time (in minutes) and \(T\) is the temperature (in degrees Celsius). $$\begin{aligned} &\left(0,78.0^{\circ}\right),\left(5,66.0^{\circ}\right),\left(10,57.5^{\circ}\right),\left(15,51.2^{\circ}\right)\\\ &\left(20,46.3^{\circ}\right),\left(25,42.4^{\circ}\right),\left(30,39.6^{\circ}\right) \end{aligned}$$ (a) The graph of the model for the data should be asymptotic with the graph of the temperature of the room. Subtract the room temperature from each of the temperatures in the ordered pairs. Use a graphing utility to plot the data points \((t, T)\) and \((t, T-21)\). (b) An exponential model for the data \((t, T-21)\) is given by \(T-21=54.4(0.964)^{t} .\) Solve for \(T\) and graph the model. Compare the result with the plot of the original data. (c) Take the natural logarithms of the revised temperatures. Use the graphing utility to plot the points \((t, \ln (T-21))\) and observe that the points appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. This resulting line has the form \(\ln (T-21)=a t+b\) Solve for \(T,\) and verify that the result is equivalent to the model in part (b). (d) Fit a rational model to the data. Take the reciprocals of the \(y\) -coordinates of the revised data points to generate the points $$\left(t, \frac{1}{T-21}\right)$$. Use the graphing utility to graph these points and observe that they appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. The resulting line has the form $$\frac{1}{T-21}=a t+b$$. Solve for \(T,\) and use the graphing utility to graph the rational function and the original data points. (e) Why did taking the logarithms of the temperatures lead to a linear scatter plot? Why did taking the reciprocals of the temperatures lead to a linear scatter plot?

Use a graphing utility to graph the functions \(y_{1}=\ln x-\ln (x-3)\) and \(y_{2}=\ln \frac{x}{x-3}\) in the same viewing window. Does the graphing utility show the functions with the same domain? If so, should it? Explain your reasoning.

Determine whether the statement is true or false given that \(f(x)=\ln x .\) Justify your answer. $$\text { If } f(u)=2 f(v), \text { then } v=u^{2}$$.

Use the following information. The relationship between the number of decibels \(\boldsymbol{\beta}\) and the intensity of a sound \(I\) in watts per square meter is given by $$\boldsymbol{\beta}=10 \log \left(\frac{\boldsymbol{I}}{\mathbf{1 0}^{-12}}\right).$$ Find the difference in loudness between an average office with an intensity of \(1.26 \times 10^{-7}\) watt per square meter and a broadcast studio with an intensity of \(3.16 \times 10^{-10}\) watt per square meter.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.