/*! 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 49 use a logarithmic transformation... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

use a logarithmic transformation to find a linear relationship between the given quantities and graph the resulting linear relationship on a log-linear plot. $$ y=2 e^{-1.2 x} $$

Short Answer

Expert verified
The linear relationship is \( \ln y = -1.2x + \ln 2 \) and plots as a straight line with slope -1.2 on a log-linear plot.

Step by step solution

01

Understand the Relationship

The given equation is an exponential function: \( y = 2 e^{-1.2x} \). Our goal is to transform this into a linear relationship using logarithms.
02

Apply Logarithmic Transformation

Take the natural logarithm of both sides: \( \ln y = \ln(2 e^{-1.2x}) \). According to the logarithm product property, this simplifies to \( \ln y = \ln 2 + \ln e^{-1.2x} \).
03

Simplify Using Logarithm Properties

Know that \( \ln e^{-1.2x} = -1.2x \) (since the natural logarithm and exponential are inverse functions). The equation becomes \( \ln y = \ln 2 - 1.2x \).
04

Identify Linear Equation Format

The resulting equation \( \ln y = -1.2x + \ln 2 \) is in the form \( y = mx + c \), where \( m = -1.2 \) and \( c = \ln 2 \). This indicates a linear relationship on a log-linear plot.
05

Graph the Transformation

Plot \( \ln y \) versus \( x \) on a graph. The slope of the line is \(-1.2\), and it intercepts the \( \ln y \)-axis at \( \ln 2 \). Use a log-linear plot, where the \( y \)-axis is logarithmic, to represent this linear relationship.

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.

Log-Linear Plot
A log-linear plot is a type of graph that serves as an effective tool for visualizing data that grows exponentially. In such plots, the horizontal axis is linear, while the vertical axis is logarithmic. This means that instead of counting units equally as in a regular graph, the log scale compresses higher values relative to lower ones. As a result, exponential relationships convert into straight lines.
  • By plotting the natural logarithm of the dependent variable against the independent variable, any exponential growth or decay in the data presents itself as a linear trend.
  • This makes it much easier to interpret empirical results and understand underlying relationships.
In the exercise, the log-linear plot facilitated the transformation of the exponential equation into a linear form. By plotting \( \ln y \) against \( x \), a clear straight line with slope \(-1.2\) and intercept at \( \ln 2 \) is achieved. This visualization aids in simplifying complex dynamics into more manageable knowledge.
Linear Relationship
In mathematics, a linear relationship refers to a kind of direct association between two variables. Such a relationship is characterized by a straight line when graphed on a Cartesian coordinate system. Its general form is expressed by the equation \( y = mx + c \), where:
  • \( y \) represents the dependent variable.
  • \( x \) stands for the independent variable.
  • \( m \) is the slope of the line, indicating the rate of change of \( y \) with respect to \( x \).
  • \( c \) is the y-intercept, where the line crosses the y-axis.
The exercise illustrated how taking the natural logarithm of the exponential function transformed it into a linear equation: \( \ln y = -1.2x + \ln 2 \). This step highlighted the inverse relationship between exponential growth and linear transformations, allowing easier interpretation of the data's behavior.
Exponential Function
An exponential function describes situations where a quantity grows or decays at a rate proportional to its current value. It is typically represented by the formula \( y = a e^{bx} \), where:
  • \( y \) is the output or dependent variable.
  • \( a \) stands for the initial value, starting point, or amplitude.
  • \( b \) represents the rate of growth (positive) or decay (negative).
  • \( e \) is the base of natural logarithms, approximately 2.718.
  • \( x \) is the input or independent variable.
In the original exercise, we dealt with \( y = 2 e^{-1.2x} \), showcasing exponential decay since the exponent \(-1.2\) is negative. This means that as \( x \) increases, \( y \) decreases rapidly at first but then slows down due to the nature of the exponential function. Through logarithmic transformation, this complex behavior was simplified into a linear form, demonstrating the utility of mathematical manipulations in revealing fundamental patterns.

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

Sketch the graph of each function. Do not use a graphing calculator. (Assume the largest possible domain.) $$ y=\ln (x-3) $$

A study of Borchert's (1994) investigated the relationship between stem water storage and wood density in a number of tree species in Costa Rica. The study showed that water storage is inversely related to wood density; that is, higher wood density corresponds to lower water content. Sketch a graph of water content as a function of wood density that illustrates this situation.

The absorption of light in a uniform water column follows an exponential law; that is, the intensity \(I(z)\) at depth \(z\) is $$ I(z)=I(0) e^{-\alpha z} $$ where \(I(0)\) is the intensity at the surface (i.e., when \(z=0)\) and \(\alpha\) is the vertical attenuation coefficient. (We assume here that \(\alpha\) is constant. In reality, \(\alpha\) depends on the wavelength of the light penetrating the surface.) (a) Suppose that \(10 \%\) of the light is absorbed in the uppermost meter. Find \(\alpha\). What are the units of \(\alpha\) ? (b) What percentage of the remaining intensity at \(1 \mathrm{~m}\) is absorbed in the second meter? What percentage of the remaining intensity at \(2 \mathrm{~m}\) is absorbed in the third meter? (c) What percentage of the initial intensity remains at \(1 \mathrm{~m}\), at 2 \(\mathrm{m}\), and at \(3 \mathrm{~m} ?\) (d) Plot the light intensity as a percentage of the surface intensity on both a linear plot and a log-linear plot. (e) Relate the slope of the curve on the log-linear plot to the attenuation coefficient \(\alpha\). (f) The level at which \(1 \%\) of the surface intensity remains is of biological significance. Approximately, it is the level where algal growth ceases. The zone above this level is called the euphotic zone. Express the depth of the euphotic zone as a function of \(\alpha\). (g) Compare a very clear lake with a milky glacier stream. Is the attenuation coefficient \(\alpha\) for the clear lake greater or smaller than the attenuation coefficient \(\alpha\) for the milky stream?

Continuation of Problem 86) Estimating \(v_{\max }\) and \(K_{m}\) from the Lineweaver-Burk graph as described in Problem 86 is not always satisfactory. A different transformation typically yields better estimates (Dowd and Riggs, 1965 ). Show that the Michaelis-Menten equation can be written as $$ \frac{v_{0}}{s_{0}}=\frac{v_{\max }}{K_{m}}-\frac{1}{K_{m}} v_{0} $$ and explain why this transformation results in a straight line when you graph \(v_{0}\) on the horizontal axis and \(\frac{v_{0}}{s_{0}}\) on the vertical axis. Explain how you can estimate \(v_{\max }\) and \(K_{m}\) from the graph.

Explain how the following functions can be obtained from \(y=\ln x\) by basic transformations: (a) \(y=\ln (1-x)\) (b) \(y=\ln (2+x)-1\) (c) \(y=-\ln (2-x)+1\)

See all solutions

Recommended explanations on Biology 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.