Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Is it possible for a logarithmic equation to have more than one extraneous solution? Explain.

Short Answer

Expert verified
Yes, it is possible for a logarithmic equation to have more than one extraneous solution. This can happen when the equation has multiple terms or multiple logarithmic functions, resulting in multiple solutions. If more than one of these solutions fall outside the domains of the involved logarithmic functions, they are considered extraneous solutions.

Step by step solution

01

Define the Domain of Logarithmic Functions

First, let's recall the domain of a logarithmic function: For any logarithmic function of the form \(y = \log_b x\) , the domain is \(x > 0\). This means logarithmic functions are only defined for positive values of \(x\). Therefore, if a solution does not respect this condition, it is considered extraneous.
02

Analyze Possibility of Multiple Extraneous Solutions

Consider a logarithmic equation with multiple terms or multiple logarithms - you can potentially end up with multiple solutions. Now, depending upon the restrictions of the domains of the involved logarithm functions, more than one of these solutions can fall outside these domains, rendering them as extraneous solutions.
03

Provide an Example

As an example, consider the equation \( \log_2(x-3) + \log_2(x+3) = 2 \). Solving this equation leads to the solutions \(x = 5\) and \(x = -1\). However, substituting \(x = -1\) back into the equation doesn't work out, as it would result in a negative argument for the logarithms, which is outside of their domain. Therefore, \(x = -1\) is an extraneous solution. By similar analysis, complex equations can yield more than one extraneous solutions.

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 Vaia!

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 logarithmic equation algebraically. Approximate the result to three decimal places.\(\log _{3}(x+8)=\log _{3}(3 x+2)\)

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\ln (x+5)=\ln (x-1)-\ln (x+1)\)

Motorola The sales per share \(S\) (in dollars) for Motorola from 1992 to 2005 can be approximated by the function \(S=\left\\{\begin{array}{lr}2.33-0.909 t+10.394 \ln t, & 2 \leq t \leq 10 \\ 0.6157 t^{2}-15.597 t+110.25, & 11 \leq t \leq 15\end{array}\right.\) where \(t\) represents the year, with \(t=2\) corresponding to 1992\. (Source: Motorola) (a) Use a graphing utility to graph the function. (b) Describe the change in sales per share that occurred in 2001 .

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\log _{2}(2 x-3)=\log _{2}(x+4)\)

Automobiles are designed with crumple zones that help protect their occupants in crashes. The crumple zones allow the occupants to move short distances when the automobiles come to abrupt stops. The greater the distance moved, the fewer g's the crash victims experience. (One \(\mathrm{g}\) is equal to the acceleration due to gravity. For very short periods of time, humans have withstood as much as 40 g's.) In crash tests with vehicles moving at 90 kilometers per hour, analysts measured the numbers of g's experienced during deceleration by crash dummies that were permitted to move \(x\) meters during impact. The data are shown in the table. $$ \begin{array}{|l|l|l|l|l|l|} \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1.0 \\ \hline g \text { 's } & 158 & 80 & 53 & 40 & 32 \\ \hline \end{array} $$A model for these data is given by \(y=-3.00+11.88 \ln x+\frac{36.94}{x}\) where \(y\) is the number of g's. (a) Complete the table using the model.$$ \begin{array}{|l|l|l|l|l|l|} \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1.0 \\ \hline y & & & & & \\ \hline \end{array} $$(b) Use a graphing utility to graph the data points and the model in the same viewing window. How do they compare? (c) Use the model to estimate the least distance traveled during impact for which the passenger does not experience more than \(30 \mathrm{~g}\) 's. (d) Do you think it is practical to lower the number of g's experienced during impact to fewer than 23 ? Explain your reasoning.

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.

Sign-up for free