Solve x in [0,2π]: Logs Homework Help

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Homework Help Overview

The discussion revolves around solving equations involving logarithmic and trigonometric functions, specifically focusing on the equation 2cos(x/3) = LnX within the interval [0, 2π]. Participants also explore related questions about finding maximum values of functions involving logarithms.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to solve the equation by substituting values and applying logarithmic properties. Some participants question the correctness of the initial substitutions and interpretations of the equations. Others inquire about the process of finding relative maximum values for a different function.

Discussion Status

The discussion includes attempts to clarify the setup of the equations and explore different aspects of the problems presented. Some guidance has been offered regarding the approach to finding maximum values, but there is no explicit consensus on the correctness of the original poster's calculations.

Contextual Notes

Participants are navigating through potential misunderstandings regarding the equations and their respective domains, particularly concerning the logarithmic function and its constraints.

UrbanXrisis
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How do I solve for x in the interval [0,2pi]:
2cos(x/3)=LnX


Here's what I got:

2(.5)=LnX
1=LnX
e^(1)=x
x=2.718
is this correct?


If x=e^(Ln3)+Lne^2-5Ln1
what is x?
x=3+2-0=5
is this correct?
 
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Was that supposed to be [itex]\cos \frac {\pi}{3}[/itex]?
 
yes sorry, that is 2cos(pi/3)=LnX
 
Last edited:
In that case, good job!
 
how do I find the relative maximum value of the function y=(LnX)/X and the domain of the function Ln(X^2-1)?
 
Just find the derivative of the function and set it to zero.
 
for the domain or for the max value?
 
That will give you the location of the relative maximum (you can verify that it is a maximum as opposed to a minimum). Then use that value of x that you find to determine the value of the function at that point.
 

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