Solving 3log(x)=6-2x: A Beginner's Guide

  • Context: High School 
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Discussion Overview

The discussion revolves around solving the equation 3log(x)=6-2x, exploring various methods and approaches to find a solution. The scope includes mathematical reasoning and numerical methods.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant suggests that the solution is not expressible in terms of elementary functions.
  • Another proposes that the Lambert W function or Newton's method could be relevant for solving the equation.
  • A different participant mentions that a numerical approach, such as Newton's method, could be used, noting that x must be greater than 0 due to the logarithmic function's domain.
  • It is suggested that an approximate solution could be found by determining the point of intersection of two graphs, but an exact solution may not be feasible without trial and error.
  • One participant shares a result from MATLAB's solver involving the Wright Omega function, indicating that the Lambert W function is necessary for the solution, while also recommending a combination of bisection and Newton's methods for practical purposes.

Areas of Agreement / Disagreement

Participants express differing views on the feasibility of finding an exact solution, with some agreeing on the necessity of numerical methods while others highlight the limitations of elementary functions.

Contextual Notes

Participants note the importance of the domain restriction x>0 due to the logarithmic function and mention the complexity of the equation in relation to elementary functions.

crazylum
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Where does one begin to solve the equation 3log(x)=6-2x?
 
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The solution to that equation is not expressible in terms of elementary functions.
 
Looks like material for Lambert's W (or Newton's method).
 
you could use a numerical approach like Newton's method to find a solution. if I am not mistaken you can say that x>0 since ln(x) is undefined otherwise and then rewrite your equation as x^{3}e^{2x}-e^{6}=0
 
an approximate solution is possible through pt of intersection of the 2 graphs...but i don't think there's any way to find an exact solution save hit n trial
 
MATLAB's solver gave:

\frac{\mathrm{e}^{2}}{\mathrm{e}^{\omega\!\left(\ln\!\left(\frac{2}{3}\right) + 2\right)}}

Where \omega is given by: http://en.wikipedia.org/wiki/Wright_Omega_function.

Good call by whoever said it needed the Lambert W-function ^_^

But for practical purposes, just use bisection+Newton.
 

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