Calculus problem which requires me to isolate X

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Homework Help Overview

The original poster is working on a calculus problem involving the equation √x = e^-3x and is attempting to isolate the variable x. The discussion revolves around the challenges of solving this equation and the methods being considered.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the transformation of the original equation into logarithmic form and the implications of that process. Some express skepticism about the possibility of finding an algebraic solution, suggesting that numerical methods may be necessary. Others introduce the concept of a unique solution defined by a constant, while also noting the cyclical nature of the transformations attempted.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have suggested that while a solution may exist, it cannot be expressed in terms of standard functions. There is a recognition of the complexity involved in isolating x, and various approaches are being considered without a clear consensus on a method to proceed.

Contextual Notes

Participants are navigating the constraints of the problem, particularly the limitations of algebraic solutions and the potential need for numerical methods. The original poster's attempts to manipulate the equation have led to a point of confusion, prompting further exploration of the problem's nature.

paiway
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I'm solving a calculus problem which requires me to equate both equations, but I can't isolate X:

√x = e^-3x

What I've got so far is:

ln√x = ln e^-3x

ln√x = -3x

1/2 ln x = -3x

ln x = -6x ... but from here I can't continue, please help:frown:
 
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you mean you want to solve for x right?? well, i don't think that can be done in terms of standard functions.
 
Yeah, like said, i can only show that a sol, to this eq exists, but i don't believe there is a solution algebraically. YOu can only solve it numerically.
 
I have quite a remarkable exact solution -

The GibZ Constant - G​

where G is defined to be the unique solution to the equation [itex]log_e x + 6x = 0[/itex] lol.
 
well if you got to here , heh , just a thought

ln x = -6x

take e to both sides
and get x=e^-6x

:D
 
th3plan said:
well if you got to here , heh , just a thought

ln x = -6x

take e to both sides
and get x=e^-6x

:D


...and...end up with exactly what you started with...lol...
 

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