Intervals of Increase and Decrease for e^x = e^-2x

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    Calculus
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Homework Help Overview

The discussion revolves around finding the intervals of increase and decrease for the equation y = e^x = e^-2x. The original poster has derived the first derivative and is seeking guidance on solving for x.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to set the first derivative equal to zero but encounters difficulty in solving for x. Some participants suggest using logarithmic properties to manipulate the equation, while others question the reasoning behind specific exponent manipulations.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to the problem. Some guidance has been offered regarding logarithmic manipulation, but there is no explicit consensus on the method to solve for x.

Contextual Notes

Participants are working within the constraints of the problem as posed, with some expressing uncertainty about the steps involved in solving the derivative equation.

erik05
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Hello all. Really quick question here...What are the intervals of increase and decrease for y= e^x = e^-2x. I found the first derivative : y'= e^x-2e^-2x and set it equal to 0 but that's where I got stuck. How would you solve for x? I know that the answer is (ln2)/3 but how would you get there? Thanks.
 
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Working out the equation, and use some logarithm properties.

Apply natural logaritm

[tex](e^{x})^{3} = 2[/tex]
 
Stupid question...but why to the exponent of 3?
 
[tex]0= e^x-\frac{2}{e^{2x}}[/tex]

Just swing that fraction to the other side and multiply out.
 
I can't believe I didn't get that...thanks.
 

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