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

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The discussion focuses on finding the intervals of increase and decrease for the equation y = e^x = e^-2x. The first derivative, y' = e^x - 2e^-2x, is set to zero to identify critical points. Participants discuss using logarithmic properties to solve for x, specifically applying the natural logarithm. The solution leads to the critical point x = (ln 2)/3, with clarification on manipulating the equation. The conversation emphasizes the importance of algebraic manipulation in solving the derivative.
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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

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

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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