Factoring e^x in a Complex Equation: Tips and Tricks

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SUMMARY

The discussion focuses on solving the equation e^{2x} - 3e^{x + 1} + 2e^{2} = 0 by factoring. Participants suggest substituting u = e^x, transforming the original equation into a quadratic form. This substitution simplifies the problem, allowing for easier resolution of x. The approach emphasizes the utility of recognizing exponential functions as quadratic equations.

PREREQUISITES
  • Understanding of exponential functions and their properties
  • Familiarity with quadratic equations and factoring techniques
  • Basic knowledge of substitution methods in algebra
  • Experience with solving equations involving variables
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  • Study the method of substitution in algebraic equations
  • Learn about solving quadratic equations using the quadratic formula
  • Explore advanced techniques for factoring exponential equations
  • Investigate the applications of exponential functions in real-world scenarios
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Students, educators, and anyone interested in mastering algebraic techniques for solving complex equations, particularly those involving exponential functions.

cscott
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[tex]e^{2x} - 3e^{x + 1} + 2e^{2} = 0[/tex]

I've been staring at this for the past while I can't solve it for x...

The only thing I can think of doing is factoring out [itex]e^{x}[/itex]
 
Last edited:
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Try setting u=e^x. Then you'll have a quadratic equation for u.
 
Damn, thanks.
 

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