Rewriting the function e^-x*x^t-1

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Discussion Overview

The discussion revolves around the manipulation of the mathematical expression \( e^{-x} x^{t-1} \) and its equivalence to another form involving logarithms. Participants explore the rewriting of the function to facilitate differentiation and integration.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant proposes that \( e^{-x} x^{t-1} \) can be rewritten as \( e^{t \ln(x) - \ln(x) - x} \), providing a step-by-step reasoning for this transformation.
  • Another participant questions the clarity of the original expression due to the absence of parentheses, suggesting that the expression may have been misinterpreted.
  • A later reply clarifies that if the expression is indeed \( e^{-x} x^{t-1} \), it can be rewritten as \( e^{-x + t \ln(x) - \ln(x)} \), indicating a different approach to the manipulation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct form of the expression, and there are competing interpretations regarding the placement of parentheses and the resulting equivalences.

Contextual Notes

There are unresolved issues regarding the correct interpretation of the original expression, particularly concerning the placement of parentheses and the implications for the equivalence of the rewritten forms.

neptune12XII
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does e^-x*x^(t-1)=
e^(t*ln(x)-ln(x)-x)
heres my reasoning:
x=e^ln(x)
e^-x*x^(t-1)=
e^-x*e^(ln(x)(t-1))=
e^-x*e^(t*ln(x)-ln(x))=
e^(t*ln(x)-ln(x)-x)

I want it in the latter form so that it is easier to take derivatives and antiderivatives. did i make any mistakes?
 
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Where are your parentheses? As written, your expression is ##e^{-x}x^t-1 = e^t \ln x - \lnx -x##, which obviously isn't true.
 
youre right srry
 
If you mean [itex]e^{-x}x^{t-1}[/itex] then it is equal to [itex]e^{-x}e^{ln(x^{t-1})}=[/itex][itex]e^{-x}e^{(t- 1)ln(x)} = e^{-x+ tln(x)- ln(x)}[/itex]
 

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