How to Solve the Equation 1^x?

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Discussion Overview

The discussion revolves around solving the equation 1^x and later shifts to the equation (1 + x)^x = x. Participants explore the implications of these equations across different domains, including rational and real numbers.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that 1^x equals 1 for all rational x, and if continuity is assumed, it holds for all real x, referencing the definition involving exp{xlog1}.
  • One participant highlights an edit in the thread title, clarifying that the current question is about solving (1 + x)^x = x.
  • Another participant argues that (1 + x)^x is not defined for x < -1 and explores the implications for various intervals of x, concluding that the equation has no solution.
  • Some participants express confusion regarding the title change and the responses to it, indicating a lack of clarity in the discussion flow.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the solvability of the equation (1 + x)^x = x, with some asserting it has no solutions while others focus on the properties of 1^x. The discussion remains unresolved regarding the latter equation.

Contextual Notes

There are limitations in the discussion regarding the definitions and assumptions about the domains of x, particularly concerning the continuity and the behavior of the functions involved.

Tombow2558
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How can I solve this equation ?
Thanks
 
Last edited:
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it is one for all rational x, and hence, if we've any intent on it being continuous it is 1 for all real x, a non-rigorous explanation admittedly but it agrees with the proper exp{xlog1} definition.
 
For people who have not spotted there was an edit in the title that has not shown up on the main board. The question s/he is now asking is how to solve:

[tex](1 + x)^x = x[/tex]
 
Did Tombow2885 edit to change to question or was mattgrime just responding to the title?

The question I see is to solve (1+x)x= x.

The left hand side is not defined for x< -1. If -1< x< 0, the left hand side is positive and so cannot equal x. The left hand side is not defined for x= 0.
If 0< x< 1, the right hand side is larger than 1 and so cannot equal x. If x= 1, the equation becomes 2= 1 which is clearly not true.

If x> 1 (1+x)x> 1+x and so cannot be equal to x.

You can't solve the equation, it has no solution.
 
matt grime said:
it is one for all rational x, and hence, if we've any intent on it being continuous it is 1 for all real x, a non-rigorous explanation admittedly but it agrees with the proper exp{xlog1} definition.
Oh, I am sorry matt, because the titles still the same even when I edited, I also didn't see your post, okey :wink:, thanks for your answer. :)
 
I responded to the original question which asked what 1^x was for x a positive real number.
 

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