1^x

[Tombow2558]

How can I solve this equation ?
Thanks

[matt grime]

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.

[Zurtex]

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:

(1 + x)^x = x

[HallsofIvy]

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.

[Tombow2558]

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. :)

[matt grime]

I responded to the original question which asked what 1^x was for x a positive real number.