An Interesting Exponential Equation

[@scottleung9587]

Nice!

[@hazalouldi7130]

multiply both side by lnx and use lambert function.

[@roberttelarket4934]

Houston we have a problem that SyberMath might solve! SyberMath we have this problem but thankfully you have the solution!

[@SyberMath]

@@roberttelarket4934 😍

[@stephenshefsky5201]

Great problem! I was very close to solving it by arranging the equation into log-product form, but I couldn't quite see the path. Then I set y = x^2 - 2x + 1, from which I found the system 0 = x^y - 2x - 1, 0 = x^2 - 2x - y + 1. Subtracting, I found 0 = (x^y - x^2) - (y - 2) which is obviously solvable by y = 2, from which I obtained 0 = x^2 - 2 - 1, and finally x = 1 +/-sqrt(2). Not so elegant, but it worked.

[@duanephinney300]

This is what I did too, works nicely .

[@mystychief]

A negative x is tricky, that's why there is no graph of the left part, but both of the x-values are valid in this case.

[@jesusalej1]

Use an excel graph, it is easier.

[@rakenzarnsworld2]

x^2-2x+1=(x-1)^2 x^{(x-1)^2}=2x+1 x = 0 (-1)^4=1, -2+1=-1

[@Don-Ensley]

I can't solve.