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.
Nice!
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.
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.
This is what I did too, works nicely .
Nice and clean! Does it have other solutions? exept those 2 and why?
nice solution
multiply both side by lnx and use lambert function.
So uncivilized... 🔫🧔🤏
Пробуем решить "в лоб". Предположим, что x-1=√2, тогда х²-2х-1=0. D=4+4=8, x=(2±2√2)/2=1±√2. х=1+√2 подходит.
Houston we have a problem that SyberMath might solve!
SyberMath we have this problem but thankfully you have the solution!
@@roberttelarket4934 😍
Use an excel graph, it is easier.
Not sure if domain was x >= 0 making the second solution extraneous
WolphamAlfa gives also 2 solutions....
x^2-2x+1=(x-1)^2
x^{(x-1)^2}=2x+1
x = 0
(-1)^4=1, -2+1=-1