Can Sup Forums solve for x?

can Sup Forums solve for x?
x^x=x+1

Other urls found in this thread:

wolframalpha.com/input/?i=x^x=x+1
twitter.com/SFWRedditGifs

/-1
x^x-1=x

It's about 0.35

X=√π

about three fidy

nope
no, but kinda close

x=rape

It's 0
>0^0 = 0+1
>1 = 1

>nope
>.35^.35 = 0.6925
>0.6925/2 = 0.3462
So really the answer is 0.3463

0 works as well I suppose

You just take an x root of both sides. It will be something like x = x root (X+1)

x^x = x + 1
x^x - x - 1 = 0
x1 = (1 + (1 - 4 × 1 × -1)^1/2) / 2 = 1 + 5^1/2
x2 = (1 - (1 - 4 × 1 × -1)^1/2) / 2 = 1 - 5^1/2
Might have made a mistake somewhere

Fuck this man is a genius

>0^0=1
you are wrong

Are you retarded or what

its x^x=x+1
not x^x=x+x

Two solutions, 0 and 1.78

its not a cuadratic equation, you cant do that

0^0=1=0+1

technically the two solutions are y (lim y->0) and 1.78

Lambert W functions

no, 0^0 is undefined, if you used limits instead of 0, yes

fuk

>not x^x=x+x
shit you're right.

1.11

more digits
1.77677504...

x^x=x+1
(x^x)-x=1
((x^x)-x)^1/2 = 1^1/2
sub x = 0
((0^0)-0)=1
=1-0=1
.*. x =0

0^0 isnt 1

It is, dude. Anything with an exponent of 0 = 1.

0^n=0
n^0=1
0^0 is undefined

Can you, summer?

yes it's only a decimal approximation though

No such number.
X doesn't exist as real number.

see

True. The answer is about x=1,777 (not an infinite seq of 7s, just rounded).

You can get the result if you plot both sides of the equation and see where they intersect.

Are you retarded? It absolutely is a quadratic equation

i calculated it with newtons method

a quadratic would be x^2 = x+1, this is x^x=x+1

So you solved for equation x^x - x - 1? I think that should be doable, the equation seems differentiable. Not a 100% sure about x^x but I think so. If you got the same result, cool.

The answer isn’t less than 2 and greater than 1. Simply plot the linear function x+1 and plot x^x and find the intersect on a graphing calculator. However, one can easily determine that 1

Here's a graph of it. Note that the image is not entirely accurate as both sides of the equation are plotted with x intervals of 0.01. I didn't include the negative x part of the plot where there's another intersection point because things get iffy at that point (basically, you get negative values to the power of all kinds of values from negative to positive and it gets complex).

*Is less than 2 greater than 1

it's so close to 0, could it be .000000000001?

The limes of x^x is 1, as is the limes of x+1. However, they will never intersect at zero as x^x is not defined at x=0.

See

its the limit as it goes to 0

yup

x^x=x+1
∴x^x-x-1=0
(f(x)=x^x-x-1)
∴x=root of f(x)
f'(x)=d/dx(x^x-x-1)
=d/dx(x^x)+d/dx(-x)+d/dx(-1)
=d/dx(x^x)-1-0
=(x^x)(1+ln(x))-1
Apply newton
result=1.77677504...

its either 0 or 1.777 recurring

wolframalpha.com/input/?i=x^x=x+1

its not 1.7..., its irrational
also it isnt 0 because 0^0 is undefined