If x^x=y, how can x be expressed as a function of y?

Question

This is if x>0
It can also be expressed as x=log y/log x,which also =xth root of y
I think its unsolvable maybe someone can figure it out

Answer 1

You are correct. the inverse of y = x^x is transcendental. It can be solved numerically using iteration, but not algebraically.

Answer 2

examine no matter if the equation is properly: enable M(x, y) = x + y enable N(x, y) = x - y ?M(x, y)/?y = a million ?N(x, y)/?x = a million The equation is properly. for this reason, there exists a function, F(x, y) the partial by-product with appreciate to x is M(x, y) and the partial by-product with appreciate to y is N(x, y). we commence through integrating M(x, y) with appreciate to x: ?(x + y)dx = (a million/2)x² + yx + h(y) the position h(y) is a function that would disappear after we differentiate with appreciate to x. We differentiate this with appreciate to y: x + h'(y) This ought to equivalent N(x, y); x + h'(y) = x - y h'(y) = -y h(y) = (-a million/2)y² + C the answer's: F(x, y) = (a million/2)(x² - y²) + xy + C

Answer 3

I'm 99.9% sure that x on it's own cannot be expressed
as a function of y, that is, you cannot have x = f(y).
This type of problem requires something like an iterative method,
such as Newton's method to solve for x, where approximations
to x are continually put back into a formula to obtain better and
better approximations.

Answer 4

x^x = y

The thing is, there's no way you can isolate x. After all, you would need to isolate x if you wanted to have x as a function of y.

It's no different from attempting to isolate x with the following equations:

x + sin(x) = 2
x - ln(x) = 4

There are approximation techniques in some advanced math courses which allow us to determine answers, but the fact of the matter is we can never get an exact answer.

Answer 5

(x+y)² = x²+2xy+y²

(x+y)² = (x+y)(x+y)
= x(x+y) + y(x+y) {distributive law)
= x²+xy + xy+y²
= x²+2xy+y²

Here is a magic trick. Use a calculator with square roots, for this:

Pick a number, any number. You may have to remember it, below.
Square it.
Add twice the original number.
Add one.
Take the square root of this.
Subtract your original number.
I bet your answer is 1 (or maybe .999999...)
The answer may be .999999..., if your number was not an integer, because of round off error.

The trick involved building up x²+2x+1, which we know is (x+1)². So, then we can take the square root, and subtract x, to get 1.

Answer 6

f(x) = x^x
and y = f(x)

I dont see the problem, iy is already explicitly written as function of x

ah !
ok x(y) is what you want ..
[let me see] u coud take the log on both sides ,.. so logy=xlogx next u can subsitute z=xlogx and then you get logy=z, no non this doesnt work mayb e you should uise primes ? and dirichelet function ...

Answer 7

y=x^x
take natural logs of each side
lny=x*lnx
raise e to the power of
each side
e^lny=e^(xlnx)
>>y=e^(x*lnx)

i hope that this helps

Answer 8

x^x=y
xlnx=lny
lnx+x*1/x=1/y*y'
lnx+1=y'/y

y'=x^x(lnx+1)

I can't wok out the x(y) only but y'.

wishing you'll be happy
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Answer 9

go to ln (neperain logarithm)

By definition x = e^lnx

You know that (e^b)^c = e^(bc)

so x^x = e^((lnx)x )= e^x(lnx)

you can write e^x(lnx) = y = e^lny

And lny = xlnx

Answer 10

I've never seen such this prob..

I'm not sure following can be the solution you want...
But just a moment

x^x=y
xlnx=lny

lnx+x*(1/x)=y' * (1/y)
lnx=(y'-y)/y

x=e^((y'-y)/y)
....

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