How would you solve for x, x^x=n?

Answer 1

First I would take the log of both sides:

x ln x = ln n

Then I would rewrite this as:

ln x e^(ln x) = ln n

Here I would apply the Lambert W fuction:

ln x = W(ln n)

And exponentiate:

x = e^(W(ln n))

Solving this algebraically without using the Lambert W function, or some related non-elementary function, is impossible.

Answer 2

Let me try by the procedure given below. We shall assume that n is a natural number.
First if n=1 then x=1
if n=4 then x=2
if n=27 then x=3
if n=64 then x=4
if n=125 then x=5 etc
Now plot a graph with n along the x-axis and x along the
y-axis. From the graph, we can read the value of x (i.e. root)for other values of n.
Clearly, for n between 4 and 27, x will lie between 2 and 3 which can be read from the graph.
I cannot think of an analytical solution for this problem.

Answer 3

What you have is known as a transcendental function. This is defined as a function which cannot be solved algebraically. Sorry, but there is no solution which you can solve for all n. You can't really "solve" it for most specific numbers.

Although sometimes the solution can be given compactly (if n=4, the x=2 for example), very often the answer cannot be described by simple numbers and functions. You might describe it as the limit of a sequence.

Answer 4

The function y = x ^ x == Exp (x*Ln(x)) is defined only for x>0. For x -> 0, the limit is (L'Hospital's rule!):

y(0) = Exp(Lim[(1/x)/(-1/x^2),x->0]) = Exp(Limit[-x, x->0]) = Exp(0) = 1

The derivative is:

y' = Exp(x*Ln(x)) (1*Ln(x) + x*(1/x)) = x^x*(Ln(x) + 1)
y' > 0 for Ln(x) + 1 > 0 <=> Ln(x) > -1 <=> x > 1/e increasing
y' < 0 for x < 1/e decreasing
The function has a minimum for x = 1/e = 0.368... and its y = Exp(-1/e), which is aproximatelly 0.692...

You have to choose one of the intervals [0, 1/e) or (1/e, Infinity) where the function is monotonic. There it certainly has an inverse function (because it is continous and monotonic). The interval (1/e, Infinity) is better because y is in the range (Exp(-1/e), Infinity). If n is a natural number, we can even take the interval [1, Infinity). Unfortunatelly, the inverse function that can not be expressed .in closed form with elementary functions. It can be expanded in a Taylor series around x = 1, which can be used for aproximate calculation of the roots from the equality:
1/x == y'*(Ln(y) + 1)

and use y = 1 + Sum[An * (x - 1)^n, {n, 1, Infinity}]

Answer 5

All answer are incorrect or partly best! The is a attempt no remember if the equation holds. because of the fact the quotient x/x is given, the equation is defined for x=0, because of the fact the certainty, that there is a genuine shrink for x goint to 0 of x/x: limit_x->0 x/x=a million. So do away with the quotient a minimum of for x unequal 0. The equation now's x+x²=x². This equation could properly be rearranged by potential of unique arithmetic operations. It supplies x=0. stupid or. What does is propose? The equation does not carry for x unequal 0. this is it. it could returned be simplified a million+x=x or a million=0, this is incorrect. For 0 is holds in the stable type and in the shrink attention: 0/0+0=a million+0 unequal 0. The equation does not be valid for any actual numbers!

Answer 6

There is no algebraic way to do this.
Write the equation as x*log x- log n = 0.
Now locate the answer between 2 values.
I.e, find a value of x for which x*log x- log n is
negative and one for which it is positive.
Such a value exists by the intermediate value
theorem. Now use Newton's method to
hone in on the answer.
See any good calculus book for details!

Answer 7

if you mean X raised to the X-power is = to N,
then the X-root of N is = X

i.e. 3^3 = 27...so the cubed root of 27 = 3.

I don't understand if X is given or N is given in your question.
Good luck.

Answer 8

i think its 0.

Apple log on both sides.

lgx^x=lgn

xlgx=lgn

xlgx-lgn=0

xlg(x/n)=0

since n cannot be 0 for a real solution, x must be zero

DOne, not sure if its correct

Answer 9

Find out what N is first? There is not enough info to answer that question.

Answer 10

by applying the lessons I have learned