I have an equation X^x=a where a is any real number.. can anyone solve it?

Question

Solve this equation please

Answer 1

I assume that you mean X and x are the same.
This equation has one real solution for a≥1 and a=(1/e)^(1/e).
It has two real solutions for (1/e)^(1/e)< a <1
(for instance, if a=0.8, x = 0.73953365... or x = 0.09464971...)
It has no real solutions for a < (1/e)^(1/e).

It cannot be solved explicitly in terms of finitely many functions of a. However, a partial solution in terms of the "infinite exponential" is:

x^x = a
x = a^(1/x) = a^((1/a)^x) = a^(((1/a)^((1/a)^x)) ...
or x = a^(((1/a)^((1/a)^((1/a)^...))))

This is a partial solution because it does not converge for all values of a for which the original equation does.

If you need to calculate exact values, try my program:
http://mudandmuck.com/str2/x^x.exe

Otherwise, you can calculate the values with some numerical interpolation method such as Newton's. Using this method, you can derive an iterative function which will converge to the solution(s) after a suitable number of iterations:

x(n+1) = x(n) + [a - x(n)^x(n)] / [x(n)^x(n) * (1 + ln(x(n))]

You must choose a "seed", (or initial guess) number for x(0). For a>1, log(a) should suffice. Then, x = lim x(n) as n → ∞ .

Answer 2

X^x = a

logX a = x

x = log a / log X or ln a / ln X

I haven't done this in a long time, so it's most likely wrong. I just tried to simplify it, because there's not enough information to solve anything.

Answer 3

X^x=a so X=(x)root of a , and if x is a even number then a can't be a negative number. so the only posibble result i can give you is these two numbers {1;0} because all roots of 1 is 1 and all roots of 0 is 0, the other is unknown.

Answer 4

If by X and x you mean the same variable, then Oracle is right.
If X is a constant, then Abdullah R's answer is correct

Answer 5

Numerically it can be solved

Answer 6

only numerically