How do you solve X ^ X = 6?

Answer 1

Unfortunately, this equation cannot be solved analytically. That is, in the end the only way to solve this equation will be to plug in values for X until you find one such that X^X = 6.

However, we can try to learn about the structure of the equation to give us a good idea of which values to guess. Take a look at Newton's method, which I give a link to in the sources. You would apply Newton's method to the expression:

X^X - 6

It will find points where X^X - 6 is approximately equal to 0. Since X^X is relatively "well behaved," then Newton's method will converge to a solution.

One such solution which results from using an iterative method like Newton's method is X=2.231828624409, where "=" is approximate. Note that:

2.231828624409^2.231828624409 = 6.000000000000

In other words, 2.231828624409 is very close to the "real" answer.

Answer 2

That's tough (to me). Someone may give you an exact answer, but in the meantime, you can take a numerical approach.
You know that 2^2 is 4, and that 3^3 is 27, so x must be slightly more than 2. Now try successive values, adding 1 decimal at a time. Always look for numbers that are just UNDER 6 (if you go over 6, you cannot correct it by adding more decimals):
x=2.5 ──►x^x = 9.9
x=2.1 ──►x^x = 4.7
x=2.2 ──►x^x = 5.7
x=2.3 ──►x^x = 6.8
x=2.25 ──►x^x = 6.2
x=2.21 ──►x^x = 5.8
x=2.22 ──►x^x = 5.9
x=2.23 ──►x^x = 5.98
x=2.235 ──►x^x = 6.03
x=2.231 ──►x^x = 5.991
x=2.232 ──►x^x = 6.002

So you already know the answer to 3 decimal places (x=2.231). You can repeat this process indefinitely to get the answer however accurate you like.

Answer 3

Well, it is between 2 and 2.5. Taking the log of both sides doesn't help that much. If you want an approximate answer, you can graph y = x^x and y =6 and see where they intersect.

Answer 4

One way would be to solve the problem numerically. You know that 2^2 = 4, and 3^3 = 27, so the answer is between 2 and 3. 2.1^2.1 ~4.75; 2.2^2.2~5.67; 2.3^2.3~6.79, so now you know the answer is between 2.2 and 2.3. Now try 2.21^2.21~5.77; 2.22^2.22~5.87; 2.23^2.23~5.98; 2.24^2.24~6.09, so now you know the answer is between 2.23 and 2.24. Keep going in this way and you can get as many digits as you need.

You've got what's called a transcendental function, which is a function which is algebraically independant of the variable.

Answer 5

solve it by iteration plug in a value and keep a chart. You get close to ther correct number