Nice way to approximate X^X = 5 without graphing calc

Discus: College Confidential Café: 2004 Archive: Nice way to approximate X^X = 5 without graphing calc
 By Thermodude (Thermodude) on Tuesday, July 06, 2004 - 10:46 pm: Edit

...read Tongos question about whether it is possible to solve X^X = 5 without graphing calc...

..for fun...I just thought of a nice way to approximate X^X = 5 without using a graphing calc. (This would work for X^X = C...with C being a constant.)

let a = x

x^a = 5

Now, with 4=2^2 being close to 5...let us first approximate a as 2. Solve x^2 = 5 for x to get a new value. Plug in this value back into a...and solve for x again. Keep on repeating until your approximation doesn't change that much. Using x^2 * a^(a-2) = 5 will also work as well. This method works...and is a way to solve X^X = 5 (or X^X = C) without a graphing calculator, thought it is rather lengthy in many cases.

 By Tongos (Tongos) on Thursday, July 08, 2004 - 02:35 pm: Edit

a cool way (tongos method)
x^x=z
input a number for x. any number. a good number is the best x-log(z)/log(x) when small. but a big number is good to.
let's say x^x=5
we input 2 for x. do log(5)/log(2) to get the exponent.
set up two equations 2^x and y^(log(5)/log(2)
do the derivative of each functions
2^x(In2) and (log(5)/log(2)y^(log(5)/log(2)-1)
plug in x as being log(5)/log(2) and y as just two. now multiply the first solution by log(5)/log(2) and the second by 2 and add them together, now divide by the (solution(1)+solution(2)) which we will call w. we do the same thing with w as we did for x and the process keeps going. w will now be very close to the real value. we should get 2.120514042^x=5
do the derivative 2.12....^x(In2.12....)-solution1 and (log5/log2.12......)y^(log5/log2.12.....-1)-solution2 plug the value of (log5/log2.12......) for x and 2.12.... for y. now we multiply solution1 by log5/log2.12..... and solution2 by just 2.12.... now we divide by solution1+solution2, and we are practically there. you don't have to use the process to many times, only two, was it correct by two decimal places, only three did my calc read the answer.
reason for the solution: the reason why it works is because it accounts the change also. because if you could just add 0.1 to the exponent it could change the answer by a whole bunch. or if you add 0.1 to the number in front, it may not change that drastically. this method tries to find a balance.
your method is pretty cool to thermodude. write back.

 By Tongos (Tongos) on Saturday, July 10, 2004 - 12:27 am: Edit

infinite complex solutions
i will treat a+bi as x+yi=(real number)
x and y could fluctuate to yield the real.

 By Thermodude (Thermodude) on Monday, July 12, 2004 - 12:51 am: Edit

Wow!...that's a nice method Tongos...totally BURNS using Newton's method.....a lot simpler this way...