How do you solve: X^X=17 without approximation or trial and error?

Question

other algorithims are okay to use.

Answer 1

Well, if you have the lambert_W function available, you can use it thus:

x^x = 17
x ln x = ln 17
ln x e^(ln x) = ln 17
ln x = W(ln 17)
x=e^(W(ln 17))

Answer 2

Ok you want an exact numerical method to compute X=√17 .

No, it doesn't need to be done by approximation.

There are probably tons of ways, here's the first one that springs to mind:
A summation by the Generalized Binomial Theorem

X = √17 = 17^½ = (1+16)^½ = 4 * [ (1/16 + 1) ^½ ]
So X/4 (call it X*) = (1/16 + 1) ^½

Then by the Generalized Binomial Theorem
(1/16 + 1) ^½ is the case x=1/16, y=1, r=½
which does converge since |x/y| = 1/16 < 1
= Σ k=0..∞ [ r_C_k * x^k * y^(r-k) ]
Noting y^(r-k) will always be 1...
= Σ k=0..∞ [ r_C_k * (1/16)^k ]
= Σ k=0..∞ [ r_C_k * 2^(-4k) ]


where the r_C_k are defined for non-integers k , and you can see the definition on the Wikipedia link below.
Let's compute ½_C_k for k=0,1,2,3,4...
½_C_0 = 1
½_C_1 = ½
½_C_2 = ½(½-1)/2! = -1/8
½_C_3 = ½(½-1)(½-2)/3! = +1/16
½_C_4 = ½(½-1)(½-2)(½-3)/4! = -5/128
...
I think you can crunch out a general result for ½_C_k with a little effort

So the first few terms:

(1/16 + 1) ^½ = Σ k=0..∞ [ ½_C_k * 2^(-4k) ]
= Σ k=0..∞ [ ½_C_k * 2^(-4k) ]
= Σ k=0..∞ [ ½_C_k * 2^(-4k) ]

= ½_C_0 * 1 + ½_C_1 * 1/16 + ½_C_2 * 1/256 + ½_C_3 * 1/4096 + ½_C_4 * 1/65536...

= (1 * 1) + (½ * 1/16) - (1/8 * 1/256) + (1/16 * 1/4096) - (5/128 * 1/65536) ...
= 1 + 1/32 - 1/2048 + 1/65536 - 5/8388608...

and finally don't forget we wanted 4 * the above quantity

X = √17 ≡ 4 + 1/8 - 1/512 + 1/16384 - 5/2097152...
= 4 + 0.125 - 0.001953125 + 0.000061035 ...
= 4.1231055 after only 5 terms i.e. we already have 7 significant figures of √17!!
But again the point is this is an *exact* method.
(As to the rate of convergence, since the sign is alternating and the ratio between successive terms ≈2^(-4) , then the rate of convergence for every two terms is ≈2^(-8) = 1/256, so you get ~two decimal points for every two terms).


(PS you can see it will be exact in decimal or binary or any base, but it's even easier when evaluated in binary; the coefficients will have very simple binary representations.)

Answer 3

I think that you would generally have to use approximation techniques to solve this sort of problem.

Answer 4

x^2 = 17
thin take log both side
2logx = log 17
logx= log17/2
logx= log17-log2

and see the value from table and pu!!!!!! answer will come out.....

Answer 5

+/-sqrt17 its decimal conversion has to be done using approximations