Solving x^x=2^2048

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#ChallengingMathProblems #ExponentialEquations

PLAYLISTS 🎵 :

[paulchapman]

I did something pretty similar to the first method, but without breaking down the powers of two all the way.
2^2048 = 2^(2*1024) = (2^2)^1024 =
4^1024 = 4^(2*512) = (4^2)^512 =
16^512 = 16^(2*256) = (16^2)^256 = 256^256.

[gregwochlik]

I found this too easy. I decomposed 2^2048 -> 4^1024 -> ... 64^64. My guess is that x = 64.

[hazartuankzltan]

this channel has 2 types of viewers
1- actually smart people
2- ppl like me dumb but still watches even they dont understand anything

[sngmn]

We can use Lambert W-function. Another way is an increasing base by decreasing exponent.

[Firefly]

To not miss out on complex solutions:

x^x = 2^2048
ln(x^x) = ln(2^2048) + 2*pi*n*i
x*ln(x) = 2048*ln(2) + 2*pi*n*i
ln(x)*e^ln(x) = 2048*ln(2) + 2*pi*n*i
ln(x) = W(2048*ln(2) + 2*pi*n*i)
x = e^W(2048*ln(2) + 2*pi*n*i)

n is an integer
2048*ln(2) + 2*pi*n*i does not equal 0

[Crestalus]

2^2048 = 2^(2^11)
We can see that every time we halves the power of 2^(2^11), the base will be squared.
2^(2^11) = (2^(2^n))^(2^(11-n))
We have
2^n = 11 - n
2^n + n = 11
By graphing this out we can see n = 3
(2^8)^(2^8)
so x = 2^8 or 256

[AyariMakrem]

By using w-lambart function( called also productlog) which is the inverse function of the function f(x) = x exp(x). The resultat is: exp(productlog( ln(2^2048))) =256

x^x = 2^2048 ; we know that x = e ^ ln x, so we obtain:
(e^ln x)^(e^ln x)=2^2048
e^(ln x e^ln x)=2^2048 ( we use the rule (a ^ b )^c = a ^(b c) )
Let's apply ln on both side
Ln(e^(ln x e^ln x))=ln(2^2048)
ln x e^lnx = ln(2^2048)
On the left we recognise the forme xe^x with x has the value ln x. So we can apply productlog on both side:
Productlog( ln x e^ln x) =productlog(ln(2^2048))
And since productlog (x e^x)=x for any x so:
Ln x = productlog(ln(2^2048))
Let's apply exp function on both sides:
e^ln x = exp(productlog(ln(2^2048)))
x = exp(productlog(ln(2^2048)))
When we give this formula to any calculator supporting productlog function we obtain 256.

[beirirangu]

OR
a^b=(a^2)^(b/2)
it won't always get you the exact answer, but in this case it will:
2^2048 = 4^1024 = 16^512 = 256^256

[БудівельникДОБРЯК]

I thought by this way:
2^2048 = 2^(8*256)=(2^8)^256 = 256^256;
Thus x^x = 256^256 => x = 2^8 = 256;
Our final answer is x=256;

[rhc]

We can also use the lambert W function: x=e^W(ln 2^2048)

[JohnRandomness]

x^x = 2^(8*256) = (2^8)^256 = 256^256. This was sort-of trial and error or guess and check.

An order-of-magnitude approximation: 2^u = 10^(.3u), so 2^2048 = 10^614.4. I'm going to bump it up, because 2^10 > 10^3, so I'll try 10^615. Valid within two or three orders of magnitude.

[ichigo_nyanko]

To find the first solution (256) find the geometric mean of 2 and 2048 and multiply by 4 (as, if you half the top power and compensate the bottom x to make it equal, you are going up by powers of four in the bottom x and up by two in the top.)

[gloystar]

I solved it in another way, I took natural log of both sides to get the exponent down in front, after that it's just Newton's method. Choose x=1 as your starting point for iterations, and in just 5 iterations you will end up with x=256. But of course you'll need calculus for that. It's amazing how Newton's method is very powerful in solving these kinds of equations.

[nintishia]

2 pow 2048 = 4 pow 1024 = 16 pow 512 = 256 pow 256. Hence x = 256.

[WahranRai]

We must find a and b divisors of 2048 which verify : x^x = 2^2048 = (2 ^a)^b
a*b = 2048 and b = 2^a then a*b = a*(2^a) = 2048 = 2 ^11 --->
let a = 2^y ---> 2^y * 2^(11-y) = 2048 = 2 ^11
With trials :
y= 3 ---> a=2^3=8 and b =2^8= 256 ---> 2^2048 = (2^8)^256 = 256^256 ---> x=256

[miniprime]

I found it by W function. x^x=2^2^11 so let make log both side then ln(x)*e^(ln(x)) = 2^11*ln(2). let make W at both side then ln(x) = W(2^11*ln(2)). Here, since ln(x) = W(2^11*ln(2)) the x = e^W(2^11 * ln(2)) = 256. Therefore, x=256. This is my guess.

[wenyilu]

another easy way:
x^x = 2^2048 = 2^{y * [(2^11)/y]} =( 2^y)^[(2^11)/y]
then two equations:
1) x = 2^y
2) x = (2^11)/y
leading to: 2^y = (2^11)/y
Now let: y = 2^t
next: 2^(2^t) = 2^(11-t)
further: 2^t = 11 - t
up to this point, it is quite noticeable that t = 3. Everything follows.

[overlordprincekhan]

BTW, you should also attempt to cover the other 4 complex solutions too.

[vladimirkaplun]

Suppose
x=2^n where n is integer
Then
x^x=((2^n )^(2^n )=2^(n*2^n )
Hence
n*2^n=2^11 or n=2^(11-n)
Thus n=2^m
Or 2^m=2^(11-2^m )
Or m=11-2^m hence m is odd and then m∈{1, 3}
Form where m=3;
The rest is the positive derivative for x>2 for example

[jmj-sq]

lb(x^x)=lb(2^2048)
xlbx=2048
x=2^a
a•2^a=2048
lb(a•2^a)=lb2048
a+lba=11
a=8
x=2^8

Ib means binary logarithm. Thank you!