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/sci/ - Science & Math

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8173009 No.8173009 [Reply] [Original]

Find the exact value of x^x = 2

Wizard mode: No Wolfram Alpha

>> No.8173011

>product log

>> No.8173015

define exact value

>> No.8173021

A value in terms of functions of the given numbers i.e. not a decimal approximation

>> No.8173203

>>8173009
x = sqrt2

>> No.8173206

>>8173009
first x is 2 and the second x is 1

>> No.8173208

Why the fuck are you idiots unable to ask a proper question?

The exact value of x^x = 2 is x^x. It is also 2.

What the fuck are you trying to ask me?

>> No.8173210

>>8173206
>>8173206
>>8173206
>>8173206

/thread

>> No.8173214

>>8173203
No, because 1.5^1.5 < 2 and 1.6^1.6 > 2 the answer exists between them. Also sqrt(2) = 1.41... doesn't work when checked.

>> No.8173218

>>8173210
> implying x =/= x
shiggy

>> No.8173219

>>8173214
Yeah, sqrt2 is between them. Nobody said sqrt(2) = 1.41...

Are you fucking retarded?

>Asks a question
>Gets the correct answer
>NO LEL IT'S NOT [WRONG ANSWER] NICE TRY THOUGH LEL I GOT YUU hue!

>> No.8173220

>>8173009
x=2^(1/x)=2^(1/(2^(1/x)))=...

>> No.8173222

>>8173218
Aww how's high school?

That's how math works when you aren't a gay engineer. x doesn't equal x unless it's defined to and there are plenty of cases in abstract algebra where it doesn't.

>> No.8173262

>>8173218
undergrad spotted

>> No.8173270

>>8173222
Show one from a text

>> No.8173278

A number such that the reciprocal of x is lg(x)
How would you even calculate that number though?

>> No.8173279

>>8173270
Let X={a,b}X={a,b}. Then the relation {(x,y),(y,x)}{(x,y),(y,x)} is symmetric but not reflexive.

Very simple example.

>> No.8173282

>>8173279
My bad. Let me rewrite that:

Let X={a,b}. Then the relation {(x,y),(y,x)} is symmetric but not reflexive.

>> No.8173297

>>8173222
>>8173262
What the fuck are you guys talking about? Are we working outside ZFC now?

>> No.8173302

>>8173279
But that's not equals. Its not even an equivallence relation.

>> No.8173303

>>8173009

x^x= 2

x= 2^(1/x)

x/(2^(1/x))=1

(x/(2^(1/x))) * ((2^x)/(2^x)) = 1

((2^x)x)/(2) = 1

(2^x) x = 2

2^x = 2x

thus the only number possible to be x is 2, which is impossible

where the fuck is my error /sci ?

>> No.8173305

>>8173282
Lol what are you even trying to do here?

You found a single relation that isn't reflexive? You just proved some relations aren't equivalence relations. What do you think you you accomplished?
Why didn't you use X in your relation?

>> No.8173306

>>8173303
> 2^(1/x)
using transcendental functions

>> No.8173307

>>8173297
No. We aren't.
>>8173302
That's the point. Congrats.
We're being pedantic about it but that was the point, yes.

>> No.8173310

>>8173307
Just because you're being retarded doesn't mean you're being pedantic. Nothing you said so far is true.

X=X always

>> No.8173313

>>8173305
>>8173310
You asked for an example where something doesn't "equal" itself. A non-reflexive relation is one such example. You're the weird one here, I just did what you or someone else told me to here: >>8173270

>> No.8173318

>>8173303
the last step you multiplied instead of dividing by x.

>> No.8173320

>>8173313
>an example where something doesn't "equal" itself
> isn't an equivallence relation or equals
coulda just say > if you were just choosing arbitrary relations

>> No.8173326

>>8173303
If I'm not mistaken [2^(1/x)]*[2^(x)] = 2^[(1+x)/x]
When multiplying terms with the same base you add the exponents.
(2^1)*(2^1) = 2^2 = 4

>> No.8173328

>>8173009
You can't solve it algebraically. Trick question I'm afraid. I think someone has a proof of that I think.

>> No.8173329

>>8173328
How would I go about solving it non-algebraically?

>> No.8173334

>>8173328
How is it solved non-algebraically? Dumb undergrad here is curious.

>> No.8173337
File: 246 KB, 1600x1200, master_race.jpg [View same] [iqdb] [saucenao] [google]
8173337

>>8173329
I'm not sure. I've only taken up to differential equations and calc 3. I haven't taken anything above that. As far as I've read on Wikipedia it has to do with the Lambert function, but I don't really understand it at all. I don't even know when you would learn that as a math undergrad desu. The fact that nobody in this thread has mentioned that might indicate that it probably isn't taught.

>> No.8173339

>>8173334
Look up Lambert function if you wanna find out.
http://2000clicks.com/mathhelp/BasicSimplifyingLambertWFunction.aspx
Here's a good place to start. From my understanding it's kind of like how the natural logarithm lnx is the inverse of the exponential function e^x, W(x) is the inverse of xe^x. Kind of weird.

>> No.8173342

>>8173337
see
>>8173011

>> No.8173344

>>8173342
Ok?

>> No.8173348

>>8173344
>The fact that nobody in this thread has mentioned that

>> No.8173352

>>8173203
sqrt(2)^2 = 2 so no

>> No.8173353

>>8173348
I hadn't realized the Lambert function and product log were the same thing.

>> No.8173359

>>8173206
loll

>> No.8173370

>>8173009
I don't think that's possible OP

>> No.8173431

It's obviously between 1 and 2, testing 3/2 got a pretty good approximation, 4/5 is too big so it's between 3/2 and 4/5. Just guessing and found 31/20 to be even better.

What (if there is one) is the general rule for getting a series approximation of n roots?

>> No.8173519

>>8173009
[math] \frac{94241}{60426} [/math]

>> No.8173521

>>8173519
Only five 9's? Pathetic. You should be ashamed of yourself.

>> No.8173522

1.559610465.....

>> No.8173525

>>8173522
1.559610469459.....

>> No.8173528

>>8173525
1.559610469462369349970388768765002993284883511843091424719

>> No.8173538

I really want to know how to do this. Somebody figure it out. Tonight I learned that all irrational numbers have an infinite continued fraction representation. Neat. Still don't know how to get this stupid problem.

>> No.8173977

>>8173538
See
>>8173339
It has an explanation for how to solve x^x= 16

>> No.8173998

>>8173218
Have you never seen a parabola that crosses the x axis twice?
X[sub]1, X[sub]2

>> No.8174047

can u tell me who is this men that i see in many memes recenlty ( i am italian, i never saw this man)

>> No.8174069

>>8173282
Equality by definition must be an equivalence relation and hence reflexive.

>> No.8174382

>>8173352
sqrt(2)^sqrt(2)=2

>> No.8174392

>>8174382
Nice troll

>> No.8174584
File: 603 KB, 984x1124, intuitive wildberger.png [View same] [iqdb] [saucenao] [google]
8174584

>>8173009
There is no solution.

>> No.8175764

>>8173009
I've gotten as far as seeing the exponential growth of x**x and decided it must have to do with e. Also I'm not in college so I'll pull out of my ass that I can log(2) this bitch. How am I doing? Something like e**log(2) with...some missing part of the log? I'm stuck unless I either already got it or am way off.

>> No.8175767

>>8174382
That's 2+(1/root2) so no.

>> No.8175768

>>8175764
Wait no I'm fucking retarded that's just =2 isn't it? My brain is melting.

>> No.8176950

So far my aproximation is e^(1-sqrt(1-ln2))

>> No.8177023
File: 9 KB, 235x236, 342.png [View same] [iqdb] [saucenao] [google]
8177023

>>8175768
it should be equal to 2 at the end so..

>> No.8177050

>>8173009
Easy
We look for [math] x \in \mathbb{Z}_3 [/math]
Then x=5 is the solution, as
[math] 5^5 = 5^3*5^2 = 5*5^2 = 5^3 = 5 = 2 [/math]

>> No.8177053

>>8173313
equal means equivalence relation, genius

>> No.8177127
File: 692 KB, 2560x1440, Message_1467410358710.jpg [View same] [iqdb] [saucenao] [google]
8177127

OP here, the answer is :

e^(W(ln (2))

Pic related shows how you get there.

>>8173339
This guy was on the right track, the W() function is the same as the Lambent function, you can see how it being the inverse of xe ^x is used in the solution. (In the link he posted there is actually a solution for x^x =16)

I love how simple this question looks and the fact that you need a transcendental function in order to solve it. It was interesting to see how everyone approached it and that the discussion moved to the reflexive property Lol.

>> No.8177137

>>8177127
Sorry about the image on mobile

>> No.8177573

>>8177050

But you've forgotten that, since 5 is prime (and hence 5^2 as well), the answer has to be nonprime, in this case either 4 or 6 (for n powers of 5).

>> No.8177614

>>8173009
The solution is almost certainly irrational which means that OP has already posted the solution in the very formulation of the question. If you can generate some sort of analytically interesting but incalculable infinite sum then I'll buy you a beer but any computer capable of floating point arithmetic can get enough digits in about x^x = 2 seconds.

Probably does seem fun though.

>> No.8177632

>>8177614
You can do square roots in 'long squares' using a pretty much identical algorithm as long division.

>> No.8177640
File: 105 KB, 1920x1080, kekus.jpg [View same] [iqdb] [saucenao] [google]
8177640

>>8173206

>> No.8177641

>>8173208
wew mate

we're looking for the value of x

>> No.8177794

>>8173222
You're wrong.

>> No.8177823

x=1.51 approx.

>> No.8177826

>>8177641
>Find the exact value of x^x = 2

>> No.8177852

>>8177127
[math]x^x=2[/math]
[math]x\ln x = \ln 2[/math]
[math]e^{\ln x} \ln x = \ln 2[/math]
[math]\ln x = W(\ln 2)[/math]
[math]x=e^{W(\ln 2)}[/math]
Alternatively,
[math]x=\frac{W(\ln 2)e^{W(\ln 2)}}{W(\ln 2)}=\frac{\ln 2}{W(\ln 2)}[/math]

>> No.8177867

>>8177852
Thanks for showing your work - from the first reply I was able to look up the W function but was curious what the process looked like as I have very little education in this area. Much more straight-forward than I anticipated.