/sci/ - Science & Math

x=2

>>2797573

Funny.. If I had 300k sitting around I'd be out doing useful, fun things, instead of trolling 4chan.

lewl

Take an x out of both sides

2=x^(x-1)
Since nothing to the power of anything is two except two to the power of one,
x-1=1
x=2

>>2797593
>dividing by x

>>2797593
*nothing to the power of itself minus one is 2
inb4 negative exponents, The only way you can get that is if you subtract one from zero and that gives you a negative one

I guess zero is an answer, too, then

>PhD. in afro american studies
>Any KFC i want
>19K starting

>>2797592
You know, I think you'd be surprised.

http://www.wolframalpha.com/input/?i=solve+2x%3Dx^x

{x->2}
x ~~ 0.3463233622785809...

x=2

log will be used to denote the logarithm with a base of x

log2 + logx = xlogx
2 + 1 = x + 1
2 = x

also, trivial x=0,

2(2) = 2^2

First post nailed it, unless Im missing something here.

>>2797699
x is also 0

>>2797630
And how do you suppose they calculated that?

>>2797690

0^0 is 1

2x=x^x
2(2)=2^2
4=4
x=4
prove me wrong!

>0^0=0
Really, guys?

This thread shows me that most people on this board know little to nothing. I'm willing to bet most of you all love to comment on the acceleration of particles near black holes and whatnot as well.

is there any way to solve this algebraically? There's two solutions, the first is trivial to solve by inspection but the second is not.

http://www.wolframalpha.com/input/?i=plot+x^x+plot+2x%2C+x%2C0%2C3

>>2797690
Wait, how the fuck?

>>2797831
Not likely since its a transcendental. You can estimate the solution using local linear approximations.

>>2797751
2(4)=(4)^(4)
8=\=256

Its 2. 2x2=2^2 4=4.

>>2797690
spot on except the x=0 bit

>ITT: OP's PhD program didn't teach him how to use logs

Lambert W function

but why is 0^0 = 1

other than the arbitrary rule mathemagicians fabricated that x^0= 1

>>2798768


2^1 = 2x(?) wtf does this even mean
2^0 = 2x(?) wtf does this even mean

draw me a picture

>>2798729
>>2797690

how the fuck is x*logx(x) equal to x + 1? and how the fuck is logx(2) equal to 2?

logx(2)+1=x

same position you were in before. hope you were trolling nigger, it fucking worked

>>2798768
>0^0 = 1
Protip: It's not. It's indeterminant. Wolfram-Alpha that bitch if you don't believe me.

I wanna say that you start by exponentiating both sides.

e(2x) = e(x^x) = e(x^2) = e(2x)...

Shit, maybe I did something wrong...

2x = x^x
ln(2x) = ln(x^x)
ln2 + lnx = xlnx
ln2 = xlnx - lnx
ln2 = lnx(x - 1)
2 = x(x - 1)
2 = x^2 - x
x^2 - x - 2 = 0
(x - 2)(x + 1) = 0
x = {-1, 2}
Discard negative, since it's a logarithm
x = 2

>>2798932

>mfw

>>2798932

How did you move from

ln2 = (lnx)(x - 1)

to

2 = x(x - 1)

If you took the exponential of either side you would get:

2 = e^((x-1)(lnx))

Now separate the exponential. You shouldn't get

2 = x(x - 1).

>>2798974
If log_b(a) = log_b(b), then a = b
It's one of the lesser known properties. I'm too lazy to derive the proof, but you can find it somewhere if you really care.

>>2799005
Actually disregard this, if you look at the first two lines of my work and go in reverse, you will see why this is true

>>2799005

But it's not just

ln(2) = ln(x(x-1))

It's

ln(2) = (x-1)ln(x)

>>2798754
you motherfuckers

>>2797573
>Madthematics in D.pH.
>job want I any
>30k0 finishing
x=2

>>2799005
It's the definition of a function that you can't provide the same input and receive a different output.

hey guys it's .34632336 and 2

>>2799020
ln2 = xlnx - lnx
ln2 = ln(x^2 - x)
2 = x^2 - x

Problem?

>>2799075

yeah, big one... you must be trollin...

xlnx =/= ln(x^2)

xlnx = ln(x^x)

>ln(2) = xlnx - lnx
>ln(2) = lnx^2 - lnx
>ln(2) = ln(x^2/x)
>ln(2) = lnx
x=2

QED bitches

>>2799075
how do natural logarithms get away with what other logarithms can't?

>>2799086
ln2 = ln(x^2) - lnx
ln2 = ln(x^2 - x)

Problem?

>>2799020
well...shit

>>2797756
0^0 is an indetermination, it can be 0.
You have to find de limits.

x = 2.

>>2797573
2x = x^x
2x = x.x.x... (with length x)
2 = x.x.x... (length x-1)
x = 2/(x.x...) (length x-2)
OH SHI -

>Degree in Philosophy
>Any job I want
>500K starting

>>2799118
0^0 is a 1 by definition.
It cannot be 0.

>>2799147

ricky gervais actually has a BA in philosophy

0^0 is exactly the same as 0/0 which which is indeterminate. So 0^0 is NOT 1! the reason anything other than 0 to the power of zero is 1 is because, anything to the power is itself, and to lower that power you must divide the base by itself.

i.e. : 2^2=4, 2^1=(2^2)/2=2, 2^0 = (2^1)/2=1.

To say 0^0 = 1, is the same thing as saying 0/0 = 1. Does it?

>>2799159
No, you misunderstand entirely
0^0 is 1 BY DEFINITION
It has been DEFINED that 0^0 is ONE. Its not something you have to show, that's like saying show that a bijective function is injective and surjective, you don't need to fucking show it because that is the DEFINITION

>>2799173
Except it hasn't been defined you fucking idiot

>>2799173
If you don't understand this, try googling "0^0" first google will simply state 0^0=1 and ask if you want to learn more about calculator, then (in your disbelief that google would tell you that so readily) you can tell it to do a google search on it, you'll be led quite quickly to the definition.

>>2799173

you're thinking of 0! = 1

0^0 is undefined.

Wolfram Alpha's calculators says indeterminate: http://www.wolframalpha.com/input/?i=0^0

Googles calculators says 1: http://www.google.com/search?q=0^0

>>2799148
Sir, are you kidding me?
2^0=1
3^0=1
1^0=1
But 0^0 is NOT 1, is a motherfucking indetermination, it knows a 14 years old kid, LOL!!!
Please, learn mathematics.

None of the attempted solutions in this thread are any good
Can you do it with infinite series maybe?

>>2799176
0^0 is an indeterminant form, so you can define it to be a variety of things, in the case above it goes without saying that it would be defined as 1 (the most common definition for 0^0) in fact unless otherwise stated this is the assumed definition in any mathematics.

>>2797573
divide both sides by x
there you go

>>2799210
Ony gives one solution. Theres a second

i

I don't need to show my work, you know I'm right, now give me a JOB!

>>2797732

k * (log(2)/W(log(2))) where k is in R

Here's is how to solve it:
2x = x^x
(2x)^(1/x) = x
(2^(1/x))*(2^(1/x))=x
2^(1/x)=(x/(x^(1/x)))*(((x^(1/x))/((x^(1/x)))
2^(1/x)=(x*(x^(1/x)))/x
(2^(1/x)^x = ((x*(x^(1/x)))/x)^x
2=((x*(x^(1/x)))/x)^x
2=(x^(1/x))^x
2=x

So Simple.

I don't think this can be solved algebraically, even with the Lambert W function

>>2799297
It can look at the post above you. I upload a pic of my work so that it is easier to read in a second. I am in Algebra 2, so this can't be too hard.

Never mind I can't take a picture. I can't find my camera. Anyway, my post above is how you solve it.

>>2799301
>>2799289
You missed a solution, and I have no idea what your doing half the time

Like how do you get from here:
2^(1/x)=(x/(x^(1/x)))*(((x^(1/x))/((x^(1/x)))
to here:
2^(1/x)=(x*(x^(1/x)))/x

shouldn't it be 2^(1/x)=x*(x^(1/x))/(x^(1/x^2))?

>>2799317
Sorry,
*shouldn't it be 2^(1/x)=x*(x^(1/x))/(x^(2/x))?

>>2799317
No.
2^(1/x)=(x/(x^(1/x)))*(((x^(1/x))/((x^(1/x)))
to:
2^(1/x)=(x*(x^(1/x)))/x

is just simplification.
(x^(1/x))*(x^(1/x)) and be simplified to just x, and that is being divided by x*(x^(1/3)).

Every time you see ^(1/x) just replace it with xroot if that helps.

>>2799341
opps i meant
>>(x^(1/x))*(x^(1/x)) CAN be simplified to just x, and that is being divided by x*(x^(1/3))

>>2799341
I'm pretty sure x^(1/x)*x^(1/x)=x^(2/x)

basic index rules

4^(1/4)*4^(1/4) != 4
3^(1/3)*3^(1/3) != 3

>>2799289
Also the 6th line in my solution is missing a parenthese sorry. Here it is fixed:
>>(2^(1/x))^x = ((x*(x^(1/x)))/x)^x