Crisis
Smash Ace
778-988-fivethreethreeeight
-
Welcome to Smashboards, the world's largest Super Smash Brothers community! Over 250,000 Smash Bros. fans from around the world have come to discuss these great games in over 19 million posts!
You are currently viewing our boards as a visitor. Click here to sign up right now and start on your path in the Smash community!
Phoenix Brawl 6 (Important Venue Change! Details inside)
- Thread starter ~Firefly~
- Start date
Tin Man
Smash Hero
I'M FLYING IN 1 MORE DAY 
Why don't you guys just type out the entire number numerically... Ya'll resort to words at some point. WC...
Ghetto a** WC Puzzle
----------------------------------------------------------------------
Why don't you guys just type out the entire number numerically... Ya'll resort to words at some point. WC...
Ghetto a** WC PuzzleWC...
Practicing my WC 
ing
ing
SinisterB
Smash Champion
i seriously have no idea where i'm going, LOL.
I threw the address into google maps and apparently it's gonna take me like 2 hours
. any directions from Lougheed Skytrain station?
also i just woke up.
Asa can probably make it there before me, lmao.
I threw the address into google maps and apparently it's gonna take me like 2 hours
probably longer
also i just woke up.
Asa can probably make it there before me, lmao.
Crisis
Smash Ace
you could skytrain to vancouver city centre, and then take the canada line down to richmond brighouse.
not sure if that's the fastest route though.
if you leave now you'll get there before asa lol
not sure if that's the fastest route though.
if you leave now you'll get there before asa lol
SinisterB
Smash Champion
Maaaaaan, i hate to be downer but i don't think i'll come out today. I didn't realize Richmond was actually that far, and I need to be home early for plans tonight. Early as in 5:30 - 6 latest. You guys should just chill or stay home until asa gets there. I'd barely get a chance to offer my wii considering it's already almost 1.
why does this always happen when i have the chance to play asa
Crisis
Smash Ace
yeaah richmond is pretty damn far from your place
Blue Yoshi
Smash Master
I heard you guys like math problems.
What values of X make X^X = X?
(X to the power of X equals X)
Also, Saturday hype!
Edit: Landon, what is your phone number? Or does anyone have his number? Just so that I can call him if anything goes wrong while I travel to his place tomorrow.
What values of X make X^X = X?
(X to the power of X equals X)
Also, Saturday hype!
Edit: Landon, what is your phone number? Or does anyone have his number? Just so that I can call him if anything goes wrong while I travel to his place tomorrow.
Sushi3
Smash Apprentice
If you can visualize the graphs of both equations you'll realize the only intersection point is at x=1.
OR, use algebra. Take the log of both sides to get
x log x= log x
(log x )(x-1)=0
The only answer is 1.
Blue Yoshi
Smash Master
Your answer is incorrect. There is a flaw in your logic.
Sushi3
Smash Apprentice
What flaw then?
Also, what's going on with Asa/Crisis today? Please text me at 484-744-4535
Also, what's going on with Asa/Crisis today? Please text me at 484-744-4535
I could come but Richmond's so far, plus I don't use MK/Snake/Falco. Driving there tomorrow and Saturday's bad enough.
[collapse=Math] Using Robian's Law (it's complicated so I won't explain, hell I don't even understand it)...
X^X = X
X = 9
[/collapse]
[collapse=Math] Using Robian's Law (it's complicated so I won't explain, hell I don't even understand it)...
X^X = X
X = 9
[/collapse]
Crisis
Smash Ace
wow this is terrible lmao
if only i had a brawl cd
if only i had a brawl cd
Blue Yoshi
Smash Master
Sam, your first step:
X^X = X
logging both sides to
X log X = log X
will only take into account values of X greater than zero. If there are any solutions greater than zero, then you will get them. However, if there were any solutions less than or equal to zero, you lose them as soon as you logged both sides.
And me saying that your answer was incorrect pretty much gives away that there is a solution that is less than or equal to zero
X^X = X
logging both sides to
X log X = log X
will only take into account values of X greater than zero. If there are any solutions greater than zero, then you will get them. However, if there were any solutions less than or equal to zero, you lose them as soon as you logged both sides.
And me saying that your answer was incorrect pretty much gives away that there is a solution that is less than or equal to zero
Sushi3
Smash Apprentice
Ok, + or -1 then.
Is there a smashfest going on today? I'll attend only if Crisis/Asa confirm that they cannot make it today.
Is there a smashfest going on today? I'll attend only if Crisis/Asa confirm that they cannot make it today.
Crisis
Smash Ace
it's a no go for me as well, if asa can't bring his wii ;_;
Blue Yoshi
Smash Master
Well, I got bored and typed out my solution to that question, so I'll just post it just because I can.
[collapse=my solution]
x^x = x
CASE 1: x IS POSITIVE:
x^(x-1) = 1 (divide both sides by x, and since x is positive, we don't change the signs).
(x-1) log (x) = log (1)
(x-1) log (x) = 0
therefore (x-1) = 0, or log(x) = 0
if x-1 = 0, x = 1
if log(x) = 0, x = 1
Therefore, the solution we get is x = 1.
This case was for if x was positive, and the solution x = 1 is positive, therefore x = 1 is a legitimate solution.
SOLUTION FOR X IS POSITIVE: x = 1
CASE 2: x is 0
This is instantly refuted:
x^x = x
0^0 is undefined
Therefore, 0 cannot be a solution.
CASE 3: x IS NEGATIVE
We are now left with three cases:
a) (-1)^x = 1 (i.e. a negative number to the power of x is positive, e.g. even numbers)
b) (-1)^x = -1 (i.e. a negative number to the power of x is negative, e.g. odd numbers)
c) (-1)^x is undefined (e.g. square root of a negative number if x = 1/2)
We can instantly ignore case c, since it will make the left side undefined.
Looking at case a, we get
x^x = x
due to the assumption of case a, the left side must be positive. However, x is a negative number, meaning the right side is negative. therefore, we have a contradiction, as a positive number cannot equal a negative number.
Therefore, the only case we have to worry about is the case b:
b) (-1)^x = -1 (i.e. a negative number to the power of x is negative, e.g. odd numbers)
Therefore:
x^x = x
-(x^x) = -x
Due to the definition of the case, we can do this:
(-x)^x = -x
Since x is negative, -x is therefore positive. Therefore, the left side and right side are both positive numbers. Therefore, we can take the logarithm:
x log (-x) = log (-x)
x log (-x) - log (-x) = 0
(x-1) log (-x) = 0
therefore, either (x-1) = 0, or log(-x) = 0
if (x-1) = 0, x = 1
if log(-x) = 0, x = -1
Case 3 is stating that x is negative. therefore, our solution of x = 1 must be discarded.
Therefore, our solution for case 3 is x = -1
SOLUTION FOR x IS NEGATIVE: x = -1
Therefore, in case 1, we have our positive solution of x = 1, while in case 3, we have our negative solution of x = -1 (no solution for case 2 of x = 0).
Therefore, our two solutions are:
x = 1, and x = -1[/collapse]
[collapse=my solution]
x^x = x
CASE 1: x IS POSITIVE:
x^(x-1) = 1 (divide both sides by x, and since x is positive, we don't change the signs).
(x-1) log (x) = log (1)
(x-1) log (x) = 0
therefore (x-1) = 0, or log(x) = 0
if x-1 = 0, x = 1
if log(x) = 0, x = 1
Therefore, the solution we get is x = 1.
This case was for if x was positive, and the solution x = 1 is positive, therefore x = 1 is a legitimate solution.
SOLUTION FOR X IS POSITIVE: x = 1
CASE 2: x is 0
This is instantly refuted:
x^x = x
0^0 is undefined
Therefore, 0 cannot be a solution.
CASE 3: x IS NEGATIVE
We are now left with three cases:
a) (-1)^x = 1 (i.e. a negative number to the power of x is positive, e.g. even numbers)
b) (-1)^x = -1 (i.e. a negative number to the power of x is negative, e.g. odd numbers)
c) (-1)^x is undefined (e.g. square root of a negative number if x = 1/2)
We can instantly ignore case c, since it will make the left side undefined.
Looking at case a, we get
x^x = x
due to the assumption of case a, the left side must be positive. However, x is a negative number, meaning the right side is negative. therefore, we have a contradiction, as a positive number cannot equal a negative number.
Therefore, the only case we have to worry about is the case b:
b) (-1)^x = -1 (i.e. a negative number to the power of x is negative, e.g. odd numbers)
Therefore:
x^x = x
-(x^x) = -x
Due to the definition of the case, we can do this:
(-x)^x = -x
Since x is negative, -x is therefore positive. Therefore, the left side and right side are both positive numbers. Therefore, we can take the logarithm:
x log (-x) = log (-x)
x log (-x) - log (-x) = 0
(x-1) log (-x) = 0
therefore, either (x-1) = 0, or log(-x) = 0
if (x-1) = 0, x = 1
if log(-x) = 0, x = -1
Case 3 is stating that x is negative. therefore, our solution of x = 1 must be discarded.
Therefore, our solution for case 3 is x = -1
SOLUTION FOR x IS NEGATIVE: x = -1
Therefore, in case 1, we have our positive solution of x = 1, while in case 3, we have our negative solution of x = -1 (no solution for case 2 of x = 0).
Therefore, our two solutions are:
x = 1, and x = -1[/collapse]
T-block
B2B TST
Lol ****
0^0 is actually 1.
| Big D |
Smash Master
48÷2(9+3) =
Blue Yoshi
Smash Master
"Anything" to the power of Zero is usually one.
Zero to the power of "Anything" is usually zero.
Therefore, 0^x is always 0, and x^0 is always 1.
But when you get 0^0, you get a contradiction. It is a divide by zero error.
It can be argued quite convincingly that 0^0 "should" equal 1, however, it is actually undefined.
Dawson, the answer to that is 2
Zero to the power of "Anything" is usually zero.
Therefore, 0^x is always 0, and x^0 is always 1.
But when you get 0^0, you get a contradiction. It is a divide by zero error.
It can be argued quite convincingly that 0^0 "should" equal 1, however, it is actually undefined.
Dawson, the answer to that is 2
88
Blue Yoshi
Smash Master
Haha now I know to never trust the Google calculator 
Kantrip
Kantplay
http://www.smashboards.com/showpost.php?p=13396421&postcount=10859
Alright Sam, let's do this. We'll see what Phoenix placings look like
Alright Sam, let's do this. We'll see what Phoenix placings look like
Drake12
Smash Ace
Oh, snap! ****s gonna go downn
Blue Yoshi
Smash Master
Wait, a "best Diddy in BC" competition? I want in. I think I have a chance
Kantrip
Kantplay
Hard to judge though. Dittos won't do.
breez
Smash Ace
Alberta people and tinaye!
Bring yours Wiis!
I already have 2 TVs and i'm trying to get a third, but I only have one wii!
I think I have a second wii somewhere else, but I dont have another copy of SSBB, so if you can't bring a wii, try and bring a disc. HYYYYYYYYYYYYYYYYYYYYYYYYPPEEEEEEEEEEEEEEE
Bring yours Wiis!
I already have 2 TVs and i'm trying to get a third, but I only have one wii!
I think I have a second wii somewhere else, but I dont have another copy of SSBB, so if you can't bring a wii, try and bring a disc. HYYYYYYYYYYYYYYYYYYYYYYYYPPEEEEEEEEEEEEEEE
breez
Smash Ace
Dittos da bess son.
breez
Smash Ace
Perfect. Try and bring a brawl disc. Im not 100% sure i'll get three setups, but it is quite likely.
_GrapE_
Smash Lord
Last post I will make before we leave.
Landon we'll be there around 5-6, as said before. Rob, be there around that time, preferably.
Dawson, please remember my controller on Saturday.
See y'all soon.
Landon we'll be there around 5-6, as said before. Rob, be there around that time, preferably.
Dawson, please remember my controller on Saturday.
See y'all soon.
Blue Yoshi
Smash Master
Caiden (and Landon), the earliest I can arrive is just before 8 (which is what I am striving for).
breez
Smash Ace
Woooooord. So hype. You all have my phone number right? And phones that can access facebook? We must have communication :D
What time do you think you will all be arriving??