i am an algebra whiz

[MunkyButt]

i can help if you have questions.

[Diamond187]

Alright, here's something interesting for you to work out:

xln(x)=x

Gimme all possible solutions, then we will all collectively call you a basic algebra whiz.

[caprarioreed25]

There are no possible solutions?

[Diamond187]

Nope, it is solvable. It really isn't that hard either. But there's one trick answer which might get some people.

[Danielle]

Darn. You had to use natural log rythm. The one thing that I only moderately understand from Alg II. Eh, I will work the problem tonight.

[Diamond187]

As a big hint I'll say that you should logically come up with 2 answers, and then rule one of them out.

[The First]

I'm not getting the question... is it x*ln(x) = x? Or does xln(x) have a different meaning?

Yes, I know. Stupid german with his different way of writing mathematical terms >_<

[Afterburner]

I'm guessing that either:

Ln(x) = 1
or
x = 0

If ln(x) = 1 then x = e

If x = 0 then ln(x) is undefined, so it's out anyways.

Am I right? x = e?

Try this, another easy one: derive the quadratic formula.

[Diamond187]

Yep, AB is right again.

And I did that once in my chemistry course. It's kinda cool when you figure out how to do it, but it really, really makes you wonder how they came up with a cubic formula.

[caprarioreed25]

Try this, another easy one: derive the quadratic formula.

x = -b ± the square root of b² - 4ac all over 2a

I could've solved the natural logarithm problem at one time, but I've forgotten that stuff. It was too boring for me...

[eatmorchikin6464]

Im not going to derive the quadratic formula, that takes lines and lines and lines.

[Diamond187]

He said derive, not state.

And yeah, it might be hard to do in computer math, and confusing to read.

[Vagrant]

Wait... looking at the natural log function...

ln(x^x * x) would be simplified. And now I'm officially confused as to how AB got that answer.

[Diamond187]

Woah, I've got no idea how you got that. I think you might be confused about some of the notation or something. The equation is, in words:

x times the natural log of x equals x

I hope that clears up something of your misunderstanding.

[Vagrant]

I might be misunderstanding this. I'm treating it like a log function, where the equals is the power to which the log is multiplied.

Or, I screwed my orders.

Simplified, is it: ln(x^2x) ?

[Diamond187]

OK, I see how you get the x^x part inside the log but where the 2 comes from I don't see. And exactly what would that be equal to when you simplify the left side to ln(x^2x)?

[The First]

T'woud be ln(x^x) = x
Which would be e^x = x^x, would it not?
Which means x = e. At least that's following Vagrants line of thought. The only way I get a second answer is the way Afterburner did.

And exactly how do you want someone to derive the Quadratic Formula? It has three variables (if you mean the entire -b +- The root of... blah blah blah). If you mean f(x)=ax²+bx+c, the derivative is 2ax+b

[Diamond187]

lol

He means derive the -b +- blah blah.

And it doesn't have 3 variables. It has 1 variable and 3 constants. Set up a generic polynomial and you'll see.

ax^2+bx+c=0

a,b&c are all constants. x is your only variable.

[The First]

Silly german understood something completely different under deriving. Apparently. I think. Anyways, I googled, it, I get what he meant. Yay for commas!

[Vagrant]

Bleh, sorry, having trouble remembering.

ln(x^x) = x

Log base e (x^x) = x

Now my memory completely died here, and I completely forgot what to do.

[Diamond187]

You need to anti-log both sides to get rid of the ln.

[Vagrant]

Err, nevermind. I got it now. I remember.

E^x = X^x

E=X

I was having problems 'cuz I'm not used to doing natural log functions. I'm used to doing normal log functions. I forgot the power rule and the number across the equal sign thingy.

[Vince]

or you can just make x=0 that leaves it 0=0
HA! I WIN! :p

[Afterburner]

...

This was discussed earlier......ln (0) is undefined.

[Diamond187]

Well, theretically, ln(0) is -infinite, and does -infinite*0=0? Iunno, if you can prove it does, you belong at NASA.

[The First]

Isn't it defined that anything multiplied with 0 will equal 0? Be it a real or imaginary number? I can't say I've spent any time at all thinking about the mathematical properties of infinity, but if you take infinity zero times, then you never took infinite... but I'm going to go off on a limb here and assume that it isn't that simple.

[Vince]

then i dunno *shrug* i tryed

[Diamond187]

infinite*0 isn't that simple, and if you think about it, it does make sense. We all know that 0/0 is completely unusable stated like that, well, we can expand that form to something like this 0*(1/0) and obviously 1/0=infinite, so

infinte*0=0/0