Hi, i'm new, and i wanted to share a very simple algebra problem i like.

 

[math]x^{2x}=2x[/math]

 

It's very simple, but not as easy as it looks, if you know the answer right away, give others a chance.

Hihi. To find an exact value, is calculus knowledge required?

wheres the problem? lol

 

edit:

nevermind... it showed up after i replied... wierd.

The answer is so simple yet I can't figure out a way of calculating it algebraicly, unless I'm missing something stupid.

No, it's not calculus, it's algebra and it is pretty simple, but not easy. You probably realized it already but you can't isolate the x, so it's not done by regular ecuation solving. Think.

i got .25 and 1.4361702.

i got .26595745 and 1.4361702.

i'm not sure about that' date=' but the gradient is [math']y=x^{2x}-2x\rightarrow\frac{dy}{dx}=2x^{2x+1}-2[/math].

 

You're complicating yourself, it's 100% algebra.

I got so far as x^2x-1=2

 

I got there using logarithms

 

edit

 

just got to x 1n(x)= 1n (3)/-2

 

edit slight error in math

 

1n(x)= 1n(3)/2x-1

I know one answer is 1/4.... the other is approximately 1.4437374... I have no clue how to solve it algebraically though, good problem!

where did you get those answers from?

i got my answers by graphing it.

I know one answer is 1/4....

 

umm, yeah, the answer is 1/4, but off course the beauty lies in solving it algebraically.

so what is the algebraic method then?

i got .26595745 and 1.4361702.

i'm not sure about that' date=' but the gradient is [math']y=x^{2x}-2x\rightarrow\frac{dy}{dx}=2x^{2x+1}-2[/math].

 

I just had to point out that the above is not correct. The derivative of [math]x^{2x}[/math] is [math]2(ln(x)+1)x^{2x}[/math]

so what is the algebraic method then?

That's what I'm asking you guys. It took me and 2 other dudes like 3 weeks to solve it, and this is the simplest problem of it's type.

i solved it graphically. i used my calculator this time and got closer values than before. they are .25 and aproxamately 1.443737. i set it =0 and assumed y=0. my answers are where the function crosses the x-axis.

that's how I did it for my answers... I gave it to my highschool calculus teacher and she can't figure it out either... I haven't given up though! I'm still trying

so razor, how did u do it?

so razor, how did u do it?

You're giving up that easy? Really, it's just a matter of imagination.

i already got the anwers, i want to know your simple yet hard algebraic method.

I know you got the answer. are you insinuating there's no other way?

no. did you not read my post? i just want to know your way.

Oh, ok. Actually there is no algebraic solution and... i'm joking, here it is:

 

[math]x^{2x}=2x[/math]

[math]x^{2x-1}=2[/math]

[math]x^{\frac{2x-1}{2}}=2^{\frac{1}{2}}[/math]

[math]x^{x-\frac{1}{2}}=2^{\frac{1}{2}}[/math]

[math]x^{x-\frac{1}{2}}=\left(\tfrac{1}{2}\right)^{-\frac{1}{2}}[/math]

[math]x^{x-\frac{1}{2}}=\left(\tfrac{1}{4}\right)^{-\frac{1}{4}}[/math]

[math]x^{x-\frac{1}{2}}=\left(\tfrac{1}{4}\right)^{\frac{1}{4}-\frac{1}{2}}[/math]

Since both members are structurally the same, you deduct:

[math]X=\frac{1}{4}[/math]

okay, I understand everything up until this point.

[math]x^{x-\frac{1}{2}}=\left(\tfrac{1}{4}\right)^{\frac{1}{4}-\frac{1}{2}}[/math]

 

Will you please explain how you got from the previous step to that one? Also, what about the other answer, 1.4437374? Is there a way to get that one?

nevermind, I didn't see the negative in the previous one and that completely messed me up until I did it on paper. I get it now, -1/4 is equal to 1/4-1/2. I'm still curious about the other answer (1.4437374) though