# Math Help!

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My Pre-Calculus teacher gave us the following problem as a challenge:

x^2x=2x

the ^ is raised to the power of (just in case you didn't know)

I found that 1/4 was the answer through guess and check but I don't know how to work it! I'm sure it's all just algebra, but I haven't got a clue. Any help would be appriciated.

Thanks

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Maybe you want to do logs of each side.

So maybe something like this:

2xlogx = log2x

Don't know if that is how you do it. (probably not)

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Hmm, my maths application (Derive) calculated for about 200 seconds with no result - therefore, I doubt it's possible to get a numeric solution for it using normal ways of calculating Sure you got the starting formula (x^2x = 2x) right?

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I gave it a quick hack, but it looks like some intuition is involved anyway.

x^(2x) = 2x

x^(2x) - 2x = 0

x*(x^(2x-1)-2)=0

thus...

either x = 0

or x^(2x-1)=2

which means...

log2(x)*(2x-1) = 1

that makes the guessing a bit easier...

but I can't go further from that point.

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One of my friends (who is in Calculus), sits around and thinks up math problems. He gave this to his calculus teacher and she couldn't solve it so she gave it to our Pre-Cal teacher and she couldn't solve it either. So she gave it to us! (as if we could solve it!) Anyway he said my answer was right (1/4), but refused to show me the work.

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There's another solution at x=1.44373742709798..., which is fairly close to 3^(1/3) (to about one part in a thousand), but I don't know how to get an exact answer.

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Mathematica is refusing to give an answer lol.

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• 2 weeks later...

Hmm.... little problem with x=0.

0^(2*0) = undefined

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That's why I didn't accept x=0 as a solution... did I mention that?

oh... whoops...

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