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Algebra Level 2

If x > 2 x>2 and 2 x = x 2 , 2^x=x^2, what is the value of x 2 ? x^2?


The answer is 16.

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2 solutions

  • The equation in general has two solutions given by x = e W ( ln 2 2 ) \displaystyle x=e^{-W(\frac{-\ln 2}{2})} which gives x = 2 , 4 x=2,4

W(.) - Lambert W function

Why is there no other solution? I am just curious.

Mohammad Farhat - 2 years, 9 months ago
Chase Marangu
May 14, 2018

So we don't worry so much about x>2 to start with so then what we do is

2 x = x 2 2^x=x^2

so

x = 2 x x=\sqrt{2^x}

so

x = 2 2 2 x x=\sqrt{2^{\sqrt{2^{\sqrt{2^x}}}}}

and we find solutions to this and one of them is 4 4

and 4 > 2 4>2

and 4 2 = 16 4^2=16

so the answer is 16

Is this the only solution?

Pi Han Goh - 3 years, 1 month ago

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@Pi Han Goh Yes, it is the only real solution for x>2. There are no imaginary solutions. There are also no complex solutions.

chase marangu - 3 years, 1 month ago

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@Pi Han Goh I forgot to add, ∞ but not -∞ because the limit as x→∞ 2^x = lim x→∞ x^2

chase marangu - 3 years, 1 month ago

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@Chase Marangu The question already stated that x > 2 x>2 , so it's a real number. But how do you know that there isn't any other (real) solution?

Pi Han Goh - 3 years, 1 month ago

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@Pi Han Goh The function is continuous and erm...simple(?) I guess? Don't know if that's a actual mathematical thing. The apparent pattern is that √(2ˣ) > x for all x>4. Then at infinity 2^∞ = ∞^2 so you get another solution there. And you can say that infinity isnt a real number but sometimes I like to pretend. I mean it's not a number anyways but still.

chase marangu - 3 years ago

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