Pre-RMO 2014/7

If x(x4)=4x^{(x^4)} = 4, what is the value of x(x2)+x(x8)x^{(x^2)} + x^{(x^8)}?


This note is part of the set Pre-RMO 2014

Note by Pranshu Gaba
5 years ago

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Let x=y14x=y^{\frac{1}{4}}

(y4)y=4 (\sqrt[4]{y})^{y}=4

yy=44 y^{y}=4^{4}

Comparing both sides we get y=4x=2y=4 \Rightarrow x=\sqrt{2}

Pluging the value of x in the expression we get 258 \boxed{258}

Note: I like keeping it real (Its a joke)

Aneesh Kundu - 5 years ago

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The equation xx4x^{x^{4}} have 4 roots

±2,±i2\pm \sqrt2, \pm i \sqrt2

It should be mentioned in the question that x is real because answer can vary

Krishna Sharma - 5 years ago

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Totally agree

Aneesh Kundu - 5 years ago

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Although this is right, such a comparison may not give all answers in some cases. e.g. x1/x=21/2x^{1/x}=2^{1/2} suggests x=2x=2 as a trivial answer, but observe that x=4x=4 is also a nontrivial answer. One needs to be careful thus. Try to justify therefore that in the above case however, y=4y=4 is the only real answer (It may be very tough to prove/disprove that there may be an imaginary solution to this.)

Siddhesh Naik - 4 years, 11 months ago

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The function xx=y\sqrt[x]{x}=y always gives 2 real solutions for yey\leq e, but function xx=yx^x=y gives 2 real values only for 1ey1\frac{1}{e} \leq y \leq 1 and since 44>1 4^4>1 we have only one real solution.

Aneesh Kundu - 4 years, 11 months ago

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@Aneesh Kundu Well although you got most of it right, it is not that easy to summarize it like this.

Firstly what you said about the function x1/xx^{1/x} should be modified as x1/x=yx^{1/x}=y always gives two positive real solutions for xx if 1<y<e1/e1<y<e^{1/e}. There may still be some negative solutions (Yes the function is defined for some subset of negative numbers, try to find it as an exercise).

Talking about xx=yx^x=y, it has only one positive solution in xx if y1y\ge 1. This equation may still have a negative solution in xx. Why I'm saying here it may have a negative solution is because the solution may or may not exist and it may be very hard to prove existence. But in this question, we need not worry about the negative values of yy as otherwise xx would become complex and the question would not have a unique solution which will also go beyond the scope of the exam. For it to make sense in the exam point of view, we make an implicit assumption that xx is real.

Siddhesh Naik - 4 years, 11 months ago

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The answer is 258. I'm still not clear on whether we just have to substitute the values in the actual paper or solve it using some other method.

Bhanu Bhandari - 5 years ago

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x=2x = \sqrt2

Ans 258

Krishna Sharma - 5 years ago

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Can you tell which method did you use to get x=2x=\sqrt{2}?

Pranshu Gaba - 5 years ago

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18

BidehRanjan Mahunta - 5 years ago

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258

Sahil Nare - 4 years, 5 months ago

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258

Akshat Sharda - 4 years, 2 months ago

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258

gaurav singh - 4 years, 2 months ago

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Can easily be solved by taking log

AYUSH JAIN - 3 years, 5 months ago

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How?

Anish Manna - 1 year, 2 months ago

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