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Pre-RMO 2014/7

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


This note is part of the set Pre-RMO 2014

Note by Pranshu Gaba
2 years, 3 months ago

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Let \(x=y^{\frac{1}{4}}\)

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

\( y^{y}=4^{4}\)

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

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

Note: I like keeping it real (Its a joke) Aneesh Kundu · 2 years, 3 months ago

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@Aneesh Kundu The equation \(x^{x^{4}}\) have 4 roots

\(\pm \sqrt2, \pm i \sqrt2\)

It should be mentioned in the question that x is real because answer can vary Krishna Sharma · 2 years, 3 months ago

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@Krishna Sharma Totally agree Aneesh Kundu · 2 years, 3 months ago

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

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@Siddhesh Naik The function \(\sqrt[x]{x}=y\) always gives 2 real solutions for \(y\leq e\), but function \(x^x=y\) gives 2 real values only for \(\frac{1}{e} \leq y \leq 1\) and since \( 4^4>1\) we have only one real solution. Aneesh Kundu · 2 years, 2 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 \(x^{1/x}\) should be modified as \(x^{1/x}=y\) always gives two positive real solutions for \(x\) if \(1<y<e^{1/e}\). 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 \(x^x=y\), it has only one positive solution in \(x\) if \(y\ge 1\). This equation may still have a negative solution in \(x\). 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 \(y\) as otherwise \(x\) 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 \(x\) is real. Siddhesh Naik · 2 years, 2 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 · 2 years, 3 months ago

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Can easily be solved by taking log Ayush Jain · 8 months ago

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258 Gaurav Singh · 1 year, 5 months ago

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258 Akshat Sharda · 1 year, 5 months ago

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258 Sahil Nare · 1 year, 8 months ago

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18 BidehRanjan Mahunta · 2 years, 3 months ago

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\(x = \sqrt2\)

Ans 258 Krishna Sharma · 2 years, 3 months ago

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@Krishna Sharma Can you tell which method did you use to get \(x=\sqrt{2}\)? Pranshu Gaba · 2 years, 3 months ago

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Comment deleted Oct 13, 2014

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@BidehRanjan Mahunta Root of \(\displaystyle x^{x^{4}} is x^{\frac{x^{4}}{2}} \)

Your argument is invalid Krishna Sharma · 2 years, 3 months ago

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@Krishna Sharma Sorry my mistake BidehRanjan Mahunta · 2 years, 3 months ago

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@BidehRanjan Mahunta Sorry my mistake BidehRanjan Mahunta · 2 years, 3 months ago

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