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

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

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

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

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

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, 11 months agoLog in to reply

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

It should be mentioned in the question that x is real because answer can vary – Krishna Sharma · 2 years, 11 months ago

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– Aneesh Kundu · 2 years, 11 months ago

Totally agreeLog in to reply

– Siddhesh Naik · 2 years, 11 months ago

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.)Log in to reply

– Aneesh Kundu · 2 years, 11 months ago

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.Log in to reply

Firstly what you said about the function \(x^{1/x}\) should be modified as \(x^{1/x}=y\) always gives two

positivereal 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 positivesolution 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, 11 months agoLog in to reply

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, 11 months ago

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Can easily be solved by taking log – Ayush Jain · 1 year, 4 months ago

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258 – Gaurav Singh · 2 years, 1 month ago

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258 – Akshat Sharda · 2 years, 1 month ago

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258 – Sahil Nare · 2 years, 5 months ago

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18 – BidehRanjan Mahunta · 2 years, 11 months ago

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\(x = \sqrt2\)

Ans 258 – Krishna Sharma · 2 years, 11 months ago

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– Pranshu Gaba · 2 years, 11 months ago

Can you tell which method did you use to get \(x=\sqrt{2}\)?Log in to reply

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Your argument is invalid – Krishna Sharma · 2 years, 11 months ago

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– BidehRanjan Mahunta · 2 years, 11 months ago

Sorry my mistakeLog in to reply

– BidehRanjan Mahunta · 2 years, 11 months ago

Sorry my mistakeLog in to reply