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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.)

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
–
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.

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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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)Log in to reply

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

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Totally agree

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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.)

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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.

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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.Log 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.

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$x = \sqrt2$

Ans 258

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

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18

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258

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258

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258

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Can easily be solved by taking log

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How?

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