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If $x^{x^4}=4$. Find $x^{x^2}+x^{x^8}$. I found this one in a competitive exam paper and found it interesting. Thanks for any help.

Soham
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  • What about $x$? –  Aug 24 '15 at 16:42
  • Nothing specific is given about $x$. Hope it refers to real numbers. – Soham Aug 24 '15 at 16:44
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    Note that any number satisfying $x^4=4$ will fit the condition since then $x^{x^4}=x^4=4$ holds. So $x=\pm \sqrt{2}$ are solutions. You may use the derivative of the function $f(x)=x^{x^4}$ to show that these are the only (real) solutions. – Tintarn Aug 24 '15 at 16:51

1 Answers1

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Let $4=x^{(x^2)}=\left((x^2)^{(x^2)}\right)^{x^2/2}$

$$x^2=y\implies y^{(y^2)}=16$$

One of the value of $y$ is $2$

$$x^{(x^2)}+x^{(x^8)}=x^2+x^{16}=2+(2)^8$$

lab bhattacharjee
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