10
$\begingroup$

If $x^{x^4}=4$, then what is the value of $x^{x^2}+x^{x^8}$?

I tried to simplify it using exponentiation and logs, and even just algebraic manipulation..But I don't know how to do this.

$\endgroup$
  • 1
    $\begingroup$ Linear algebra? To clarify, the expression is $x^{(x^4)}$ and so on? If so, then note that the real roots of $x^{(x^4)}=4$ are $±\sqrt 2$. $\endgroup$ – lulu Oct 6 '15 at 10:35
  • $\begingroup$ Isn't the domain of $x^{x^4}$ should be $x>0$? $\endgroup$ – Quang Hoang Oct 6 '15 at 10:46
  • $\begingroup$ Yes. @lulu That's what the question means. $\endgroup$ – Kugelblitz Oct 6 '15 at 10:50
  • 1
    $\begingroup$ Note that if a number $x$ satisfies $x^4=4$, then it also satisfies $x^{x^4}=4$ $\endgroup$ – nospoon Oct 6 '15 at 10:54
  • 1
    $\begingroup$ A cute trick for this kind of thing: if $x^{x^4}=4$ then $(x^4)^{x^4}=x^{4x^4}=(x^{x^4})^4=4^4$. This makes it easy to spot the solution $x^4=4$. (similar) $\endgroup$ – user21467 Oct 6 '15 at 11:52
8
$\begingroup$

Just note that $x^{x^4}$ is increasing for $x>1$ and is $<1$ for $x<1$. The unique positive solution to $x^{x^4}=4$ is $x=\sqrt2$.

$\endgroup$
  • $\begingroup$ Yea.... got it. $\endgroup$ – Kugelblitz Oct 6 '15 at 10:49
4
$\begingroup$

$x^{x^{4}}=4$. Let us replace 4 of the exponent of left hand side by $x^{x^{4}}$ then equation becomes $x^{x^{x^{x^{4}}}}=4$. Repeating this process infinitely we get $x^{x^{x^{x^{x..}}}}=4$. Now We replace the exponent of left hand side by 4 and equation now become $x^{4}=4$ hence $x=\sqrt{2}$ or $x=-\sqrt{2}$ . So $x^{x^{2}}+x^{x^{8}} =258$

$\endgroup$
  • $\begingroup$ Nice observation and method. $\endgroup$ – rugi Feb 26 '16 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.