If $x^{x^{4}} = 4$ $x^(x^4) = 4$ , then $x^{x^{2}} + x^{x^{8}}$ $x^(x^2) + x^(x^8)$ will be equal to what?

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If $x^{x^{4}} = 4$ $x^(x^4) = 4$ , then $x^{x^{2}} + x^{x^{8}}$ $x^(x^2) + x^(x^8)$ will be equal to what?

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Answer 1 · 2016-11-16T07:45:05

$258$ $258$

Explanation:

In general equations of the form $x^{x^{4}} = a$ $x^(x^4) = a$ are difficult to solve, but in the case of $x^{x^{4}} = 4$ $x^(x^4) = 4$ there is a guessable solution.

${\sqrt{2}}^{{\sqrt{2}}^{4}} = {\sqrt{2}}^{4} = 4$ $sqrt(2)^(sqrt(2)^4) = sqrt(2)^4 = 4$

So $x = \sqrt{2}$ $x = sqrt(2)$ is one solution and $x = - \sqrt{2}$ $x = -sqrt(2)$ also works.

Then we find:

${\sqrt{2}}^{{\sqrt{2}}^{2}} + {\sqrt{2}}^{{\sqrt{2}}^{8}} = {\sqrt{2}}^{2} + {\sqrt{2}}^{16} = 2 + 256 = 258$ $sqrt(2)^(sqrt(2)^2) + sqrt(2)^(sqrt(2)^8) = sqrt(2)^2 + sqrt(2)^16 = 2+256 = 258$

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Footnote

Concerning the "guessing", notice that if:

$x^{a} = a$ $x^a = a$

then:

$x^{x^{a}} = x^{a} = a$ $x^(x^a) = x^a = a" "$ too.

So in general, $x = \sqrt[a]{a}$ $x=root(a)(a)$ is a solution of $x^{x^{a}} = a$ $x^(x^a) = a$

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If $x^{x^{4}} = 4$ $x^(x^4) = 4$ , then $x^{x^{2}} + x^{x^{8}}$ $x^(x^2) + x^(x^8)$ will be equal to what?

1 Answer

Explanation:

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If $x^{x^{4}} = 4$ $x^(x^4) = 4$ , then $x^{x^{2}} + x^{x^{8}}$ $x^(x^2) + x^(x^8)$ will be equal to what?