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Question Number 17989 by alex041103 last updated on 13/Jul/17

Solve for x  x^x^x^x^(....)    = 4

$${Solve}\:{for}\:{x} \\ $$$${x}^{{x}^{{x}^{{x}^{....} } } } =\:\mathrm{4} \\ $$

Commented by alex041103 last updated on 13/Jul/17

just for fun

$${just}\:{for}\:{fun} \\ $$

Commented by mrW1 last updated on 13/Jul/17

x^4 =4  x=^4 (√4)=(√2)

$$\mathrm{x}^{\mathrm{4}} =\mathrm{4} \\ $$$$\mathrm{x}=\:^{\mathrm{4}} \sqrt{\mathrm{4}}=\sqrt{\mathrm{2}} \\ $$

Commented by alex041103 last updated on 13/Jul/17

good

$${good} \\ $$

Commented by mrW1 last updated on 13/Jul/17

((√2))^(((√2))^(((√2))^((.....)) ) ) =?

$$\left(\sqrt{\mathrm{2}}\right)^{\left(\sqrt{\mathrm{2}}\right)^{\left(\sqrt{\mathrm{2}}\right)^{\left(.....\right)} } } =? \\ $$

Commented by mrW1 last updated on 13/Jul/17

2^(k/2)  = k  ⇒k=2 or 4  i.e.  ((√2))^(((√2))^(((√2))^((...)) ) )  = 2 or 4  but how can it be two different values?

$$\mathrm{2}^{\frac{{k}}{\mathrm{2}}} \:=\:{k} \\ $$$$\Rightarrow\mathrm{k}=\mathrm{2}\:\mathrm{or}\:\mathrm{4} \\ $$$$\mathrm{i}.\mathrm{e}. \\ $$$$\left(\sqrt{\mathrm{2}}\right)^{\left(\sqrt{\mathrm{2}}\right)^{\left(\sqrt{\mathrm{2}}\right)^{\left(...\right)} } } \:=\:\mathrm{2}\:\mathrm{or}\:\mathrm{4} \\ $$$$\mathrm{but}\:\mathrm{how}\:\mathrm{can}\:\mathrm{it}\:\mathrm{be}\:\mathrm{two}\:\mathrm{different}\:\mathrm{values}? \\ $$

Commented by alex041103 last updated on 13/Jul/17

also −(√2) ia solution  (−(√2))^4 =(−1)^4 ((√2))^4 =4

$${also}\:−\sqrt{\mathrm{2}}\:{ia}\:{solution} \\ $$$$\left(−\sqrt{\mathrm{2}}\right)^{\mathrm{4}} =\left(−\mathrm{1}\right)^{\mathrm{4}} \left(\sqrt{\mathrm{2}}\right)^{\mathrm{4}} =\mathrm{4} \\ $$

Commented by prakash jain last updated on 13/Jul/17

x^x^x^(⋰∞)   =4  (√2) is not a valid solution.

$${x}^{{x}^{{x}^{\iddots\infty} } } =\mathrm{4} \\ $$$$\sqrt{\mathrm{2}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{valid}\:\mathrm{solution}. \\ $$

Commented by alex041103 last updated on 13/Jul/17

True

$${True} \\ $$

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