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Question Number 216266 by mathlove last updated on 02/Feb/25

(√2)^((√2)^((√2)^⋰ ) ) =?

$$\sqrt{\mathrm{2}}\:^{\sqrt{\mathrm{2}}\:^{\sqrt{\mathrm{2}}\:^{\iddots} } } =? \\ $$

Answered by BaliramKumar last updated on 02/Feb/25

((√2))^x = x 2^(x/2) = x 2^x = x^2 x = 2 or 4

$$\left(\sqrt{\mathrm{2}}\right)^{{x}} \:=\:{x} \\ $$$$\mathrm{2}^{\frac{{x}}{\mathrm{2}}} \:=\:{x} \\ $$$$\mathrm{2}^{{x}} \:=\:{x}^{\mathrm{2}} \\ $$$${x}\:=\:\mathrm{2}\:{or}\:\mathrm{4} \\ $$

Commented by Frix last updated on 02/Feb/25

I don′t think the value can be 4 y=x^x^(x...) y=x^y ((ln y)/y)=ln x This has 2 solutions for 0<x<e^(1/e) ≈1.44467 exactly 1 solution for x=e^(1/e) and no solution for x>e^(1/e) : x→0^+ ⇒ y_1 →1∧y_2 →∞ which means y_2 is obviously wrong. Using a good calculator we see that 1.1^(1.1^(...) ) ≈1.11178 how could it be 38.2287? 1.01^(1.01^(...) ) ≈1.01010 how could it be 651.100? ... ⇒ y_2 is always wrong

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{the}\:\mathrm{value}\:\mathrm{can}\:\mathrm{be}\:\mathrm{4} \\ $$$${y}={x}^{{x}^{{x}...} } \\ $$$${y}={x}^{{y}} \\ $$$$\frac{\mathrm{ln}\:{y}}{{y}}=\mathrm{ln}\:{x} \\ $$$$\mathrm{This}\:\mathrm{has}\:\mathrm{2}\:\mathrm{solutions}\:\mathrm{for}\:\mathrm{0}<{x}<\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{e}}} \approx\mathrm{1}.\mathrm{44467} \\ $$$$\mathrm{exactly}\:\mathrm{1}\:\mathrm{solution}\:\mathrm{for}\:{x}=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{e}}} \:\mathrm{and}\:\mathrm{no}\:\mathrm{solution} \\ $$$$\mathrm{for}\:{x}>\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{e}}} :\:{x}\rightarrow\mathrm{0}^{+} \:\Rightarrow\:{y}_{\mathrm{1}} \rightarrow\mathrm{1}\wedge{y}_{\mathrm{2}} \rightarrow\infty\:\mathrm{which} \\ $$$$\mathrm{means}\:{y}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{obviously}\:\mathrm{wrong}.\:\mathrm{Using}\:\mathrm{a}\:\mathrm{good} \\ $$$$\mathrm{calculator}\:\mathrm{we}\:\mathrm{see}\:\mathrm{that} \\ $$$$\mathrm{1}.\mathrm{1}^{\mathrm{1}.\mathrm{1}^{...} } \approx\mathrm{1}.\mathrm{11178}\:\mathrm{how}\:\mathrm{could}\:\mathrm{it}\:\mathrm{be}\:\mathrm{38}.\mathrm{2287}? \\ $$$$\mathrm{1}.\mathrm{01}^{\mathrm{1}.\mathrm{01}^{...} } \approx\mathrm{1}.\mathrm{01010}\:\mathrm{how}\:\mathrm{could}\:\mathrm{it}\:\mathrm{be}\:\mathrm{651}.\mathrm{100}? \\ $$$$... \\ $$$$\Rightarrow \\ $$$${y}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{always}\:\mathrm{wrong} \\ $$

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