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if-x-x-x-x-2020-2020-Solve-for-x-




Question Number 98968 by Brokris last updated on 17/Jun/20
if x^x^x^x^(2020)    = 2020.  Solve for x
$$\mathrm{if}\:\mathrm{x}^{\mathrm{x}^{\mathrm{x}^{\mathrm{x}^{\mathrm{2020}} } } } =\:\mathrm{2020}.\:\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{x} \\ $$$$ \\ $$
Commented by mr W last updated on 17/Jun/20
x^x^(...x^(2020) )  =2020  has the same root as  x^(2020) =2020  x=±((2020))^(1/(2020)) ≈±1.00377
$${x}^{{x}^{…{x}^{\mathrm{2020}} } } =\mathrm{2020} \\ $$$${has}\:{the}\:{same}\:{root}\:{as} \\ $$$${x}^{\mathrm{2020}} =\mathrm{2020} \\ $$$${x}=\pm\sqrt[{\mathrm{2020}}]{\mathrm{2020}}\approx\pm\mathrm{1}.\mathrm{00377} \\ $$

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