Question Number 63175 by Tawa1 last updated on 30/Jun/19
$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{x}^{\mathrm{x}} } \:=\:\:\mathrm{16} \\ $$$$\mathrm{x}\:=\:\mathrm{2},\:\:\:\:\:\mathrm{but}\:\mathrm{how}\:\mathrm{to}\:\mathrm{use}\:\mathrm{Lambert}\:\mathrm{W}\:\mathrm{function} \\ $$
Commented by mr W last updated on 30/Jun/19
$${i}\:{remember}\:{you}\:{have}\:{asked}\:{the}\:{same} \\ $$$${question}\:{in}\:{an}\:{earlier}\:{time}.\:{the}\:{answer} \\ $$$${is}\:{you}\:{can}\:{not}\:{use}\:{Lambert}\:{function} \\ $$$${to}\:{solve}\:{it}.\:{besides},\:{it}\:{is}\:{not}\:{clear} \\ $$$${what}\:{is}\:{meant}\:{concretely}\:{with}\:{x}^{{x}^{{x}} } . \\ $$$$\mathrm{3}^{\mathrm{3}^{\mathrm{3}} } =\mathrm{3}^{\mathrm{9}} \:{or}\:\mathrm{3}^{\mathrm{27}} \:{or}\:\mathrm{27}^{\mathrm{3}} \:? \\ $$
Commented by Tawa1 last updated on 30/Jun/19
$$\mathrm{Alright}\:\mathrm{sir},\:\:\mathrm{hint}\:\mathrm{me}\:\mathrm{on}\:\mathrm{the}\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} } } \:=\:\:\mathrm{49},\:\:\:\mathrm{again}\:\mathrm{sir}.\: \\ $$