Question Number 183769 by Michaelfaraday last updated on 29/Dec/22
$${solve}\:{for}\:{x}: \\ $$$${x}^{{x}^{{x}} } =\mathrm{2}^{\mathrm{2048}} \\ $$$${by}\:{using}\:{lambert}\:{function} \\ $$
Commented by mr W last updated on 30/Dec/22
$${it}\:{has}\:{no}\:{solution}\:{by}\:{lambert}! \\ $$$${no}\:{solution}\:{by}\:{anything}! \\ $$
Commented by mr W last updated on 30/Dec/22
$$\mathrm{4}^{\mathrm{4}^{\mathrm{4}} } =\mathrm{2}^{\mathrm{512}} <\mathrm{2}^{\mathrm{2048}} \\ $$$$\mathrm{5}^{\mathrm{5}^{\mathrm{5}} } =\mathrm{5}^{\mathrm{3125}} >\mathrm{2}^{\mathrm{2048}} \\ $$$$\Rightarrow{the}\:{root}\:{of}\:{x}^{{x}^{{x}} } =\mathrm{2}^{\mathrm{2048}} \:{lies}\:{between} \\ $$$$\mathrm{4}\:{and}\:\mathrm{5}. \\ $$
Commented by Michaelfaraday last updated on 30/Dec/22
$${okay}\:{thanks}\:{sir} \\ $$
Commented by paul2222 last updated on 03/Jan/23
$$\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{use}}\:\boldsymbol{\mathrm{newton}}\:\boldsymbol{\mathrm{raphson}}\:\boldsymbol{\mathrm{iteration}} \\ $$