Question Number 168383 by cortano1 last updated on 09/Apr/22
Commented by mr W last updated on 09/Apr/22
$${x}=\mathrm{2}^{\mathrm{192}} \\ $$
Answered by greogoury55 last updated on 09/Apr/22
![{ ((t^(256) = log _t (x)⇒x=t^t^(256) )),((t^(128) = log _t (log _t (2))+log _t (24)−128)) :} ⇒t^(128) = log _t (24 log _t (2))−128 ⇒[ log _t (24.log _t (2))−128 ]^2 = log _t (x)](Q168385.png)
$$\:\begin{cases}{{t}^{\mathrm{256}} =\:\mathrm{log}\:_{{t}} \left({x}\right)\Rightarrow{x}={t}^{{t}^{\mathrm{256}} } }\\{{t}^{\mathrm{128}} \:=\:\mathrm{log}\:_{{t}} \left(\mathrm{log}\:_{{t}} \left(\mathrm{2}\right)\right)+\mathrm{log}\:_{{t}} \left(\mathrm{24}\right)−\mathrm{128}}\end{cases} \\ $$$$\:\Rightarrow{t}^{\mathrm{128}} =\:\mathrm{log}\:_{{t}} \left(\mathrm{24}\:\mathrm{log}\:_{{t}} \left(\mathrm{2}\right)\right)−\mathrm{128} \\ $$$$\Rightarrow\left[\:\mathrm{log}\:_{{t}} \left(\mathrm{24}.\mathrm{log}\:_{{t}} \left(\mathrm{2}\right)\right)−\mathrm{128}\:\right]^{\mathrm{2}} =\:\mathrm{log}\:_{{t}} \left({x}\right) \\ $$
Answered by mr W last updated on 10/Apr/22
$$\mathrm{log}_{{t}} \:\left(\mathrm{24}\:\mathrm{log}_{{t}} \:\mathrm{2}\right)−\mathrm{128}={t}^{\mathrm{128}} \\ $$$$\mathrm{log}_{{t}} \:\left(\mathrm{log}_{{t}} \:\mathrm{2}^{\mathrm{24}} \right)=\mathrm{128}+{t}^{\mathrm{128}} \\ $$$$\mathrm{log}_{{t}} \:\mathrm{2}^{\mathrm{24}} ={t}^{\mathrm{128}+{t}^{\mathrm{128}} } ={t}^{\mathrm{128}} {t}^{{t}^{\mathrm{128}} } \\ $$$$\mathrm{2}^{\mathrm{24}} ={t}^{{t}^{\mathrm{128}} {t}^{{t}^{\mathrm{128}} } } =\left({t}^{{t}^{\mathrm{128}} } \right)^{{t}^{{t}^{\mathrm{128}} } } \\ $$$${let}\:{a}={t}^{{t}^{\mathrm{128}} } \\ $$$$\Rightarrow{a}^{{a}} =\mathrm{2}^{\mathrm{24}} =\mathrm{2}^{\mathrm{3}×\mathrm{8}} =\mathrm{8}^{\mathrm{8}} \:\Rightarrow{a}=\mathrm{8} \\ $$$${or} \\ $$$${e}^{\mathrm{ln}\:{a}} \mathrm{ln}\:{a}=\mathrm{24ln}\:\mathrm{2} \\ $$$$\Rightarrow\mathrm{ln}\:{a}={W}\left(\mathrm{24ln}\:\mathrm{2}\right) \\ $$$$\Rightarrow{a}={e}^{{W}\left(\mathrm{24ln}\:\mathrm{2}\right)} =\frac{\mathrm{24ln}\:\mathrm{2}}{{W}\left(\mathrm{24ln}\:\mathrm{2}\right)}=\mathrm{8} \\ $$$${t}^{{t}^{\mathrm{128}} } ={a} \\ $$$$\left({t}^{\mathrm{128}} \right)^{{t}^{\mathrm{128}} } ={a}^{\mathrm{128}} =\mathrm{8}^{\mathrm{128}} \\ $$$${let}\:{b}={t}^{\mathrm{128}} \\ $$$${b}^{{b}} =\mathrm{8}^{\mathrm{2}×\mathrm{64}} =\mathrm{64}^{\mathrm{64}} \:\Rightarrow{b}=\mathrm{64} \\ $$$${or} \\ $$$$\Rightarrow{b}={t}^{\mathrm{128}} =\frac{\mathrm{384ln}\:\mathrm{2}}{{W}\left(\mathrm{384ln}\:\mathrm{2}\right)}=\mathrm{64} \\ $$$$\Rightarrow{t}=\mathrm{64}^{\frac{\mathrm{1}}{\mathrm{128}}} =\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{64}}} \\ $$$$ \\ $$$$\mathrm{log}_{{t}} \:{x}={t}^{\mathrm{256}} \\ $$$$\Rightarrow{x}={t}^{{t}^{\mathrm{256}} } ={t}^{{t}^{\mathrm{128}} {t}^{\mathrm{128}} } =\left({t}^{{t}^{\mathrm{128}} } \right)^{{t}^{\mathrm{128}} } ={a}^{{b}} =\mathrm{8}^{\mathrm{64}} =\mathrm{2}^{\mathrm{192}} \\ $$
Commented by Tawa11 last updated on 09/Apr/22
$$\mathrm{Great}\:\mathrm{sir}. \\ $$