Question Number 149770 by mathdanisur last updated on 07/Aug/21
$$\mathrm{6}\boldsymbol{{z}}^{\mathrm{4}} \:=\:\boldsymbol{{z}}^{\boldsymbol{{z}}^{\mathrm{2}} } \:\:\Rightarrow\:\:\mathrm{1}\:+\:\boldsymbol{{z}}^{\mathrm{2}} \:+\:\boldsymbol{{z}}^{\mathrm{4}} \:=\:? \\ $$
Commented by amin96 last updated on 07/Aug/21
$${z}^{\mathrm{2}} ={log}_{{z}} \mathrm{6}{z}^{\mathrm{4}} \:\:\:\Rightarrow\:\:{z}^{\mathrm{2}} ={log}_{{z}} \mathrm{6}+\mathrm{4} \\ $$$${z}^{\mathrm{2}} ={log}_{\mathrm{6}} {z}\:\:\:{log}_{\mathrm{6}} {z}={log}_{{z}} \mathrm{6}+\mathrm{4} \\ $$$${log}_{\mathrm{6}} {z}−\frac{\mathrm{1}}{{log}_{\mathrm{6}} {z}}=\mathrm{4}\:\:\:\:\:{log}_{\mathrm{6}} {z}={x}\:\:\:{z}>\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}=\mathrm{0}\:\:\:\:\:{x}=\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$${log}_{\mathrm{6}} {z}=\mathrm{2}+\sqrt{\mathrm{3}}\:\:\:\:{z}^{\mathrm{2}} =\mathrm{2}+\sqrt{\mathrm{3}}\:\:{z}^{\mathrm{4}} =\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\mathrm{1}+{z}^{\mathrm{2}} +{z}^{\mathrm{4}} =\mathrm{1}+\mathrm{2}+\sqrt{\mathrm{3}}+\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}=\mathrm{10}+\mathrm{5}\sqrt{\mathrm{3}} \\ $$
Commented by mathdanisur last updated on 07/Aug/21
$$\mathrm{Thanks}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$