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Question-209062




Question Number 209062 by hardmath last updated on 01/Jul/24
Answered by som(math1967) last updated on 01/Jul/24
 x^y^z  =1   x,y,z ∈N ⇒1^y^z  =1 ⇒x=1   y^z^x  =125⇒y^z =5^3 ⇒y=5⇒z=3   z^y^x  =3^5^1  =243   9x+10y−18z=9+10×5−18×3  =5
$$\:{x}^{{y}^{{z}} } =\mathrm{1} \\ $$$$\:{x},{y},{z}\:\in{N}\:\Rightarrow\mathrm{1}^{{y}^{{z}} } =\mathrm{1}\:\Rightarrow{x}=\mathrm{1} \\ $$$$\:{y}^{{z}^{{x}} } =\mathrm{125}\Rightarrow{y}^{{z}} =\mathrm{5}^{\mathrm{3}} \Rightarrow{y}=\mathrm{5}\Rightarrow{z}=\mathrm{3} \\ $$$$\:{z}^{{y}^{{x}} } =\mathrm{3}^{\mathrm{5}^{\mathrm{1}} } =\mathrm{243} \\ $$$$\:\mathrm{9}{x}+\mathrm{10}{y}−\mathrm{18}{z}=\mathrm{9}+\mathrm{10}×\mathrm{5}−\mathrm{18}×\mathrm{3} \\ $$$$=\mathrm{5} \\ $$

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