Question Number 78732 by Pratah last updated on 20/Jan/20
Commented by john santu last updated on 20/Jan/20
$$\left(\mathrm{1}\right)\:\left(\mathrm{6}−{y}\right)^{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{6}}} \:=\:\mathrm{3}^{{x}} \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{6}−{z}\right)^{\sqrt{{y}^{\mathrm{2}} +\mathrm{2}{y}+\mathrm{6}}} \:=\:\mathrm{3}^{{y}} \\ $$$$\left(\mathrm{3}\right)\:\left(\mathrm{6}−{x}\right)^{\sqrt{{z}^{\mathrm{2}} +\mathrm{2}{z}+\mathrm{6}}} \:=\:\mathrm{3}^{{z}} \\ $$$$ \\ $$
Commented by mr W last updated on 20/Jan/20
$${a}\:{tormenting}\:{question}\:\left({again}\right)... \\ $$$${answer}\:{is}\:{x}={y}={z}={something}\:{which} \\ $$$${can}\:{not}\:{be}\:{exactly}\:{calculated}\approx\mathrm{3}.\mathrm{800135}. \\ $$