Question Number 113886 by Aina Samuel Temidayo last updated on 16/Sep/20 | ||
Answered by MJS_new last updated on 16/Sep/20 | ||
$$\left({a}\right) \\ $$ | ||
Commented by Aina Samuel Temidayo last updated on 16/Sep/20 | ||
$$\mathrm{Solution}\:\mathrm{please}? \\ $$ | ||
Commented by MJS_new last updated on 16/Sep/20 | ||
$$\mathrm{lol}...\:\mathrm{just}\:\mathrm{use}\:\mathrm{your}\:\mathrm{brain} \\ $$ | ||
Answered by Rasheed.Sindhi last updated on 16/Sep/20 | ||
$${a}^{{x}} =\left({x}+{y}+{z}\right)^{{y}} ,{a}^{{y}} =\left({x}+{y}+{z}\right)^{{z}} ,{a}^{{z}} =\left({x}+{y}+{z}\right)^{{x}} \\ $$$${a}=\left({x}+{y}+{z}\right)^{{y}/{x}} =\left({x}+{y}+{z}\right)^{{z}/{y}} =\left({x}+{y}+{z}\right)^{{x}/{z}} \\ $$$$\:\:\:\:\:\:\:\:\frac{{y}}{{x}}=\frac{{z}}{{y}}=\frac{{x}}{{z}}=\frac{{x}+{y}+{z}}{{x}+{y}+{z}}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:{y}={x},{z}={y},{x}={z} \\ $$$$\:\:\:\:\:\:{x}={y}={z}={k}\left({say}\right) \\ $$$${a}^{{k}} =\left({k}+{k}+{k}\right)^{{k}} \\ $$$${a}=\mathrm{3}{k} \\ $$$${k}={a}/\mathrm{3}={x}={y}={z} \\ $$ | ||
Answered by Rasheed.Sindhi last updated on 17/Sep/20 | ||
$${Multiplying}\:{all}\:{the}\:{three} \\ $$$${a}^{{x}+{y}+{z}} =\left({x}+{y}+{z}\right)^{{x}+{y}+{z}} \\ $$$${a}={x}+{y}+{z}...................\mathrm{A} \\ $$$${a}^{{x}} ={a}^{{y}} ,{a}^{{y}} ={a}^{{z}} ,{a}^{{z}} ={a}^{{x}} \\ $$$$\frac{{a}^{{x}} }{{a}^{{y}} }=\frac{{a}^{{y}} }{{a}^{{z}} }=\frac{{a}^{{z}} }{{a}^{{x}} }=\mathrm{1} \\ $$$${a}^{{x}−{y}} ={a}^{{y}−{z}} ={a}^{{z}−{x}} ={a}^{\mathrm{0}} \\ $$$${x}−{y}={y}−{z}={z}−{x}=\mathrm{0} \\ $$$${x}={y}={z}={k}\left({say}\right) \\ $$$${In}\:\:\mathrm{A} \\ $$$${a}={k}+{k}+{k}=\mathrm{3}{k}\Rightarrow{k}={a}/\mathrm{3} \\ $$$${k}={x}={y}={z}={a}/\mathrm{3} \\ $$ | ||