Question Number 218504 by hardmath last updated on 10/Apr/25
$$\mathrm{Let}\:\:\:\mathrm{a}>\mathrm{1}\:\:\:\mathrm{fixed} \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{the}\:\mathrm{system} \\ $$$$\begin{cases}{\mathrm{a}^{\boldsymbol{\mathrm{x}}} \:\:+\:\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:\:=\:\:\mathrm{ay}\:+\:\mathrm{2}}\\{\mathrm{a}^{\boldsymbol{\mathrm{y}}} \:\:+\:\:\mathrm{2}^{\boldsymbol{\mathrm{y}}} \:\:=\:\:\mathrm{az}\:\:+\:\:\mathrm{2}}\\{\mathrm{a}^{\boldsymbol{\mathrm{z}}} \:\:+\:\:\mathrm{2}^{\boldsymbol{\mathrm{z}}} \:\:=\:\:\mathrm{ax}\:\:+\:\:\mathrm{2}}\end{cases}\: \\ $$
Answered by vnm last updated on 11/Apr/25
$${Suppose}\:{y}>{x}. \\ $$$${Then}\:{a}^{{y}} +\mathrm{2}^{{y}} >{a}^{{x}} +\mathrm{2}^{{x}} \Rightarrow \\ $$$${az}+\mathrm{2}>{ay}+\mathrm{2}\Rightarrow{z}>{y}\Rightarrow \\ $$$${a}^{{z}} +\mathrm{2}^{{z}} >{a}^{{y}} +\mathrm{2}^{{y}} >{a}^{{x}} +\mathrm{2}^{{x}} >{ax}+\mathrm{2} \\ $$$${because}\:{a}^{{x}} +\mathrm{2}^{{x}} ={ay}+\mathrm{2}\:\&\:{y}>{x}. \\ $$$${Contradiction}\:{with}\:{the}\:{third}\:{equation}. \\ $$$${Similar}\:{reasoning}\:{for}\:{the}\:{case}\:{y}<{x},\: \\ $$$${so}\:{the}\:{only}\:{possible}\:{case}\:{is}\:{x}={y}={z}. \\ $$$$\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }\left({a}^{{x}} +\mathrm{2}^{{x}} \right)>\mathrm{0}\:\forall{x}\:\Rightarrow\:{equation}\:{a}^{{x}} +\mathrm{2}^{{x}} ={ax}+\mathrm{2}\:{cannot}\:{have}\:{more}\:{than}\:{two}\:{roots}. \\ $$$${x}_{\mathrm{1}} ={y}_{\mathrm{1}} ={z}_{\mathrm{1}} =\mathrm{0},\:{x}_{\mathrm{2}} ={y}_{\mathrm{2}} ={z}_{\mathrm{2}} =\mathrm{1} \\ $$
Commented by hardmath last updated on 11/Apr/25
$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professor} \\ $$