Question Number 203424 by Calculusboy last updated on 18/Jan/24
Answered by Rasheed.Sindhi last updated on 19/Jan/24
$${x}+{y}=\mathrm{2}^{{x}−{y}} \:…\left({i}\right) \\ $$$$\left({x}+{y}\right)^{{x}−{y}} =\mathrm{2}…\left({ii}\right) \\ $$$$\left({ii}\right)/\left({i}\right):\:\left({x}+{y}\right)^{{x}−{y}−\mathrm{1}} =\mathrm{2}^{\mathrm{1}−{x}+{y}} \\ $$$$\Rightarrow{x}+{y}=\mathrm{2}\:\wedge\:{x}−{y}−\mathrm{1}=\mathrm{1}−{x}+{y} \\ $$$$\Rightarrow{x}+{y}=\mathrm{2}\:\wedge\:\mathrm{2}{x}−\mathrm{2}{y}=\mathrm{2}\Rightarrow{x}−{y}=\mathrm{1} \\ $$$$\:\:\:\:{x}=\mathrm{3}/\mathrm{2}\:\:\wedge\:\:{y}=\mathrm{1}/\mathrm{2} \\ $$
Commented by mr W last updated on 19/Jan/24
$${for}\:\left({x}+{y}\right)^{{x}−{y}−\mathrm{1}} =\mathrm{2}^{\mathrm{1}−{x}+{y}} =\mathrm{2}^{−\left({x}−{y}−\mathrm{1}\right)} \\ $$$${there}\:{is}\:{an}\:{other}\:{possibility}: \\ $$$${x}+{y}=\mathrm{2}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Rasheed.Sindhi last updated on 19/Jan/24
$$\mathcal{N}{ice}\:\boldsymbol{{sir}},\:\mathcal{T}{hanks}! \\ $$
Commented by Rasheed.Sindhi last updated on 19/Jan/24
$$\begin{cases}{{x}+{y}=\:\:\:\:\:\:\mathrm{2}^{{x}−{y}} ….{i}}\\{\:\:\:\:\mathrm{2}\:\:\:=\left({x}+{y}\right)^{{x}−{y}} …{ii}}\end{cases} \\ $$$${i}/{ii}:\:\frac{{x}+{y}}{\mathrm{2}}=\left(\frac{\mathrm{2}}{{x}+{y}}\right)^{{x}−{y}} \\ $$$$\:\:\:\:\:\:\frac{{x}+{y}}{\mathrm{2}}=\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{{y}−{x}} \\ $$$$\:\:\:\:\:\:\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{{y}−{x}−\mathrm{1}} =\mathrm{1}=\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{0}} \\ $$$${x}−{y}=−\mathrm{1} \\ $$$$\begin{cases}{\frac{{x}+{y}}{\mathrm{2}}=\mathrm{1}\Rightarrow{x}+{y}=\mathrm{2}}\\{\:\:\:\:\:\mathrm{or}}\\{{x}−{y}=−\mathrm{1}}\end{cases}\: \\ $$$${i}×{ii}:\:\mathrm{2}\left({x}+{y}\right)=\mathrm{2}^{{x}−{y}} \left({x}+{y}\right)^{{x}−{y}} \\ $$$$\frac{\mathrm{2}}{\mathrm{2}^{{x}−{y}} }=\frac{\left({x}+{y}\right)^{{x}−{y}} }{{x}+{y}} \\ $$$$\left({x}+{y}\right)^{{x}−{y}−\mathrm{1}} =\mathrm{2}^{\mathrm{1}−{x}+{y}} \\ $$$$\left({x}+{y}\right)^{{x}−{y}−\mathrm{1}} =\mathrm{2}^{−\left({x}−{y}−\mathrm{1}\right)} \\ $$$$\begin{cases}{{x}+{y}=\mathrm{2}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}}\\{{x}+{y}=\mathrm{2}\:\wedge\:{x}−{y}−\mathrm{1}=−{x}+{y}+\mathrm{1}}\end{cases} \\ $$$$ \\ $$
Commented by Calculusboy last updated on 21/Jan/24
$$\boldsymbol{\mathrm{thanks}}\:\boldsymbol{\mathrm{sir}} \\ $$
Answered by mr W last updated on 19/Jan/24
$$\mathrm{2}^{\left({x}−{y}\right)^{\mathrm{2}} } =\mathrm{2} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${x}−{y}=\pm\mathrm{1} \\ $$$${x}+{y}=\mathrm{2}^{\pm\mathrm{1}} =\mathrm{2}\:{or}\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\pm\mathrm{1}+\mathrm{2}^{\pm\mathrm{1}} }{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}\:{or}\:−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${y}=\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}\:{or}\:−\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\left({x},\:{y}\right)=\left(\frac{\mathrm{3}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{2}}\right)\:{or}\:\left(−\frac{\mathrm{1}}{\mathrm{4}},\:\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$
Commented by Calculusboy last updated on 21/Jan/24
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Answered by esmaeil last updated on 18/Jan/24
$$\left(\mathrm{2}\left({x}+{y}\right)\right)^{{x}−{y}} =\mathrm{2}\left({x}+{y}\right)\rightarrow \\ $$$${x}={y}+\mathrm{1}\rightarrow \\ $$$$\mathrm{2}{y}+\mathrm{1}=\mathrm{2}\rightarrow{y}=\frac{\mathrm{1}}{\mathrm{2}}\rightarrow \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by Calculusboy last updated on 21/Jan/24
$$\boldsymbol{{nice}}\:\boldsymbol{{sir}} \\ $$