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Question-203424




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

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