Question Number 115858 by bemath last updated on 29/Sep/20
$${What}\:{are}\:{all}\:{ordered}\:{pairs}\:{of}\:{real} \\ $$$${number}\:\left({x},{y}\right)\:{for}\:{which}\: \\ $$$$\mathrm{5}^{{y}−{x}} \:\left({x}+{y}\right)\:=\:\mathrm{1}\:{and}\:\left({x}+{y}\right)^{{x}−{y}} \:=\:\mathrm{5} \\ $$
Answered by floor(10²Eta[1]) last updated on 29/Sep/20
$$\left(\mathrm{x}+\mathrm{y}\right)=\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{y}−\mathrm{x}} }=\mathrm{5}^{\mathrm{x}−\mathrm{y}} \Rightarrow\left(\mathrm{5}^{\left(\mathrm{x}−\mathrm{y}\right)} \right)^{\mathrm{x}−\mathrm{y}} =\mathrm{5} \\ $$$$\Rightarrow\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} =\mathrm{1}\therefore\mathrm{x}−\mathrm{y}=\pm\mathrm{1} \\ $$$$\mathrm{x}=\mathrm{y}\pm\mathrm{1} \\ $$$$\mathrm{y}=\mathrm{2}\Rightarrow\mathrm{x}=\mathrm{3} \\ $$$$\mathrm{y}=\frac{\mathrm{3}}{\mathrm{5}}\Rightarrow\mathrm{x}=\frac{−\mathrm{2}}{\mathrm{5}} \\ $$
Answered by bobhans last updated on 29/Sep/20
![⇔ (x+y) = 5^(x−y) ∧ (x+y)^(x−y) = 5 ⇒ [ (x+y)^(x−y) ]^(x−y) = 5^(x−y) ⇒ (x+y)^((x−y)^2 ) = (x+y)⇒ (x−y)^2 =1 → { ((x=y+1 )),((x=y−1)) :} case(1) →x=y+1 (2y+1)^1 = 5 ⇒y = 2 ∧x=3 case(2)→x=y−1 (2y−1)^(−1) = 5 ⇒2y−1=(1/5) y = (3/5) ∧ x=−(2/5) solution : (−(2/5),(3/5)) and (3,2)](https://www.tinkutara.com/question/Q115868.png)
$$\Leftrightarrow\:\left({x}+{y}\right)\:=\:\mathrm{5}^{{x}−{y}} \:\wedge\:\left({x}+{y}\right)^{{x}−{y}} \:=\:\mathrm{5} \\ $$$$\Rightarrow\:\left[\:\left({x}+{y}\right)^{{x}−{y}} \:\right]^{{x}−{y}} \:=\:\mathrm{5}^{{x}−{y}} \\ $$$$\Rightarrow\:\left({x}+{y}\right)^{\left({x}−{y}\right)^{\mathrm{2}} } \:=\:\left({x}+{y}\right)\Rightarrow\:\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\rightarrow\begin{cases}{{x}={y}+\mathrm{1}\:}\\{{x}={y}−\mathrm{1}}\end{cases} \\ $$$${case}\left(\mathrm{1}\right)\:\rightarrow{x}={y}+\mathrm{1} \\ $$$$\:\left(\mathrm{2}{y}+\mathrm{1}\right)^{\mathrm{1}} \:=\:\mathrm{5}\:\Rightarrow{y}\:=\:\mathrm{2}\:\wedge{x}=\mathrm{3} \\ $$$${case}\left(\mathrm{2}\right)\rightarrow{x}={y}−\mathrm{1} \\ $$$$\left(\mathrm{2}{y}−\mathrm{1}\right)^{−\mathrm{1}} \:=\:\mathrm{5}\:\Rightarrow\mathrm{2}{y}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\:{y}\:=\:\frac{\mathrm{3}}{\mathrm{5}}\:\wedge\:{x}=−\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${solution}\::\:\left(−\frac{\mathrm{2}}{\mathrm{5}},\frac{\mathrm{3}}{\mathrm{5}}\right)\:{and}\:\left(\mathrm{3},\mathrm{2}\right) \\ $$