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Question Number 73191 by Mr. K last updated on 07/Nov/19

find x and y: { ((2x^y −x^(−y) =1)),((log_2 y=(√x))) :}

$${find}\:{x}\:{and}\:{y}: \\ $$$$\begin{cases}{\mathrm{2}{x}^{{y}} −{x}^{−{y}} =\mathrm{1}}\\{{log}_{\mathrm{2}} {y}=\sqrt{{x}}}\end{cases} \\ $$

Answered by mind is power last updated on 07/Nov/19

⇔ { ((2x^y −(1/x^y )=1....1)),((ln(y)=(√x)ln(2)...2⇒y∈C^∗ )) :} let z=x^y ⇒2z−(1/z)=1⇔2z^2 −z−1=0 ⇒z∈{1,−(1/2)} z=1⇒x^y =1⇒yln(x)=0⇒y=0 or x=1 y=0 not possibl ⇒x=1 ln(y)=1.ln(2)=⇒y=2⇒(x,y)=(1,2) x^y =−(1/2)is over C its very hard to solve it too bee continued

$$\Leftrightarrow\begin{cases}{\mathrm{2x}^{\mathrm{y}} −\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{y}} }=\mathrm{1}....\mathrm{1}}\\{\mathrm{ln}\left(\mathrm{y}\right)=\sqrt{\mathrm{x}}\mathrm{ln}\left(\mathrm{2}\right)...\mathrm{2}\Rightarrow\mathrm{y}\in\mathbb{C}^{\ast} }\end{cases} \\ $$$$\mathrm{let}\:\mathrm{z}=\mathrm{x}^{\mathrm{y}} \\ $$$$\Rightarrow\mathrm{2z}−\frac{\mathrm{1}}{\mathrm{z}}=\mathrm{1}\Leftrightarrow\mathrm{2z}^{\mathrm{2}} −\mathrm{z}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{z}\in\left\{\mathrm{1},−\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$\mathrm{z}=\mathrm{1}\Rightarrow\mathrm{x}^{\mathrm{y}} =\mathrm{1}\Rightarrow\mathrm{yln}\left(\mathrm{x}\right)=\mathrm{0}\Rightarrow\mathrm{y}=\mathrm{0}\:\mathrm{or}\:\mathrm{x}=\mathrm{1} \\ $$$$\mathrm{y}=\mathrm{0}\:\mathrm{not}\:\mathrm{possibl}\:\Rightarrow\mathrm{x}=\mathrm{1} \\ $$$$\mathrm{ln}\left(\mathrm{y}\right)=\mathrm{1}.\mathrm{ln}\left(\mathrm{2}\right)=\Rightarrow\mathrm{y}=\mathrm{2}\Rightarrow\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{1},\mathrm{2}\right) \\ $$$$\mathrm{x}^{\mathrm{y}} =−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{is}\:\mathrm{over}\:\mathrm{C}\:\mathrm{its}\:\mathrm{very}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{too}\:\mathrm{bee}\:\mathrm{continued} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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