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

$$findxandy:$$$$\{\begin{array}{l}2x^y-x^{-y}=1\\ {log}_{2}y=\sqrt{x}\end{array}$$

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

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

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