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Question Number 143811 by liberty last updated on 18/Jun/21

$$\{\begin{array}{l}5(\log {}_{\mathrm{y}}\left(\mathrm{x}\right)+\log {}_{\mathrm{x}}\left(\mathrm{y}\right))=26\\ \mathrm{xy}=64\end{array}\mathrm{then}$$$${\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{xy}=?$$

Answered by liberty last updated on 18/Jun/21

$$\{\begin{array}{l}\log {}_{\mathrm{y}}\left(\mathrm{x}\right)+\log {}_{\mathrm{x}}\left(\mathrm{y}\right)=\frac{26}{5}\\ \mathrm{xy}=64\end{array}$$$$\mathrm{let}\log {}_{\mathrm{y}}\left(\mathrm{x}\right)=\mathrm{u}\to \mathrm{u}+\frac{1}{\mathrm{u}}=\frac{26}{5}$$$$\Rightarrow \mathrm{it}\mathrm{follows}\mathrm{that}\mathrm{u}=5$$$$\mathrm{and}\mathrm{x}={\mathrm{y}}^5\wedge {\mathrm{y}}^6=64$$$$\Rightarrow \mathrm{y}=2\mathrm{\&}\mathrm{x}=32$$$$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{xy}={(\mathrm{x}+\mathrm{y})}^2-\mathrm{xy}$$$$={34}^2-64$$

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