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Question-213639




Question Number 213639 by efronzo1 last updated on 12/Nov/24
Answered by golsendro last updated on 12/Nov/24
   8^x =25^y = t ⇒ { ((x= log _8 t)),((y= log _(25) t)) :}    ⇒8((1/x))+25((1/y))= 1     8. log _t  8 + 25. log _t 25=1      log _t (8^8 .25^5 )=1⇒t=8^8 .25^5        { ((8^x = 8^8 .25^(25) )),((25^y = 8^8 .25^(25) )) :}    ⇒((8^x +25^y ))^(1/(25)) = (2.8^8 .25^(25) )^(1/(25))          = (2^(25) .25^(25) )^(1/(25))  = 50
$$\:\:\:\mathrm{8}^{\mathrm{x}} =\mathrm{25}^{\mathrm{y}} =\:\mathrm{t}\:\Rightarrow\begin{cases}{\mathrm{x}=\:\mathrm{log}\:_{\mathrm{8}} \mathrm{t}}\\{\mathrm{y}=\:\mathrm{log}\:_{\mathrm{25}} \mathrm{t}}\end{cases} \\ $$$$\:\:\Rightarrow\mathrm{8}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)+\mathrm{25}\left(\frac{\mathrm{1}}{\mathrm{y}}\right)=\:\mathrm{1} \\ $$$$\:\:\:\mathrm{8}.\:\mathrm{log}\:_{\mathrm{t}} \:\mathrm{8}\:+\:\mathrm{25}.\:\mathrm{log}\:_{\mathrm{t}} \mathrm{25}=\mathrm{1} \\ $$$$\:\:\:\:\mathrm{log}\:_{\mathrm{t}} \left(\mathrm{8}^{\mathrm{8}} .\mathrm{25}^{\mathrm{5}} \right)=\mathrm{1}\Rightarrow\mathrm{t}=\mathrm{8}^{\mathrm{8}} .\mathrm{25}^{\mathrm{5}} \\ $$$$\:\:\:\:\begin{cases}{\mathrm{8}^{\mathrm{x}} =\:\mathrm{8}^{\mathrm{8}} .\mathrm{25}^{\mathrm{25}} }\\{\mathrm{25}^{\mathrm{y}} =\:\mathrm{8}^{\mathrm{8}} .\mathrm{25}^{\mathrm{25}} }\end{cases} \\ $$$$\:\:\Rightarrow\sqrt[{\mathrm{25}}]{\mathrm{8}^{\mathrm{x}} +\mathrm{25}^{\mathrm{y}} }=\:\left(\mathrm{2}.\mathrm{8}^{\mathrm{8}} .\mathrm{25}^{\mathrm{25}} \right)^{\frac{\mathrm{1}}{\mathrm{25}}} \\ $$$$\:\:\:\:\:\:\:=\:\left(\mathrm{2}^{\mathrm{25}} .\mathrm{25}^{\mathrm{25}} \right)^{\frac{\mathrm{1}}{\mathrm{25}}} \:=\:\mathrm{50} \\ $$

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