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Question Number 196288 by mathlove last updated on 22/Aug/23

4^x =(√5^y )=400 ((xy)/(2x+y))=?

$$\mathrm{4}^{{x}} =\sqrt{\mathrm{5}^{{y}} }=\mathrm{400} \\ $$$$\frac{{xy}}{\mathrm{2}{x}+{y}}=? \\ $$

Commented by hardmath last updated on 22/Aug/23

→ x = log_2 (20) → y = 4 log_2 (20) ⇒ (1/((xy)/(2x + y))) = log_(20) ((√(20))) = (1/2) ⇒ ((xy)/(2x + y)) = 2 ✓

$$\rightarrow\:\mathrm{x}\:=\:\mathrm{log}_{\mathrm{2}} \left(\mathrm{20}\right) \\ $$$$\rightarrow\:\mathrm{y}\:=\:\mathrm{4}\:\mathrm{log}_{\mathrm{2}} \left(\mathrm{20}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\frac{\mathrm{xy}}{\mathrm{2x}\:+\:\mathrm{y}}}\:=\:\mathrm{log}_{\mathrm{20}} \left(\sqrt{\mathrm{20}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\frac{\mathrm{xy}}{\mathrm{2x}\:+\:\mathrm{y}}\:=\:\mathrm{2}\:\checkmark \\ $$

Answered by Rasheed.Sindhi last updated on 22/Aug/23

4^x =(√5^y )=400 ; ((xy)/(2x+y))=? x=((log 400)/(log 4)) (y/2)=((log 400)/(log 5))⇒y=((2log 400)/(log 5)) xy=(((log 400)/(log 4)))(((2log 400)/(log 5))) =((2(log 400)^2 )/(log 4∙log 5)) 2x+y=2(((log 400)/(log 4)))+((2log 400)/(log 5)) =((2log 400∙log 5+2log 400∙log 4)/(log 4∙log 5)) ((xy)/(2x+y))=((2(log 400)^2 )/(2log 400∙log 5+2log 400∙log 4)) =((2log 400)/(2log 5+2log 4)) =((log(20^2 ))/(log 5+log 4)) =((2log 20)/(log 20))=2

$$\mathrm{4}^{{x}} =\sqrt{\mathrm{5}^{{y}} }=\mathrm{400}\:;\:\frac{{xy}}{\mathrm{2}{x}+{y}}=? \\ $$$${x}=\frac{\mathrm{log}\:\mathrm{400}}{\mathrm{log}\:\mathrm{4}} \\ $$$$\frac{{y}}{\mathrm{2}}=\frac{\mathrm{log}\:\mathrm{400}}{\mathrm{log}\:\mathrm{5}}\Rightarrow{y}=\frac{\mathrm{2log}\:\mathrm{400}}{\mathrm{log}\:\mathrm{5}} \\ $$$${xy}=\left(\frac{\mathrm{log}\:\mathrm{400}}{\mathrm{log}\:\mathrm{4}}\right)\left(\frac{\mathrm{2log}\:\mathrm{400}}{\mathrm{log}\:\mathrm{5}}\right) \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{2}\left(\mathrm{log}\:\mathrm{400}\right)^{\mathrm{2}} }{\mathrm{log}\:\mathrm{4}\centerdot\mathrm{log}\:\mathrm{5}} \\ $$$$\mathrm{2}{x}+{y}=\mathrm{2}\left(\frac{\mathrm{log}\:\mathrm{400}}{\mathrm{log}\:\mathrm{4}}\right)+\frac{\mathrm{2log}\:\mathrm{400}}{\mathrm{log}\:\mathrm{5}} \\ $$$$=\frac{\mathrm{2log}\:\mathrm{400}\centerdot\mathrm{log}\:\mathrm{5}+\mathrm{2log}\:\mathrm{400}\centerdot\mathrm{log}\:\mathrm{4}}{\mathrm{log}\:\mathrm{4}\centerdot\mathrm{log}\:\mathrm{5}} \\ $$$$\frac{{xy}}{\mathrm{2}{x}+{y}}=\frac{\mathrm{2}\left(\mathrm{log}\:\mathrm{400}\right)^{\mathrm{2}} }{\mathrm{2log}\:\mathrm{400}\centerdot\mathrm{log}\:\mathrm{5}+\mathrm{2log}\:\mathrm{400}\centerdot\mathrm{log}\:\mathrm{4}} \\ $$$$\:\:=\frac{\mathrm{2log}\:\mathrm{400}}{\mathrm{2log}\:\mathrm{5}+\mathrm{2log}\:\mathrm{4}} \\ $$$$\:\:\:\:=\frac{\mathrm{log}\left(\mathrm{20}^{\mathrm{2}} \right)}{\mathrm{log}\:\mathrm{5}+\mathrm{log}\:\mathrm{4}} \\ $$$$\:\:\:=\frac{\mathrm{2log}\:\mathrm{20}}{\mathrm{log}\:\mathrm{20}}=\mathrm{2} \\ $$

Commented by mathlove last updated on 22/Aug/23

thanks

$${thanks} \\ $$

Answered by Rasheed.Sindhi last updated on 22/Aug/23

4^x =(√5^y )=400;((xy)/(2x+y))=? { ( { ((2^(2x) =20^2 ⇒2^x =20)),((5^(y/2) =400⇒5^y =400^2 =20^4 )) :}),((((xy)/(2x+y))=(1/(2((1/y))+(1/x))))) :} { ((2^x =20⇒x=((ln 20)/(ln 2))⇒(1/x)=((ln 2)/(ln 20)))),((5^y =20^4 ⇒y=((4ln 20)/(ln 5))⇒(1/y)=((ln 5)/(4ln 20)))) :} ((xy)/(2x+y))=(1/(2((1/y))+(1/x))) =(1/(2(((ln 5)/(4ln 20)))+((ln 2)/(ln 20)))) =(1/(((ln 5)/(2ln 20))+((ln 2)/(ln 20)))) =(1/((ln 5+2ln 2)/(2ln 20))) =(1/((ln(5∙2^2 ))/(2ln 20)))=(1/(1/2))=2✓

$$\mathrm{4}^{{x}} =\sqrt{\mathrm{5}^{{y}} }=\mathrm{400};\frac{{xy}}{\mathrm{2}{x}+{y}}=? \\ $$$$\begin{cases}{\begin{cases}{\mathrm{2}^{\mathrm{2}{x}} =\mathrm{20}^{\mathrm{2}} \Rightarrow\mathrm{2}^{{x}} =\mathrm{20}}\\{\mathrm{5}^{{y}/\mathrm{2}} =\mathrm{400}\Rightarrow\mathrm{5}^{{y}} =\mathrm{400}^{\mathrm{2}} =\mathrm{20}^{\mathrm{4}} }\end{cases}}\\{\frac{{xy}}{\mathrm{2}{x}+{y}}=\frac{\mathrm{1}}{\mathrm{2}\left(\frac{\mathrm{1}}{{y}}\right)+\frac{\mathrm{1}}{{x}}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{2}^{{x}} =\mathrm{20}\Rightarrow{x}=\frac{\mathrm{ln}\:\mathrm{20}}{\mathrm{ln}\:\mathrm{2}}\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{20}}}\\{\mathrm{5}^{{y}} =\mathrm{20}^{\mathrm{4}} \Rightarrow{y}=\frac{\mathrm{4ln}\:\mathrm{20}}{\mathrm{ln}\:\mathrm{5}}\Rightarrow\frac{\mathrm{1}}{{y}}=\frac{\mathrm{ln}\:\mathrm{5}}{\mathrm{4ln}\:\mathrm{20}}}\end{cases} \\ $$$$\frac{{xy}}{\mathrm{2}{x}+{y}}=\frac{\mathrm{1}}{\mathrm{2}\left(\frac{\mathrm{1}}{{y}}\right)+\frac{\mathrm{1}}{{x}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left(\frac{\mathrm{ln}\:\mathrm{5}}{\mathrm{4ln}\:\mathrm{20}}\right)+\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{20}}} \\ $$$$=\frac{\mathrm{1}}{\frac{\mathrm{ln}\:\mathrm{5}}{\mathrm{2ln}\:\mathrm{20}}+\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{20}}} \\ $$$$=\frac{\mathrm{1}}{\frac{\mathrm{ln}\:\mathrm{5}+\mathrm{2ln}\:\mathrm{2}}{\mathrm{2ln}\:\mathrm{20}}} \\ $$$$=\frac{\mathrm{1}}{\frac{\mathrm{ln}\left(\mathrm{5}\centerdot\mathrm{2}^{\mathrm{2}} \right)}{\mathrm{2ln}\:\mathrm{20}}}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{2}\checkmark \\ $$

Answered by Rasheed.Sindhi last updated on 22/Aug/23

4^x =(√5^y )=400; ((xy)/(2x+y))=? 2^(2x) =5^(y/2) =2^4 .5^2 ⇒ { ((2^(2x−4) =5^2 ⇒2^(x−2) =5)),((5^y =2^8 .5^4 ⇒5^(y−4) =2^8 )) :} (2^(x−2) )^(y−4) =2^8 xy−4x−2y+8=8 xy=4x+2y=2(2x+y) ⇒((xy)/(2x+y))=2

$$\mathrm{4}^{{x}} =\sqrt{\mathrm{5}^{{y}} }=\mathrm{400};\:\:\frac{{xy}}{\mathrm{2}{x}+{y}}=? \\ $$$$\mathrm{2}^{\mathrm{2}{x}} =\mathrm{5}^{\frac{{y}}{\mathrm{2}}} =\mathrm{2}^{\mathrm{4}} .\mathrm{5}^{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{\mathrm{2}^{\mathrm{2}{x}−\mathrm{4}} =\mathrm{5}^{\mathrm{2}} \Rightarrow\mathrm{2}^{{x}−\mathrm{2}} =\mathrm{5}}\\{\mathrm{5}^{{y}} =\mathrm{2}^{\mathrm{8}} .\mathrm{5}^{\mathrm{4}} \Rightarrow\mathrm{5}^{{y}−\mathrm{4}} =\mathrm{2}^{\mathrm{8}} }\end{cases} \\ $$$$\left(\mathrm{2}^{{x}−\mathrm{2}} \right)^{{y}−\mathrm{4}} =\mathrm{2}^{\mathrm{8}} \\ $$$${xy}−\mathrm{4}{x}−\mathrm{2}{y}+\mathrm{8}=\mathrm{8} \\ $$$${xy}=\mathrm{4}{x}+\mathrm{2}{y}=\mathrm{2}\left(\mathrm{2}{x}+{y}\right) \\ $$$$\Rightarrow\frac{{xy}}{\mathrm{2}{x}+{y}}=\mathrm{2} \\ $$

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