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Question-200224

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Question Number 200224 by cortano12 last updated on 15/Nov/23
Answered by Rasheed.Sindhi last updated on 16/Nov/23
{ ((yx^(log_y x ) =x^(5/2) ...............(i))),((log_4 y.log_y (y−3x)=1...(ii) )) :} (ii)⇒ (1/(log_y 4 ))=(1/(log_y (y−3x) )) y−3x=4 y=x^a (say) log_y y=alog_y x a=(1/(log_y x)) y=x^(1/(log_y x)) (i)⇒x^(1/(log_y x)) ∙x^(log_y x ) =x^(5/2) x^(log_y x +(1/(log_y x))) =x^(5/2) log_y x +(1/(log_y x))=(5/2) 2(log_y x)^2 −5log_y x +2=0 (2log_y x−1)(log_y x−2)=0 log_y x=(1/2),2 ∧ y=3x+4 y^2 =x ∣ y^(1/2) =x (3x+4)^2 =x ∣ (3x+4)^(1/2) =x 9x^2 +23x+16=0_(No real roots) ∣ 3x+4=x^2 x^2 −3x−4=0 (x−4)(x+1)=0 x=4 ,x=−1 y=3(4)+4=16,3(−1)+4=1 y=16,1 (x,y)=(4,16),(−1,1)^(invalid)
$$\begin{cases}{{yx}^{\mathrm{log}_{{y}} {x}\:} ={x}^{\frac{\mathrm{5}}{\mathrm{2}}} ……………\left({i}\right)}\\{\mathrm{log}_{\mathrm{4}} {y}.\mathrm{log}_{{y}} \left({y}−\mathrm{3}{x}\right)=\mathrm{1}…\left({ii}\right)\:\:}\end{cases} \\ $$$$\left({ii}\right)\Rightarrow\:\frac{\mathrm{1}}{\mathrm{log}_{{y}} \mathrm{4}\:}=\frac{\mathrm{1}}{\mathrm{log}_{{y}} \left({y}−\mathrm{3}{x}\right)\:} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{y}−\mathrm{3}{x}=\mathrm{4} \\ $$$${y}={x}^{{a}} \:\:\left({say}\right) \\ $$$$\mathrm{log}_{{y}} {y}={a}\mathrm{log}_{{y}} {x} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{log}_{{y}} {x}}\: \\ $$$${y}={x}^{\frac{\mathrm{1}}{\mathrm{log}_{{y}} {x}}} \\ $$$$\left({i}\right)\Rightarrow{x}^{\frac{\mathrm{1}}{\mathrm{log}_{{y}} {x}}} \centerdot{x}^{\mathrm{log}_{{y}} {x}\:} ={x}^{\frac{\mathrm{5}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{log}_{{y}} {x}\:+\frac{\mathrm{1}}{\mathrm{log}_{{y}} {x}}} ={x}^{\frac{\mathrm{5}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\mathrm{log}_{{y}} {x}\:+\frac{\mathrm{1}}{\mathrm{log}_{{y}} {x}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{2}\left(\mathrm{log}_{{y}} {x}\right)^{\mathrm{2}} −\mathrm{5log}_{{y}} {x}\:+\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\left(\mathrm{2log}_{{y}} {x}−\mathrm{1}\right)\left(\mathrm{log}_{{y}} {x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\mathrm{log}_{{y}} {x}=\frac{\mathrm{1}}{\mathrm{2}},\mathrm{2}\:\:\wedge\:\:{y}=\mathrm{3}{x}+\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} ={x}\:\mid\:{y}^{\mathrm{1}/\mathrm{2}} ={x} \\ $$$$\:\:\:\:\:\left(\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{2}} ={x}\:\mid\:\left(\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{1}/\mathrm{2}} ={x} \\ $$$$\:\:\:\:\underset{{No}\:{real}\:{roots}} {\:\mathrm{9}{x}^{\mathrm{2}} +\mathrm{23}{x}+\mathrm{16}=\mathrm{0}}\:\mid\:\mathrm{3}{x}+\mathrm{4}={x}^{\mathrm{2}} \: \\ $$$$\:\:\:{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}=\mathrm{0} \\ $$$$\left({x}−\mathrm{4}\right)\left({x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{4}\:,{x}=−\mathrm{1} \\ $$$${y}=\mathrm{3}\left(\mathrm{4}\right)+\mathrm{4}=\mathrm{16},\mathrm{3}\left(−\mathrm{1}\right)+\mathrm{4}=\mathrm{1} \\ $$$${y}=\mathrm{16},\mathrm{1} \\ $$$$\left({x},{y}\right)=\left(\mathrm{4},\mathrm{16}\right),\overset{{invalid}} {\left(−\mathrm{1},\mathrm{1}\right)} \\ $$
Answered by Frix last updated on 16/Nov/23
With x, y >0... ...the 1^(st) equation is equal to (ln x −2ln y)(2ln x −ln y)=0 ⇒ y=(√x)∨y=x^2 ...the 2^(nd) equation is equal to 3x−y+4=0 ⇒ y=3x+4 (1) (√x)=3x+4 ⇒ x∉R (2) x^2 =3x+4 ⇒ x=4∧y=16
$$\mathrm{With}\:{x},\:{y}\:>\mathrm{0}… \\ $$$$…\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{equation}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left(\mathrm{ln}\:{x}\:−\mathrm{2ln}\:{y}\right)\left(\mathrm{2ln}\:{x}\:−\mathrm{ln}\:{y}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{y}=\sqrt{{x}}\vee{y}={x}^{\mathrm{2}} \\ $$$$…\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{equation}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\mathrm{3}{x}−{y}+\mathrm{4}=\mathrm{0}\:\Rightarrow\:{y}=\mathrm{3}{x}+\mathrm{4} \\ $$$$\left(\mathrm{1}\right)\:\sqrt{{x}}=\mathrm{3}{x}+\mathrm{4}\:\Rightarrow\:{x}\notin\mathbb{R} \\ $$$$\left(\mathrm{2}\right)\:{x}^{\mathrm{2}} =\mathrm{3}{x}+\mathrm{4}\:\Rightarrow\:{x}=\mathrm{4}\wedge{y}=\mathrm{16} \\ $$
Answered by Rasheed.Sindhi last updated on 16/Nov/23
{ ((y x^(log_y x ) =x^(5/2) .................(i))),((log_4 y.log_y (y−3x)=1....(ii) )) :} (i)⇒y=x^((5/2)−log_y x ) ⇒y^2 =x^(5−2log_y x ) ⇒log_y y^2 =(5−2log_y x)log_y x 2=5log_y x−2(log_y x)^2 2(log_y x)^2 −5log_y x+2=0 log_y x=((5±(√(25−16)))/4)=2,(1/2) y^2 =x ∣ y^(1/2) =x⇒y=x^2 .........(iii) (ii)⇒(1/(log_y 4 ))=(1/(log_y (y−3x) )) ⇒y−3x=4 ⇒y=3x+4...................(iv) (iii) & (iv) : (3x+4)^2 =x ∣ 3x+4=x^2 9x^2 +23x+16=0_(No real roots) ∣ x^2 −3x−4=0 (x−4)(x+1)=0⇒x=4,−1 ⇒y=3(4)+4 , 3(−1)+4=16,1 but y≠1 ∴ (x,y)=(16,4) ✓
$$\begin{cases}{{y}\:{x}^{\mathrm{log}_{{y}} {x}\:} ={x}^{\frac{\mathrm{5}}{\mathrm{2}}} ……………..\left({i}\right)}\\{\mathrm{log}_{\mathrm{4}} {y}.\mathrm{log}_{{y}} \left({y}−\mathrm{3}{x}\right)=\mathrm{1}….\left({ii}\right)\:\:}\end{cases} \\ $$$$\left({i}\right)\Rightarrow{y}={x}^{\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{log}_{{y}} {x}\:} \Rightarrow{y}^{\mathrm{2}} ={x}^{\mathrm{5}−\mathrm{2log}_{{y}} {x}\:} \\ $$$$\Rightarrow\mathrm{log}_{{y}} {y}^{\mathrm{2}} =\left(\mathrm{5}−\mathrm{2log}_{{y}} {x}\right)\mathrm{log}_{{y}} {x} \\ $$$$\mathrm{2}=\mathrm{5log}_{{y}} {x}−\mathrm{2}\left(\mathrm{log}_{{y}} {x}\right)^{\mathrm{2}} \:\:\: \\ $$$$\mathrm{2}\left(\mathrm{log}_{{y}} {x}\right)^{\mathrm{2}} −\mathrm{5log}_{{y}} {x}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{log}_{{y}} {x}=\frac{\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{16}}}{\mathrm{4}}=\mathrm{2},\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} ={x}\:\mid\:{y}^{\mathrm{1}/\mathrm{2}} ={x}\Rightarrow{y}={x}^{\mathrm{2}} ………\left({iii}\right) \\ $$$$\left({ii}\right)\Rightarrow\frac{\mathrm{1}}{\mathrm{log}_{{y}} \mathrm{4}\:}=\frac{\mathrm{1}}{\mathrm{log}_{{y}} \left({y}−\mathrm{3}{x}\right)\:} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow{y}−\mathrm{3}{x}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow{y}=\mathrm{3}{x}+\mathrm{4}……………….\left({iv}\right) \\ $$$$\: \\ $$$$\left({iii}\right)\:\&\:\left({iv}\right)\::\:\:\:\:\left(\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{2}} ={x}\:\:\mid\:\:\mathrm{3}{x}+\mathrm{4}={x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{No}\:{real}\:{roots}} {\mathrm{9}{x}^{\mathrm{2}} +\mathrm{23}{x}+\mathrm{16}=\mathrm{0}}\:\:\mid\:\:{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}=\mathrm{0} \\ $$$$\left({x}−\mathrm{4}\right)\left({x}+\mathrm{1}\right)=\mathrm{0}\Rightarrow{x}=\mathrm{4},−\mathrm{1} \\ $$$$\:\:\:\:\:\:\Rightarrow{y}=\mathrm{3}\left(\mathrm{4}\right)+\mathrm{4}\:,\:\mathrm{3}\left(−\mathrm{1}\right)+\mathrm{4}=\mathrm{16},\mathrm{1}\:{but}\:{y}\neq\mathrm{1} \\ $$$$\therefore\:\left({x},{y}\right)=\left(\mathrm{16},\mathrm{4}\right)\:\checkmark \\ $$

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