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Given-2log-5-x-2-log-5-y-1-2-0-log-5-x-2-2x-2-0-find-the-value-of-y-




Question Number 105994 by bemath last updated on 02/Aug/20
Given  { ((2log _5 (x+2)−log _5 (y)+(1/2)=0)),((log _5 (x^2 +2x−2)=0 )) :}  find the value of y
$$\mathbb{G}{iven}\:\begin{cases}{\mathrm{2log}\:_{\mathrm{5}} \left({x}+\mathrm{2}\right)−\mathrm{log}\:_{\mathrm{5}} \left({y}\right)+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}}\\{\mathrm{log}\:_{\mathrm{5}} \left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{2}\right)=\mathrm{0}\:}\end{cases} \\ $$$${find}\:{the}\:{value}\:{of}\:{y}\: \\ $$
Commented by PRITHWISH SEN 2 last updated on 02/Aug/20
x^2 +2x−2=1⇒x=1,−3  (((x+2)^2 )/y) =(1/( (√5))) ⇒x=1,y=9(√5),  x=−3, y= (√5)  (x,y) = (1,9(√5)),(−3,(√5))
$$\mathrm{x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{2}=\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{1},−\mathrm{3} \\ $$$$\frac{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{y}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\mathrm{x}=\mathrm{1},\mathrm{y}=\mathrm{9}\sqrt{\mathrm{5}},\:\:\mathrm{x}=−\mathrm{3},\:\mathrm{y}=\:\sqrt{\mathrm{5}} \\ $$$$\left(\mathrm{x},\mathrm{y}\right)\:=\:\left(\mathrm{1},\mathrm{9}\sqrt{\mathrm{5}}\right),\left(−\mathrm{3},\sqrt{\mathrm{5}}\right) \\ $$
Answered by bobhans last updated on 02/Aug/20
(1) x^2 +2x−2=1 ⇒ x^2 +2x−3=0 ∧x>−2  ⇒(x+3)(x−1)=0⇒x=1, x=−3(rejected)  (2) 2log _5 (x+2)−log _5 (y)=−(1/2)  log _5 ((((x+2)^2 )/y))=log _5 ((1/( (√5))))  → y = (√5) (x+2)^2  = 9(√5) = ■
$$\left(\mathrm{1}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{2}=\mathrm{1}\:\Rightarrow\:\mathrm{x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{3}=\mathrm{0}\:\wedge\mathrm{x}>−\mathrm{2} \\ $$$$\Rightarrow\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}−\mathrm{1}\right)=\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{1},\:\mathrm{x}=−\mathrm{3}\left(\mathrm{rejected}\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2log}\:_{\mathrm{5}} \left(\mathrm{x}+\mathrm{2}\right)−\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{y}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{log}\:_{\mathrm{5}} \left(\frac{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{y}}\right)=\mathrm{log}\:_{\mathrm{5}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right) \\ $$$$\rightarrow\:\mathrm{y}\:=\:\sqrt{\mathrm{5}}\:\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} \:=\:\mathrm{9}\sqrt{\mathrm{5}}\:=\:\blacksquare \\ $$$$ \\ $$

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