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Question-74589




Question Number 74589 by Aditya789 last updated on 27/Nov/19
Answered by MJS last updated on 27/Nov/19
a(b−c)x^2 +b(c−a)xy+c(a−b)y^2 =0  x^2 +((b(c−a)y)/(a(b−c)))x+((c(a−b)y^2 )/(a(b−c)))=0  x=t−((b(c−a)y)/(2a(b−c)))  t^2 −((((ab−2ac+bc)y)/(2a(b−c))))^2 =0  to get a perfect square t=0  ⇒  y=0∨b=((2ac)/(a+c))  b=((2ac)/(a+c)) ⇒ ((log (a+c) +log (a−2b+c))/(log (a−c)))=  =((log (a+c) +log (((a−c)^2 )/(a+c)))/(log (a−c)))=  =((log (a+c) +2log (a−c) −log (a+c))/(log (a−c)))=2
$${a}\left({b}−{c}\right){x}^{\mathrm{2}} +{b}\left({c}−{a}\right){xy}+{c}\left({a}−{b}\right){y}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\frac{{b}\left({c}−{a}\right){y}}{{a}\left({b}−{c}\right)}{x}+\frac{{c}\left({a}−{b}\right){y}^{\mathrm{2}} }{{a}\left({b}−{c}\right)}=\mathrm{0} \\ $$$${x}={t}−\frac{{b}\left({c}−{a}\right){y}}{\mathrm{2}{a}\left({b}−{c}\right)} \\ $$$${t}^{\mathrm{2}} −\left(\frac{\left({ab}−\mathrm{2}{ac}+{bc}\right){y}}{\mathrm{2}{a}\left({b}−{c}\right)}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}\:{t}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${y}=\mathrm{0}\vee{b}=\frac{\mathrm{2}{ac}}{{a}+{c}} \\ $$$${b}=\frac{\mathrm{2}{ac}}{{a}+{c}}\:\Rightarrow\:\frac{\mathrm{log}\:\left({a}+{c}\right)\:+\mathrm{log}\:\left({a}−\mathrm{2}{b}+{c}\right)}{\mathrm{log}\:\left({a}−{c}\right)}= \\ $$$$=\frac{\mathrm{log}\:\left({a}+{c}\right)\:+\mathrm{log}\:\frac{\left({a}−{c}\right)^{\mathrm{2}} }{{a}+{c}}}{\mathrm{log}\:\left({a}−{c}\right)}= \\ $$$$=\frac{\mathrm{log}\:\left({a}+{c}\right)\:+\mathrm{2log}\:\left({a}−{c}\right)\:−\mathrm{log}\:\left({a}+{c}\right)}{\mathrm{log}\:\left({a}−{c}\right)}=\mathrm{2} \\ $$

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