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Question Number 167539 by Tawa11 last updated on 19/Mar/22

Suppose  x and y are real, such that,   log_4 x   =   log_6 y   =   log_9 (x + y),  find   (y/x)

$$\mathrm{Suppose}\:\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{are}\:\mathrm{real},\:\mathrm{such}\:\mathrm{that},\:\:\:\mathrm{log}_{\mathrm{4}} \mathrm{x}\:\:\:=\:\:\:\mathrm{log}_{\mathrm{6}} \mathrm{y}\:\:\:=\:\:\:\mathrm{log}_{\mathrm{9}} \left(\mathrm{x}\:+\:\mathrm{y}\right), \\ $$$$\mathrm{find}\:\:\:\frac{\mathrm{y}}{\mathrm{x}} \\ $$

Answered by Rasheed.Sindhi last updated on 19/Mar/22

 log_4 x = log_6 y = log_9 (x + y)=k(say)  x=4^k  ,y=6^k , x+y=9^k   4^k +6^k =9^k   ((4/9))^k +((6/9))^k =1  ((2/3))^(2k) +((2/3))^k =1  ((2/3))^k =u  u^2 +u−1=0  u=((−1±(√5) )/2)  ((2/3))^k =((−1±(√5) )/2)  (y/x)=(6^k /4^k )=((3/2))^k =(((2/3))^k )^(−1) =(((−1±(√5) )/2))^(−1)   =(2/(−1±(√5) ))∙((−1∓(√5) )/(−1∓(√5) ))=((−2(1±(√5) ))/(1−5))  =((1±(√5))/2)

$$\:\mathrm{log}_{\mathrm{4}} \mathrm{x}\:=\:\mathrm{log}_{\mathrm{6}} \mathrm{y}\:=\:\mathrm{log}_{\mathrm{9}} \left(\mathrm{x}\:+\:\mathrm{y}\right)=\mathrm{k}\left({say}\right) \\ $$$$\mathrm{x}=\mathrm{4}^{\mathrm{k}} \:,\mathrm{y}=\mathrm{6}^{\mathrm{k}} ,\:\mathrm{x}+\mathrm{y}=\mathrm{9}^{\mathrm{k}} \\ $$$$\mathrm{4}^{\mathrm{k}} +\mathrm{6}^{\mathrm{k}} =\mathrm{9}^{\mathrm{k}} \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{\mathrm{k}} +\left(\frac{\mathrm{6}}{\mathrm{9}}\right)^{\mathrm{k}} =\mathrm{1} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2k}} +\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{k}} =\mathrm{1} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{k}} =\mathrm{u} \\ $$$$\mathrm{u}^{\mathrm{2}} +\mathrm{u}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{u}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{k}} =\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$$\frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{6}^{\mathrm{k}} }{\mathrm{4}^{\mathrm{k}} }=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{k}} =\left(\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{k}} \right)^{−\mathrm{1}} =\left(\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{2}}\right)^{−\mathrm{1}} \\ $$$$=\frac{\mathrm{2}}{−\mathrm{1}\pm\sqrt{\mathrm{5}}\:}\centerdot\frac{−\mathrm{1}\mp\sqrt{\mathrm{5}}\:}{−\mathrm{1}\mp\sqrt{\mathrm{5}}\:}=\frac{−\mathrm{2}\left(\mathrm{1}\pm\sqrt{\mathrm{5}}\:\right)}{\mathrm{1}−\mathrm{5}} \\ $$$$=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 19/Mar/22

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by Rasheed.Sindhi last updated on 19/Mar/22

   log_4 x = log_6 y = log_9 (x + y)=k (say)  x=4^k  ,y=6^k ,    x+y=9^k ⇒4^k +6^k =9^k   1+(6^k /4^k )=(9^k /4^k )  1+((3/2))^k =((3/2))^(2k)   ((3/2))^(2k) −((3/2))^k −1=0  ((3/2))^k =u  u^2 −u−1=0  u=((1±(√5) )/2)  ((3/2))^k =((1±(√5) )/2)  (y/x)=(6^k /4^k )=((3/2))^k =((1±(√5) )/2)

$$ \\ $$$$\:\mathrm{log}_{\mathrm{4}} \mathrm{x}\:=\:\mathrm{log}_{\mathrm{6}} \mathrm{y}\:=\:\mathrm{log}_{\mathrm{9}} \left(\mathrm{x}\:+\:\mathrm{y}\right)=\mathrm{k}\:\left({say}\right) \\ $$$$\mathrm{x}=\mathrm{4}^{\mathrm{k}} \:,\mathrm{y}=\mathrm{6}^{\mathrm{k}} ,\:\:\:\:\mathrm{x}+\mathrm{y}=\mathrm{9}^{\mathrm{k}} \Rightarrow\mathrm{4}^{\mathrm{k}} +\mathrm{6}^{\mathrm{k}} =\mathrm{9}^{\mathrm{k}} \\ $$$$\mathrm{1}+\frac{\mathrm{6}^{\mathrm{k}} }{\mathrm{4}^{\mathrm{k}} }=\frac{\mathrm{9}^{\mathrm{k}} }{\mathrm{4}^{\mathrm{k}} } \\ $$$$\mathrm{1}+\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{k}} =\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2k}} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2k}} −\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{k}} −\mathrm{1}=\mathrm{0} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{k}} =\mathrm{u} \\ $$$$\mathrm{u}^{\mathrm{2}} −\mathrm{u}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{u}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{k}} =\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$$\frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{6}^{\mathrm{k}} }{\mathrm{4}^{\mathrm{k}} }=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{k}} =\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$

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