If x² +y² =13xy and 2 log ( x + y) = log k + log l+ log x + log y where k and l are real, then the value of (k+I) is​

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1. Correct question:

   If x² + y² = 13xy and 2log(x + y) = log(k) + log(l)+ log(x) + log(y) where k and l are integers, then the value of (k+I) is:

   (Real numbers will have infinite solutions as that would include non integers, as well as integers.)

   [tex]\\[/tex]

   Answer:

   Given that,

   [tex]x^2+y^2=13xy[/tex]

   Adding [tex]2xy[/tex] to both sides,

   [tex]\longrightarrow x^2+y^2+2xy=13xy+2xy[/tex]

   [tex]\longrightarrow (x+y)^2=15xy\quad\quad\dots(1)[/tex]

   And it is also given that,

   [tex]2\log(x+y)=\log(k)+\log(l)+\log(x)+\log(y)[/tex]

   [tex]\longrightarrow \log\{(x+y)^2\}=\log(kl)+\log(xy)[/tex]

   From (1),

   [tex]\longrightarrow \log(15xy)=\log(kl)+\log(xy)[/tex]

   [tex]\longrightarrow \log(15xy)-\log(xy)=\log(kl)[/tex]

   [tex]\longrightarrow \log\bigg(\dfrac{15xy}{xy}\bigg)=\log(kl)[/tex]

   [tex]\longrightarrow \log(15)=\log(kl)[/tex]

   Taking antilog,

   [tex]\longrightarrow kl=15[/tex]

   Now we have a Diophantine Equation:

   [tex]kl=15\:\:\forall\:\:k,l\in\mathbb{Z}[/tex]

   So, possible pairs include,

   • [tex]k=5, l=3[/tex]
   • [tex]k=-5, l=-3[/tex]

   Hence, the value of [tex](k+l)[/tex] can be,

   • [tex]k+l=5+3=8[/tex]
   • [tex]k+l=(-5)+(-3)=-8[/tex]

   So, the answer is,

   [tex]\longrightarrow \underline{\underline{k+l=8\:or\:-8}}[/tex]

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