Question

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​

Answer 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.)

Answer:

Given that,

$x^2+y^2=13xy$

Adding $2xy$ to both sides,

$\longrightarrow x^2+y^2+2xy=13xy+2xy$

$\longrightarrow (x+y)^2=15xy\quad\quad\dots(1)$

And it is also given that,

$2\log(x+y)=\log(k)+\log(l)+\log(x)+\log(y)$

$\longrightarrow \log\{(x+y)^2\}=\log(kl)+\log(xy)$

From (1),

$\longrightarrow \log(15xy)=\log(kl)+\log(xy)$

$\longrightarrow \log(15xy)-\log(xy)=\log(kl)$

$\longrightarrow \log\bigg(\dfrac{15xy}{xy}\bigg)=\log(kl)$

$\longrightarrow \log(15)=\log(kl)$

Taking antilog,

$\longrightarrow kl=15$

Now we have a Diophantine Equation:

$kl=15\:\:\forall\:\:k,l\in\mathbb{Z}$

So, possible pairs include,

• $k=5, l=3$
• $k=-5, l=-3$

Hence, the value of $(k+l)$ can be,

• $k+l=5+3=8$
• $k+l=(-5)+(-3)=-8$

So, the answer is,

$\longrightarrow \underline{\underline{k+l=8\:or\:-8}}$