Question

Find the positive solution of the system of equations $x^{x+y}=y^n$ and $y^{x+y}=x^{2n}.y^n$, where n > 0

• A. $x=1$
• B. $y=1$
• C. $y=\frac{-1+\sqrt{(1+8n)}}{2}$
• D. $x={\left(\frac{-1+\sqrt{(1+8n)}}{2}\right)}^2$

Answer 1

Answer: (C), (D)

The correct options are
C $y=\frac{-1+\sqrt{(1+8n)}}{2}$
D $x={\left(\frac{-1+\sqrt{(1+8n)}}{2}\right)}^2$

Given equations
$x^{x+y}=y^n$
Taking log on both sides, we get
$(x+y)\log x=n\log y$ .....(1)
Other eqn is $y^{x+y}=x^{2n}.y^n$
Taking log on both sides, we get
$(x+y)\log y=2n\log x+n\log y$
$\Rightarrow (x+y-n)\log y=2n\log x$ .....(2)
Solving these eqns , we get
$\left[\right(x+y{)}^2-n(x+y)-2n^2]\log x=0$
either $\left[\right(x+y{)}^2-n(x+y)-2n^2]=0$ or $\log x=0$
$\log x=0$ when $x=1$
Using this in (1), we get
$n\log y=0$
Since, $n>0$ so $\log y=0$
which is possible when $y=1$
Now, we will solve the "either part " equation
$\left[\right(x+y{)}^2-n(x+y)-2n^2]=0$
$\Rightarrow x+y=\frac{n\pm 3n}{2}$
$\Rightarrow x+y=2n$ or $x+y=-n$
Option C and D satisfies the equation $x+2y=n$
Hence, option C and D are also correct