Can I find the value of $x & y$

Question

Find x,y from N such as $x^{(2y)}=1560-x^{y}$.Is it possible to find the value of x and y only from one equation. please help me.I approached in different ways.But all my attempt went in vain.

Answer 1

No, you cannot (without restrictions) solve a system of equations with only 1 equation. You can however, solve for $x^y$ (as Edward Jiang mentioned in the comments):

to solve for $x^y$

lets say $x^y$ is $u$

$u^2 = 1560 -u$

$u^2 + u - 1560=0$

look familiar?

(beware of one of your answers: $(x^y)^2$ must be positive)

Answer 2

Let $z=x^y$. The equation then becomes $z^2=1560-z$ or $z^2+z-1560=0$. This factors as $(z-39 )(z+ 40)=0$. Which gives $z=39$ and $z=-40$. Re-substitute the $x^y$. A logarithm will help here to get $y$ in terms of $x$.

Answer 3

Quadratic equation: \begin{align*} x^{2y}-1560+x^{y} & =0\\ \left(x^{y}\right)^{2}+x^{y}-1560 & =0 \end{align*} This is just a quadratic equation with solutions $$x^{y}=-40,39.$$ Since you cannot take the log of a negative number (in $\mathbb{R}$), the only real solution is $$y=\log_{x}39.$$