Exponential systems of equations

Question

I'm preparing for uni entrance exam. I've been struggling with this problem for about 90 minutes, tried everything I could think of. Can anybody explain how to solve this step by step?

$$x^{x+y}\:=\:y^{12},\:y^{x+y}=x^3$$

Answer 1

First taking the log of both sides of both equations and isolating $x+y$ $$x^{x+y}\:=\:y^{12},\:y^{x+y}=x^3 \\ \implies x+y = 12 \frac{log(y) }{log(x)}=3 \frac{log(x) }{log(y)}$$

Now let $z \equiv \frac{log(x) }{log(y)}$ so the last equation becomes $$ \frac{12}z=3z \implies z=\pm2$$ so $$ x = y^{\pm 2}$$ plugging $ x = y^{ 2}$into the original equations gives

$$ y^{2(x+y)} = y^{12} \text{ and } y^{x+y}=y^6 \implies x+y=6$$ SInce $ x = y^{ 2}$ we have ... $$y^2+y=6 \implies (y+3)(y-2)=0$$

Which gives us two solutions for $(x,y)$, namely $(9,-3)$ and $(4,2)$

If we consider $x=y^{-2}$ we get $y^2+y=-6$ which has no real solutions.

So it looks like the only two solutions are $(9,-3)$ and $(4,2)$

Answer 2

Assuming that $x,y >0$, $(x+y)\log\, x =12 \log\, y$ and $(x+y) \log\, y=3\log\, x$. Dividing one by the other we get $\frac {log\, x} {log\, y}=4 \frac {log\, y} {log\, x}$ which gives $\frac {log\, x} {log\, y}=\pm 2$. Now use the first equation to get $x+y=\pm 6$. Since $x,y>0$ we must have $x+y=6$ and $\frac {log\, x} {log\, y}= 2$. This gives $x+y=6$ and $x=y^{2}$. I leave the rest to you.

Answer 3

Hint

$$x^3=y^{x+y}$$

$$(x^3)^{12}=(x^{x+y})^{x+y}$$

Either $x=1$

Or $x=-1$ and $36-(x+y)^2$ is even

Or $36-(x+y)^2=0$

Or $x=0$ and $(x+y)^2\ge0$ which holds true for all real $y$

Plug the values of $x$ in $$y^{12}=x^{x+y}$$

See also : Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$