Solving the system with logarithms

Question

I tried solving the system $ \begin{cases} (4x)^{\log_2 (2y)} = 64 \\ (8y)^{\log_2 (2y)} = 256 \end{cases} $ several times but still keep getting wrong solutions.

Answer 1

$$\log_2 4x \log_2 2y= \log_2 64=6$$ $$\log_2 2y \log_2 8y= \log_2 256=8$$

Some simplifications leads to:

$$ (\log_2 4 + \log_2 x)(\log_2 2 + \log_2 y)=6$$ $$ (\log_2 2 + \log_2 y)(\log_2 8 + \log_2 y)=8$$

Call $u=\log_2 x$, $v=\log_2 y$:

$$(2+u)(1+v)=6$$ $$(1+v)(3+v)=8$$

The solutions are: $$\left\{u=1,v=1 \right\} \lor \left\{u=-\frac{7}{2} v=-5\right\}$$

Then:

$$\left\{x=2,y=2 \right\} \lor \left\{x=2^{-\frac{7}{2}}, y=2^{-5}\right\}$$