Solving a system of equations with logarithms in exponents I am interested in what approach could be applied to solve such a system:
$$ 2^{x-\log_2{y}} - 8^{x+\log_2{y}} = 5y - 2^{x-\log_2{y}} $$
$$ \frac{2^{x-\log_2{y}}}{8^{x+\log_2{y}}} = \frac{5y}{2^{x-\log_2{y}}} $$

Notice that the first equation is
$$\frac{2^x}y-8^x\cdot y^3=5\cdot y-\frac{2^x}y$$
Multiply this equation for $y$ and get
$$8^x\cdot y^4+5\cdot y^2-2^{x+1}=0$$
so you have a way to express $y$ purely in function of $x$. Now, the second equation is
$$\frac{2^x}{8^x\cdot y^4}=\frac{5\cdot y^2}{2^x}$$
so you can rewrite it as $1=5\cdot 2^x\cdot y^6$. Now you know that $y^6=5^{-1}\cdot 2^{-x}$ so you have the system
$$\left\{\begin{array}{l} (2^x)^3\cdot y^4+5\cdot y^2-2^{x+1}=0;\\
y^6=5^{-1}\cdot 2^{-x}
\end{array}\right.$$
that is easier than the original one. For example, you could call $z:=2^x$ and get $z=1/(5y^6)$ from the second equation, and plug it into the first to get
$$\frac{y^4}{5^3y^{18}}+5y^2-\frac2{5y^6}=0$$
from which you get $y^4+5^4y^{20}-50y^{12}=0$ so $5^4y^{16}-50y^8+1=0$ that is $(25y^8-1)^2=0$ from which you get $y=\pm1/\sqrt[4]5$. Now, $z=\sqrt5$ and so $x=\log_2(5)/2$.
