Solving a system of equations with logarithms in exponents

Question

I am interested in what approach could be applied to solve such a system:

$$ 2^{x-\log_2{y}} - 8^{x+\log_2{y}} = 5y - 2^{x-\log_2{y}} $$

$$ \frac{2^{x-\log_2{y}}}{8^{x+\log_2{y}}} = \frac{5y}{2^{x-\log_2{y}}} $$

Answer 1

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 $x=\log_2(5)/2$.