<s>[INST] Let $\log x=\log 2+\log y$ and $2^x+8^y=4$ then find the $x$ 
Let $\log x=\log 2+\log y$ and $2^x+8^y=4$ then find the $x$

My Try :
$$\log x=\log 2+\log y \\ \log x=\log 2y \\x=2y $$
So we have :
$$2^x+8^y=4\\2^{2y}+2^{3y}=4 \\t=2^y \\t^2+t^3=4$$
now what ?
 [/INST] from the first equation we have $$\log(x)=\log(2y)$$ and we get $$x=2x$$ plugging this in your second equation we have
$$2^{2y}+8^y=4$$
this can be written as $$(2^y)^2+(2^y)^3=4$$ with $t=2^y$ we have to solve
$$t^3+t^2-4=0$$
one solution is given by $$t=\frac{1}{3} \left(-1+\sqrt[3]{53-6
   \sqrt{78}}+\sqrt[3]{53+6 \sqrt{78}}\right)$$ after the cubic formulas. This is the only real solution
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