<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] I will assume you want a real solution. $t^2+t^3=4$ can be rewritten as $t^3+t^2-4=0$. Using the cubic formula, the real solution for $t$ is $$t=\frac{1}{3} \left(-1+\sqrt[3]{53-6\sqrt{78}}+\sqrt[3]{53+6 \sqrt{78}}\right)$$
We also know that $y=log_2(t)$ and that $x=2y$. So, using that, 
$$x=2\cdot log_2\left(\frac{1}{3} \left(-1+\sqrt[3]{53-6\sqrt{78}}+\sqrt[3]{53+6 \sqrt{78}}\right)\right)$$
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