Given the following equation:$$2^{x^{\frac{1}{12}}}+2^{x^{\frac{1}{4}}}=2\cdot2^{x^{\frac{1}{6}}}$$
I have to solve for $x$, I tried working a bit with logarithms but the plus sign is quite annoying so I didn't think that was the correct way, I saw the '$2\cdot$' on the RHS and that suggested me that maybe I should bring it to the other side and the equation will turn out to be a perfect square, but that wasn't the case... I didn't get much further (the problem doesn't specify so I assumed that $x$ is a real number)
By Am Gm inequality (we use it twice) we have $$2\cdot2^{x^{\frac{1}{6}}} =2^{x^{\frac{1}{12}}}+2^{x^{\frac{1}{4}}}\geq 2\sqrt{\cdot2^{x^{\frac{1}{12}}+x^{\frac{1}{4}}}}\geq 2\cdot \sqrt{2^{2{\sqrt{x^{\frac{1}{12}+\frac{1}{4}}}}}}=2\cdot2^{{x^{\frac{1}{6}}}}$$
So we have equality case which is achieved iff $$ x^{\frac{1}{12}}=x^{\frac{1}{4}}\implies x^3=x\implies x\in\{0,1,-1\}$$
Since radicand $x$ must be nonegative we have $x=0$ or $x=1$