Solving an equation with radicals in the exponent

Question

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)

@Ertxiem Yes, but I'm not sure if I'm missing out on non-trivial solutions — Spasoje Durovic, Mar 16 '19 at 19:26

nonuser · Accepted Answer · 2019-03-16T19:29:45.360

3

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 $x=1$

edited Mar 16 '19 at 19:29

answered Mar 16 '19 at 19:16

nonuser

90,026

1

It always surprises me how clever manipulations of the simple Am-Gm inequality can lead to these results... How did you come to this idea? – Dr. Mathva Mar 16 '19 at 19:47
I solved some similar problems before. – nonuser Mar 16 '19 at 19:48

Solving an equation with radicals in the exponent

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