Prove that this number is irrational it's involve the exponential

Question

Prove that $2^{\sqrt{3}}$ is irrational or even it is transcendental? I am wondering if there exists the method that describes its property and its irrationality of these kinds (or even with some numbers of the form like $5^{\sqrt{2}+\dfrac{1}{1+\sqrt[4]{2}}}$), which involves only with the algebraic number

Answer 1

Gelfond-Schneider theorem: if $a,b$ are algebraic, $a \ne 0,1$, $b$ irrational, then all values of $a^b$ are transcendental.

So: $2^{\sqrt{3}}$ is transcendental.

To show $$ 5^{\displaystyle \left(\sqrt{2}+\frac{1}{1+\sqrt[4]{2}}\right)} \tag1$$ is transcendental, it suffices to show $$ \sqrt{2}+\frac{1}{1+\sqrt[4]{2}} \tag2$$ is irrational. I leave that to you, using the rational root theorem.

Note: Numerically evaluating $(2)$ we get $1.8709999$ so we may mistakenly think it is rational. In fact with more digits it is $1.8709999455101501528$.