 Solution

The Gelfond–Schneider theorem says that if $a\ne 0,1$ is algebraic and $b$ is algebraic irrational, every number $e^{b\lambda }$ with $e^{\lambda }=a$ is transcendental. Thus the statement includes every determination of the power, not merely a chosen principal branch.

Here is an auxiliary-function outline with the arithmetic and analytic steps specified. Assume $v=e^{b\lambda }$ were algebraic, and put $K=\mathbb{Q}(a,b,v)$, $d=[K:\mathbb{Q}]$. For a large integer $T$, choose degrees $L_1\asymp T\log T$ and $L_2\asymp T/\log T$, with constants large enough that $(L_1+1)(L_2+1)>2d(T+1{)}^2$. Apply Siegel lemma to imposeon a nonzero polynomial $P(X,Y)\in \mathbb{Z}[X,Y]$ of these two degrees. Each field equation gives at most $d$ rational equations. Clearing denominators and bounding conjugates gives equation-coefficient logarithmic size $O(L_1\log T+L_{2}T)$; the surplus of variables therefore permits coefficient height $H$ with

The entire function $F(z)=P(z,e^{\lambda z})$ is nonzero: its polynomial-exponential terms have distinct exponents $j\lambda$, and their independence follows by the differential-operator argument in question 2. Here $\lambda \ne 0$. Its maximum modulus satisfies $\log M_F(R)=O(\log H+L_1\log R+L_{2}R)$. The grid points $s+tb$ are distinct because $b$ is irrational, and lie in a disk of radius $(1+|b|)T$.

Suppose the vanishing grid has been extended to $0\le s,t\le B$, for some $B\ge T$. Divide $F$ by the product of its $(B+1{)}^2$ known linear zero factors and use the maximum principle on a disk of radius $CB$, with $C$ large and fixed. At a point of the doubled grid, the quotient estimate givesif the value is nonzero. But this value equals the algebraic number $P(s+tb,a^s{v}^t)$. Clearing its denominators and taking its norm gives the opposite boundFor sufficiently large initial $T$, the terms on the right are $o(B^2)$ uniformly for every $B\ge T$: $L_2/B=O(1/\log T)$ and $L_1\log B/B^2=O((\log T{)}^2/T)$. Thus every doubled-grid value must be zero. Iterating doubles the grid indefinitely. This supplies quadratically many distinct zeros in disks of linearly growing radius, whereas Jensen's formula and the exponential-polynomial growth bound allow only $O(R)$ zeros for a nonzero $F$. The contradiction proves the theorem. This is the Schneider lattice proof of the Gelfond–Schneider theorem.

Now fix any logarithms of algebraic $\alpha ,\beta \ne 0,1$. Their values are nonzero. If $q=\log \alpha /\log \beta$ were algebraic irrational, then $e^{q\log \beta }=\alpha$ would be algebraic, contradicting the theorem with base $\beta$. HenceFor arbitrarily many logarithms the qualitative Baker theorem on algebraic linear combinations of logarithms states that logarithms of algebraic numbers which are linearly independent over $\mathbb{Q}$ remain independent over $\overline{\mathbb{Q}}$, even after adjoining one to the list. Equivalently a nonzero homogeneous algebraic-coefficient linear combination of such logarithms is transcendental. Rational relations are therefore the only source of homogeneous algebraic relations among the chosen logarithms.