Simple Question:
Is the proof given below correct?
Let $t$ satisfy $2^t+3^t=1$. We have $t \approx -0.787884911025869783628555917298434738269083137354182194199 \dots$ according to wolfram alpha.
Then $t$ is demonstrably transcendental assuming Schanuel's Conjecture. Which says that for linearly independent numbers $\alpha_1, \dots ,\alpha_n$ over the rationals there exists a subset of $\{\alpha_1, \dots \alpha_n, e^\alpha_1, \dots, e^\alpha_n\}$ with cardinality $n$ which is algebraically independent over the rational numbers.
Proof (Using Schanuel's Conjecture)
The $t$ that satisfies $2^t+3^t=1$ must be an irrational value. This is shown here.
Assuming by way of contradiction $t$ an irrational algebraic. Then we let $\alpha_1=\ln(2), \alpha_2=\ln(3)$
$$\{\alpha_1,\alpha_2,\alpha_1 t, \alpha_2t \}=\{\ln(2),\ln(3),\ln(2)t,\ln(3)t \}$$
Otherwise, $\ln(2)/\ln(3)$ proves to be algebraic. $$\sum_{n=1}^4 a_n \alpha_n=0 \implies (a_1+a_3t)\ln(2)+(a_2+a_4t)\ln(3)=0\implies \frac{\ln(2)}{\ln(3)}=-\frac{a_2+a_4t}{a_1+a_3t}$$ But then the righthand side is algebraic but $\ln(2)/\ln(3)$ is transcendental by Gelfond–Schneider theorem. We conclude that $\{\alpha_1,\alpha_2,\alpha_1 t, \alpha_2t\}$ meets the hypothesis of the conjecture and arrive at the conclusion that $\left\{\ln2,\ln3,\ln2^{t},\ln3^{t},2,3,2^{t},3^{t}\right\}$ has transcendence degree greater than $4$. This means we can select an algebraically independent subset of cardinality $4$. But then we can remove the algebraic numbers like $2$ and $3$. And we can pick only one from the pair $\{\ln(2), t\ln(2)\}$ and likewise $\{\ln(3), t\ln(3) \}$. I am justifying this argument using this post: $S$ is algebraically independent over $\mathbb{Q}$ implies $S$ algebraically independent over $\overline{\mathbb{Q}}$. And we conclude that
$\left\{\ln2,\ln3,2^{t},3^{t}\right\}$ has transcendence degree greater than 4.
But this violates our construction of $t$. Namely, $2^t+3^t=1$ by assumption and therefore $2^t$ and $3^t$ aren't algebraically independent and we have landed at our contradiction. This demonstrates the transcendality of $t$.
Motivations I am hoping to work out an example of an application of Schanuel's Conjecture.
In the link above appears the quote.
"Macintyre (1991) proved that the truth of Schanuel's conjecture also guarantees that there are no unexpected exponential-algebraic relations on the integers Z (Marker 1996)." I'm hoping this gives a small example of what is meant by this quote.
Harder Question It would be interesting to know if $t$ can proven transcendental using weaker tools than Schanuel's conjecture. I suspect that this is open but maybe I am missing a method.
