Are there rational $a$ and $b$ with $$a+be = \frac{1}{\ln 2}\ ?$$

The absence of rational solutions follows from Schanuel's conjecture, as shown in this answer where I used it repeatedly. Is there a proof without that conjecture?

Are there rational $a$ and $b$ with $$a+be = \frac{1}{\ln 2}\ ?$$

I used the absence of rational solutions repeatedly in this answer.

Here is a proof using Schanuel's conjecture: $e^q$ is transcendental for any algebraic $q$ by the Lindemann-Weierstrass theorem. In particular $e^q\neq 2$ and $\ln 2$ is irrational. So $1$ and $\ln 2$ are linearly independent over the rationals. Then by Schanuel's conjecture, the set $\{1, \ln 2, e, 2\}$ must have transcendence degree at least $2$ over the rationals, while the above equation would imply that the transcendence degree is only $1$. So there are no rational solutions to the above equation.

Is there a proof without that conjecture?

Are there rational $a$ and $b$ with $$a+be = \frac{1}{\ln 2}\ ?$$

This is a special case of Schanuel's conjecture, which I used repeatedly in this answer -- it seems simple enough that it might have an independent proof.

Are there rational $a$ and $b$ with $$a+be = \frac{1}{\ln 2}\ ?$$

The absence of rational solutions follows from Schanuel's conjecture, as shown in this answer where I used it repeatedly. Is there a proof without that conjecture?