Schanuel's conjecture implies that if $x$ is an algebraic irrational, $n^x$ for natural numbers $n$ are linearly independent over the rationals: see Will Sawin's answer to this recent question for a proof. It's easy to extend this to positive rationals. Thus Schanuel implies that if the $a_i$ and $b_i$ are rational, $x$ will either be rational or transcendental. So (if you're willing to accept Schanuel) you just have to check for rational solutions, which for something like $(1/3)^x + (3/4)^x = 1$ is not hard.

Schanuel's conjecture implies that if $x$ is an algebraic irrational, $n^x$ for natural numbers $n$ are linearly independent over the rationals: see Will Sawin's answer to this recent question for a proof. It's easy to extend this to positive rationals. Thus Schanuel implies that if the $a_i$ and $b_i$ are rational, $x$ will either be rational or transcendental. So (if you're willing to accept Schanuel) you just have to check for rational solutions, which for something like $(1/3)^x + (3/4)^x = 1$ is not hard.

Schanuel's conjecture implies that if $x$ is an algebraic irrational, $n^x$ for natural numbers $n$ are linearly independent over the rationals: see Will Sawin's answer to this recent question for a proof. It's easy to extend this to positive rationals. Thus Schanuel implies that if the $a_i$ and $b_i$ are rational, $x$ will either be rational or transcendental.

Schanuel's conjecture implies that if $x$ is an algebraic irrational, $n^x$ for natural numbers $n$ are linearly independent over the rationals: see Will Sawin's answer to this recent question for a proof. It's easy to extend this to positive rationals. Thus Schanuel implies that if the $a_i$ and $b_i$ are rational, $x$ will either be rational or transcendental. So (if you're willing to accept Schanuel) you just have to check for rational solutions, which for something like $(1/3)^x + (3/4)^x = 1$ is not hard.