 Solution

$\boxed{BS(3,4)\text{ is not residually finite}.}$ We exhibit a nonidentity element killed by every finite quotient.

In any finite image, let $r$ be the order of the image of $a$. The relation conjugating $a^3$ to $a^4$ givesThus $\gcd (r,3)=\gcd (r,4)=1$. Since $3$ is invertible modulo $r$, the relation implies that the image of $ba{b}^{-1}$ is a power of the image of $a$. Therefore every finite image kills the group commutatorusing $[s,t]=s^{-1}{t}^{-1}st$.

View $BS(3,4)$ as an HNN extension of $⟨a⟩\cong \mathbb{Z}$, with associated subgroups $⟨a^3⟩$ and $⟨a^4⟩$. A pinch in an HNN extension would be $b{a}^{3j}{b}^{-1}$ or $b^{-1}{a}^{4j}b$. The word $w$ has none: its intervening exponents are $-1,1,1$, incompatible with the required divisibilities $3,4,3$. By Britton's lemma, $w\ne 1$. Hence finite quotients fail to separate this nonidentity element, proving the conclusion.