= Solution
Using base-two logarithms, define the <quantum conditional entropy> by
$$
\boxed{S(A\mid B)=S(\rho_{AB})-S(\rho_B).}
$$
The <Subadditivity of Von Neumann entropy>, $S(AB)\leq S(A)+S(B)$, immediately gives $S(A\mid B)\leq S(A)$.
For the lower bound, purify $\rho_{AB}$ by an auxiliary system $C$. Complementary subsystems of a <pure state> have the same nonzero reduced-state <eigenvalues>, by the <Schmidt decomposition>. Hence $S(BC)=S(A)$ and $S(C)=S(AB)$. Applying <Subadditivity of Von Neumann entropy> to $BC$ gives
$$
S(A)=S(BC)\leq S(B)+S(C)=S(B)+S(AB).
$$
Rearranging yields $S(A\mid B)\geq-S(A)$. This is the relevant side of the <Araki–Lieb inequality>, here derived using purification.
Finally the <maximum entropy of a quantum state> gives $S(A)\leq\log_2\dim\mathcal H_A$. For example, this follows directly from
$$
D\left(\rho_A\middle\|\frac{I_A}{\dim\mathcal H_A}\right)
=\log_2\dim\mathcal H_A-S(A)\geq0
$$
by <nonnegativity of quantum relative entropy>. Combining the two bounds proves
$$
\boxed{|S(A\mid B)|\leq\log_2\dim\mathcal H_A.}
$$
Both signs can be attained: a maximally mixed $A$ independent of $B$ has the positive extreme, while a <maximally entangled state> with a sufficiently large $B$ has the negative extreme.
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