Solution. (a). Bring the coefficient inside as a power, then combine: \[\log _a 4+\log _a 3^{2}-\log _a 6=\log _a\frac {4\times 9}{6}=\log _a 6 .\]

(b). A quotient becomes a difference and a product becomes a sum, and every exponent comes down in front: \[\log _a\frac {x^{3}}{y^{2}z}=\log _a x^{3}-\log _a\left (y^{2}z\right ) =3\log _a x-2\log _a y-\log _a z .\] Note the minus applying to both terms of the denominator — a bracket worth writing down, since dropping it is the usual error here.

Solution. (a). Write everything in terms of \(u=2^{x}\). Since \(2^{2x+1}=2\cdot 2^{2x}=2u^{2}\), \[2u^{2}=3u-1 \implies 2u^{2}-3u+1=0 \implies (2u-1)(u-1)=0 ,\] so \(u=\frac {1}{2}\) or \(u=1\). Then \(2^{x}=\frac {1}{2}\) gives \(x=-1\), and \(2^{x}=1\) gives \(x=0\).

(b). By the change of base result, \(\log _z 3=\frac {1}{\log _3 z}\). Putting \(w=\log _3 z\), \[w=\frac {4}{w} \implies w^{2}=4 \implies w=\pm 2 .\] So \(\log _3 z=2\), giving \(z=9\), or \(\log _3 z=-2\), giving \(z=\frac {1}{9}\). Both are positive and neither is \(1\), so both are admissible bases and both are solutions.