Answer: (a) \(a=32\); (b) \(x=3\sqrt3\) or \(x=\frac{\sqrt3}{9}\).

We start with the main method. Use logarithm laws and index laws first so that the equation is reduced to an algebraic equation.

(a) Use the logarithm laws:

\(\log_p a+\log_p12-\log_p6=\log_p\left(\frac{12a}{6}\right)=\log_p(2a).\)

Also

\(3\log_p4=\log_p4^3=\log_p64.\)

Therefore

\(\log_p(2a)=\log_p64,\)

so

\(2a=64.\)

Hence

\(a=32.\)

(b) Let

\(u=\log_3x.\)

By change of base,

\(\log_x3=\frac1{\log_3x}=\frac1u.\)

The equation becomes

\(4u=\frac9u.\)

Thus

\(4u^2=9,\)

so

\(u=\pm\frac32.\)

Therefore

\(x=3^{3/2}\quad\text{or}\quad x=3^{-3/2}.\)

So

\(x=3\sqrt3\quad\text{or}\quad x=\frac{\sqrt3}{9}.\)

This completes the solution and gives the required result.