Answer: (a) \(p=16\); (b) \(x=-1\); (c) \(y=4\).

(a) Use the laws of logarithms:

Hence

\(\log_a\left(\frac{5p}{4}\right)=\log_a20.\)

So

\(\frac{5p}{4}=20,\)

and therefore

\(p=16.\)

(b) Let \(u=3^x\). Then \(3^{2x+1}=3(3^x)^2=3u^2\), so the equation becomes

\(3u^2+8u-3=0.\)

Factorising,

\((3u-1)(u+3)=0.\)

Since \(u=3^x\) is positive, \(u=-3\) is rejected. Thus

\(3^x=\frac13=3^{-1},\)

so

\(x=-1.\)

(c) Let

\(t=\log_y2.\)

Then

\(\log_2y=\frac1t.\)

The equation becomes

\(4t+\frac1t=4.\)

Multiplying by \(t\),

\(4t^2+1=4t.\)

So

\(4t^2-4t+1=0,\)

which gives

\((2t-1)^2=0.\)

Hence \(t=\frac12\). Therefore

\(\log_y2=\frac12,\)

so

\(y^{\frac12}=2.\)

Therefore

\(y=4.\)