Answer: (a) \(p=\frac{125}{8}\) or \(p=-\frac1{27}\); (b) \(x=\frac{-5+\sqrt5}{4}\) or \(x=\frac{-5-\sqrt5}{4}\).

Use the laws of logarithms and exponentials first, then check that any logarithm arguments are positive where required.

(a) Let

\(u=p^{\frac13}.\)

Then

\(p^{-\frac13}=\frac1u.\)

The equation becomes

\(6u-\frac5u-13=0.\)

Multiply by \(u\):

\(6u^2-13u-5=0.\)

Factorise:

\((2u-5)(3u+1)=0.\)

So

\(u=\frac52 \quad\text{or}\quad u=-\frac13.\)

Since \(u=p^{1/3}\),

\(p=\left(\frac52\right)^3=\frac{125}{8}\)

or

\(p=\left(-\frac13\right)^3=-\frac1{27}.\)

(b) Use the laws of logarithms:

\(2\lg(2x+5)-\lg(x+2)=1\)

gives

\(\lg\left(\frac{(2x+5)^2}{x+2}\right)=1.\)

Therefore

\(\frac{(2x+5)^2}{x+2}=10.\)

So

\((2x+5)^2=10(x+2).\)

Expand and simplify:

\(4x^2+20x+25=10x+20,\)

so

\(4x^2+10x+5=0.\)

Using the quadratic formula,

\(x=\frac{-10\pm\sqrt{100-80}}{8} =\frac{-10\pm2\sqrt5}{8}.\)

Thus

Both values satisfy the logarithm domain \(x\gt -2\).