Answer: \(\displaystyle y=p^{5/2}\) or \(y=p^2\).

Answer: \(\displaystyle y=p^{5/2}\) or \(y=p^2\).

Let

\(u=\log_p y.\)

Then

\(\log_y p=\frac{1}{\log_p y}=\frac1u.\)

The equation becomes

\(2u+\frac{10}{u}-9=0.\)

Multiply by \(u\):

\(2u^2-9u+10=0.\)

Factorise:

\((2u-5)(u-2)=0.\)

Thus

\(u=\frac52\quad\text{or}\quad u=2.\)

Since \(u=\log_p y\),

\(\log_p y=\frac52\quad\text{or}\quad \log_p y=2.\)

Therefore

\(y=p^{5/2}\quad\text{or}\quad y=p^2.\)