solving P and Q where they are integers
(25/16)^ -3/2 = 2^P * 5^Q
no need for logs, note ...
$\left(\dfrac{25}{16}\right)^{-3/2} = \left(\dfrac{16}{25}\right)^{3/2} = \left(\dfrac{4^2}{5^2}\right)^{3/2} = \left(\dfrac{4}{5}\right)^3 = 4^3 \cdot 5^{-3}$
can you finish finding the values of P and Q from here?
Substitute back into the original equation to check on your own.
$\left ( \dfrac{25}{16} \right )^{-(3/2)} = \left ( \dfrac{16}{25} \right )^{(3/2)} = \left ( \sqrt{\dfrac{16}{25} } \right )^3 =$
$\left ( \dfrac{4}{5} \right )^3 = \left ( \dfrac{2^2}{5} \right )^3 = \dfrac{2^6}{5^3} = 2^6 * 5^{-3}.$