\(\left\{ {\begin{array}{*{20}{c}}{{{\log }_n}p \ge 0,}\\{n \ne 1,\,\,\,\,\,\,\,\,\,\,\,\,}\\{p \ne 1\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{0 < p < 1,}\\{0 < n < 1,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{p > 1,\,\,\,\,\,\,\,\,\,}\\{n > 1.\,\,\,\,\,\,\,\,\,}\end{array}} \right.}\end{array}} \right.\)
\(\sqrt {{{\log }_n}p + {{\log }_p}n + 2} \cdot \left( {{{\log }_n}p-{{\log }_{np}}p} \right) \cdot \sqrt {{{\log }_n}p} = \)
\( = \sqrt {{{\log }_n}p + \frac{1}{{{{\log }_n}p}} + 2} \cdot \left( {{{\log }_n}p-\dfrac{1}{{{{\log }_p}\left( {np} \right)}}} \right) \cdot \sqrt {{{\log }_n}p} = \)
\( = \sqrt {\dfrac{{{{\left( {{{\log }_n}p + 1} \right)}^2}}}{{{{\log }_n}p}}} \cdot \left( {{{\log }_n}p-\dfrac{1}{{1 + {{\log }_p}n}}} \right) \cdot \sqrt {{{\log }_n}p} = \)
\( = \left( {{{\log }_n}p + 1} \right) \cdot \left( {{{\log }_n}p-\dfrac{1}{{1 + \dfrac{1}{{{{\log }_n}p}}}}} \right) = \left( {{{\log }_n}p + 1} \right)\left( {{{\log }_n}p-\dfrac{{{{\log }_n}p}}{{{{\log }_n}p + 1}}} \right) = \)
\( = \dfrac{{\left( {{{\log }_n}p + 1} \right)\left( {\log _n^2p + {{\log }_n}p-{{\log }_n}p} \right)}}{{{{\log }_n}p + 1}} = \log _n^2p.\)
Ответ: \(\log _n^2p\), где \(\left\{ {\begin{array}{*{20}{c}}{0 < p < 1,}\\{0 < n < 1\,\,}\end{array}} \right.\) или \(\left\{ {\begin{array}{*{20}{c}}{p > 1,}\\{n > 1.}\end{array}} \right.\)