Упрощение логарифмических выражений. Задача 11

ОТВЕТ:  \(1-{\log _a}\left( {a-b} \right),\) если  \(\left\{ {\begin{array}{*{20}{c}}  {0 < a < 1,} \\   {b < 0} \end{array}\,} \right.\)  или  \(\left\{ {\begin{array}{*{20}{c}}  {a > {\text{1}}{\text{,}}} \\   {{\text{0}} < b < a;} \end{array}} \right.\)  \({\log _a}\left( {a-b} \right)-1,\)  если  \(0 < b < a < 1\)  или  \(\left\{ {\begin{array}{*{20}{c}}  {a > 1,} \\   {b < 0.} \end{array}} \right.\)

\(I = \dfrac{{1-{{\log }_{\frac{1}{a}}}\dfrac{1}{{{{\left( {a-b} \right)}^2}}} + \log _a^2\left( {a-b} \right)}}{{{{\left( {1-{{\log }_{\sqrt a }}\left( {a-b} \right) + \log _a^2\left( {a-b} \right)} \right)}^{\frac{1}{2}}}}} = \dfrac{{1-2{{\log }_a}\left( {a-b} \right) + \log _a^2\left( {a-b} \right)}}{{\sqrt {1-2{{\log }_a}\left( {a-b} \right) + \log _a^2\left( {a-b} \right)} }} = \)

\( = \dfrac{{{{\left( {{{\log }_a}\left( {a-b} \right)-1} \right)}^2}}}{{\sqrt {{{\left( {{{\log }_a}\left( {a-b} \right)-1} \right)}^2}} }} = \dfrac{{{{\left( {{{\log }_a}\left( {a-b} \right)-1} \right)}^2}}}{{\left| {{{\log }_a}\left( {a-b} \right)-1} \right|}}.\)

Если  \({\log _a}\left( {a-b} \right)-1 > 0\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\log _a}\left( {a-b} \right) > {\log _a}a\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{a > 1,}\\{b < 0}\end{array}\,\,\,\,\,\,\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{0 < a < 1,}\\{0 < b < a}\end{array}} \right.}\end{array}} \right.\), то 

\(I = \dfrac{{{{\left( {{{\log }_a}\left( {a-b} \right)-1} \right)}^2}}}{{{{\log }_a}\left( {a-b} \right)-1}} = {\log _a}\left( {a-b} \right)-1.\)

Если  \({\log _a}\left( {a-b} \right)-1 < 0\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\log _a}\left( {a-b} \right) < {\log _a}a\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{a > 1,\,\,\,\,\,\,\,\,}\\{0 < b < a}\end{array}\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{0 < a < 1,}\\{b < 0\,\,\,\,\,\,\,\,}\end{array}} \right.}\end{array}} \right.\), то

\(I = \dfrac{{{{\left( {{{\log }_a}\left( {a-b} \right)-1} \right)}^2}}}{{-\left( {{{\log }_a}\left( {a-b} \right)-1} \right)}} = 1-{\log _a}\left( {a-b} \right).\)

Ответ:  \({\log _a}\left( {a-b} \right)-1\), если  \(\left\{ {\begin{array}{*{20}{c}}{a > 1,}\\{b < 0\,}\end{array}} \right.\)   или   \(0 < b < a < 1;\)

             \(1-{\log _a}\left( {a-b} \right)\), если  \(\left\{ {\begin{array}{*{20}{c}}{a > 1,\,\,\,\,\,\,\,\,}\\{0 < b < a}\end{array}} \right.\)   или   \(\left\{ {\begin{array}{*{20}{c}}{0 < a < 1,}\\{b < 0.\,\,\,\,\,\,\,}\end{array}} \right.\)