Задача 12. Упростите выражение \(\dfrac{{{{\log }_a}b-{{\log }_{\frac{{\sqrt a }}{{{b^3}}}}}\sqrt b }}{{{{\log }_{\frac{a}{{{b^4}}}}}b-{{\log }_{\frac{a}{{{b^6}}}}}b}}:{\log _b}\left( {{a^3}{b^{-12}}} \right)\)
Ответ
ОТВЕТ: \({\log _a}b,\) где \(a > 0,\,\;a \ne 1,\;\,b > 0,{\text{ }}\,b \ne 1,\;\,a \ne {b^4}{\text{,}}\,{\text{ }}a \ne {b^6}.\)
Решение
Запишем область допустимых значений на переменные a и b: \(\left\{ {\begin{array}{*{20}{c}}{a > 0,\,\,}\\{a \ne 1,\,\,}\\{b > 0,\,\,}\\{\begin{array}{*{20}{c}}{b \ne 1,\,\,}\\{a \ne {b^4},}\\{a \ne {b^6}.}\end{array}}\end{array}} \right.\)
\(\dfrac{{{{\log }_a}b-{{\log }_{\frac{{\sqrt a }}{{{b^3}}}}}\sqrt b }}{{{{\log }_{\frac{a}{{{b^4}}}}}b-{{\log }_{\frac{a}{{{b^6}}}}}b}}:{\log _b}\left( {{a^3} \cdot {b^{-12}}} \right) = \dfrac{{{{\log }_a}b-\dfrac{{{{\log }_a}\sqrt b }}{{{{\log }_a}\left( {\dfrac{{\sqrt a }}{{{b^3}}}} \right)}}}}{{\dfrac{{{{\log }_a}b}}{{{{\log }_a}\dfrac{a}{{{b^4}}}}}-\dfrac{{{{\log }_a}b}}{{{{\log }_a}\dfrac{a}{{{b^6}}}}}}} \cdot \dfrac{{{{\log }_a}b}}{{{{\log }_a}\left( {{a^3} \cdot {b^{-12}}} \right)}} = \)
\( = \dfrac{{{{\log }_a}b-\dfrac{{{{\log }_a}b}}{{2\left( {\dfrac{1}{2}-3{{\log }_a}b} \right)}}}}{{\dfrac{{{{\log }_a}b}}{{1-4{{\log }_a}b}}-\dfrac{{{{\log }_a}b}}{{1-6{{\log }_a}b}}}} \cdot \dfrac{{{{\log }_a}b}}{{3-12{{\log }_a}b}} = \dfrac{{1-\dfrac{1}{{1-6{{\log }_a}b}}}}{{\dfrac{1}{{1-4{{\log }_a}b}}-\dfrac{1}{{1-6{{\log }_a}b}}}} \cdot \dfrac{{{{\log }_a}b}}{{3\left( {1-4{{\log }_a}b} \right)}} = \)
\( = \dfrac{{\left( {1-6{{\log }_a}b-1} \right)}}{{1-6{{\log }_a}b}} \cdot \dfrac{{\left( {1-4{{\log }_a}b} \right)\left( {1-6{{\log }_a}b} \right)}}{{-2{{\log }_a}b}} \cdot \dfrac{{{{\log }_a}b}}{{3\left( {1-4{{\log }_a}b} \right)}} = \dfrac{{-6{{\log }_a}b}}{{-6}} = {\log _a}b.\)
Ответ: \({\log _a}b\), где \(a > 0,\,\,\,a \ne 1,\,\,\,b > 0,\,\,\,\,b \ne 1,\,\,\,\,a \ne {b^4},\,\,\,\,\,a \ne {b^6}.\)