Свойства логарифмов. Логарифмические вычисления. Задача 32math100admin44242025-03-27T21:16:16+03:00
Задача 32. Вычислите \({\left( {{5^{\log _{25}^{\,2}3}} — {{\sqrt 3 }^{\,\,{{\log }_5}\sqrt 3 }} + 2} \right)^2} + {\log _{25}}{\log _5}\sqrt[5]{{\,\sqrt {25} }}\)
Решение
\({\left( {{5^{\log _{25}^23}}-{{\sqrt 3 }^{{{\log }_5}\sqrt 3 }} + 2} \right)^2} + {\log _{25}}{\log _5}\sqrt[5]{{\sqrt {25} }} = \)
\( = {\left( {{{\left( {{5^{{{\log }_{25}}3}}} \right)}^{{{\log }_{25}}3}}-{{\sqrt 3 }^{{{\log }_5}\sqrt 3 }} + 2} \right)^2} + {\log _{25}}{\log _5}\sqrt[5]{5} = \)
\( = {\left( {{{\left( {{5^{\frac{1}{2}{{\log }_5}3}}} \right)}^{{{\log }_{25}}3}}-{{\sqrt 3 }^{{{\log }_5}{3^{\frac{1}{2}}}}} + 2} \right)^2} + {\log _{25}}{\log _5}{5^{\frac{1}{5}}} = \)
\( = {\left( {{{\left( {{5^{{{\log }_5}\sqrt 3 }}} \right)}^{{{\log }_{25}}3}}-{{\sqrt 3 }^{\frac{1}{2}{{\log }_5}3}} + 2} \right)^2} + {\log _{25}}\left( {\dfrac{1}{5}{{\log }_5}5} \right) = \)
\( = {\left( {{{\sqrt 3 }^{{{\log }_{25}}3}}-{{\sqrt 3 }^{{{\log }_{25}}3}} + 2} \right)^2} + {\log _{{5^2}}}{5^{-1}} = {2^2} + \dfrac{{-1}}{2}{\log _5}5 = 4-\dfrac{1}{2} = 3,5.\)
Ответ: 3,5.